Choose an xaxis and a yaxis (orthonormal) and let O be the origin.
A circle of radius one centered at O is called
'the' trigonometric circle or 'the' unit circle.
Turning counterclockwise is the positive orientation in trigonometry.
Angles are measured starting from the xaxis.
The units used to measure an angle are 'degree' and 'radian'.
A right angle is an angle whose measure is exactly 90 degrees or pi/2 radians.
In this theory we use mainly radians.
Each real number t corresponds to exactly one angle, and to exactly one point P on the unit circle.
We call that point the 'image point' of t.
Examples:
In an analogous manner we find that cotan(t) is the xcoordinate of the intersection point S'
of the line OP with the line y = 1.
cos^{2}(t) + sin^{2}(t) = 1 sin^{2}(t) 1 + tan^{2}(t) = 1 +  cos^{2}(t) cos^{2}(t)+sin^{2}(t) =  cos^{2}(t) 1 =  = sec^{2}(t) cos^{2}(t) In the same way : 1 + cotan^{2}(t) = 1/ sin^{2}(t) = csc^{2}(t)
cos^{2}(t) + sin^{2}(t) = 1 1 + tan^{2}(t) = sec^{2}(t) 1 + cot^{2}(t) = csc^{2}(t) 
sin^{2}(t) = 1  cos^{2}(t) cos^{2}(4t) = 1  sin^{2}(4t) 1 + tan^{2}(t/2) = sec^{2}(t/2) csc^{2}(t^{2})  cot^{2}(t^{2}) = 1Exercise:
If cos(t)=0.5 then sin^{2}(t) = ... If cos(t)=0.1 then tan^{2}(t) = ... If cot(t)=0.2 then sin^{2}(t) = ...
If t  t' = 2.k.pi (k is an integer) then sin(t) = sin(t') 
t and t' are supplementary values <=> t+t' = pi.
With the help of a unit circle we see that the corresponding image points are symmetric relative to the yaxis. Hence, we have :
sin(t) = sin(pi  t) 
sin(t + pi/2) = sin(pi/2  t) tan(2t + 0.2) =  tan(pi 0.2  2t)  tan(pi t) = tan(t) sin(pit) + cos(3pit)  sin(t+4pi) + cos(t) = sin(t) + cos(pit)  sin(t) + cos(t) = sin(t)  cos(t)  sin(t) + cos(t) = 0
t and t' are complementary values <=> t+t' = pi/2.
The corresponding image points on a unit circle are symmetric relative to the line y = x . Hence, we have :
sin(t) = cos(pi/2  t) 
tan(pi/4 +3t) = cot(pi/4 3t) cos(3pi/2 t) = sin( t  pi) = sin(t + 2pi) = sin(t) cot(3x  pi/2) = tan(3x + pi ) =  tan(3x)  cos(pi/2  2x) + sin(2x  pi)  cos(3pi  2x) =  sin(2x) + sin(pi  2x)  cos(pi  2x) =  sin(2x) + sin(2x) + cos(2x) = cos(2x)
t and t' are opposite values <=> t+t' = 0.
Now, the corresponding image points are symmetric relative to the xaxis. Hence, we have :
sin(t) = sin(t) 
cos(pi/2 + x) = cos(pi/2  x) = sin (x) sin(6x  pi) =  sin(pi  6x) =  sin(6x) cot(x + 4pi) = cot(x) =  cot(x)
t and t' are antisupplementary values <=> ( tt' = pi or t't = pi )
The corresponding image points are symmetric relative to the origin O . Hence, we have :
sin(t) = sin(t+pi) 
tan(5a + 3pi) = tan(5a + pi) = tan(5a) cot(t/2 + pi/2) = cot(t/2  pi/2) =  cot(pi/2  t/2) =  tan(t/2) sin(x + 3 pi) + sin(x) = sin(x) + sin(x) = 0
Now sin(B),cos(B) and 1 are directly proportional with b, c and a.
sin(B) cos(B) 1  =  =  b c a => sin(B) = b/a cos(B) = c/a tan(B) = b/c and since the angles B and C are complementary angles cos(C) = b/a sin(C) = c/a tan(C) = c/b
In each rightangled triangle ABC, with A as the right angle, we have
sin(B) = b/a cos(B) = c/a tan(B) = b/c cos(C) = b/a sin(C) = c/a tan(C) = c/b 

Applications:
The tangent lines, in the points A and B of a circle with center O and radius r,
intersect in point P.
The chord AB and the line OP intersect in point S. Let a = OP and k = AB. Express k as a function of r and a. 
In the rightangled triangle AOP : The leg of a right triangle is the mean proportional between the hypotenuse and the its orthogonal projection on the hypotenuse.
OA^{2} = OP OS. => OS = r^{2}/aIn the rightangled triangle OAS : The square of the hypotenuse is equal to the sum of the squares of the other two sides.
OA^{2} = OS^{2} + AS^{2} => r^{2} = OS^{2} + k^{2}/4 => r^{2} = r^{4}/a^{2} + k^{2}/4 => .... 2r ___________ => k =  \/ a^{2}  r^{2} a
Divide a given segment BC by the construction of a point D. The two parts BD and DC must have a suitable length x and y such that the product x.y is equal to a given value m^{2}. 
x+y=BC en m^{2} = x.y
m is the mean proportional between x and y
We know that the altitude to the hypotenuse BC of a right triangle ABC is the
mean proportional between the segments into which it divides the hypotenuse.
We are looking for a right triangle ABC with base the hypotenuse BC
and with m as the height.
The vertex A of the rightangled triangle is located on the circle with diameter BC.
Construction steps:
The area of the triangle is a.h/2 .
The area of a triangle ABC =(1/2) a.c.sin(B) = (1/2) b.c.sin(A) = (1/2) a.b.sin(C)

You can also use Heron's formula to calculate the area of a triangle.
Let s = half the circumference of the triangle = (a +b + c)/2. 
Exercise:
A triangle has sides with length 5, 4 and 7.
Draw accurately the triangle.
Calculate the area of the triangle with the formula of Heron and check the result
by measuring height of the triangle and calculating the area with this height.
Now measure an angle of the triangle and calculate the area a third time.
Conversely, you can calculate the angles of the triangle if you know
the area of the triangle.
The area of a triangle ABC = a.c.sin(B)/2 = b.c.sin(A)/2 = a.b.sin(C)/2 => a.c.sin(B) = b.c.sin(A) = a.b.sin(C)dividing through by a.b.c, we get
a b c  =  =  sin(A) sin(B) sin(C)This formula is called the sine rule in a triangle ABC.
Let R be the radius of the circle with center O through the points A,B and C.
Let B' be the second intersection point of BO with the circle. The angle B'
in triangle BB'C is equal to, or supplementary with, A.
In the rightangled triangle BB'C we see that a = 2R sin(B') = 2R sin(A).
Thus, the fractions in the sinus rule are all equal to 2R.
In any triangle ABC we have
a b c  =  =  = 2R sin(A) sin(B) sin(C) 
Exercise:
A triangle has sides with length a = 5, b = 4 and c = 7.
Draw accurately the triangle.
Calculate the area of the triangle with the formula of Heron.
Now calculate the angle A with the area formula (1/2).b.c.sin (A).
Now, use the angle A to find the radius R of the circumscribed circle.
Check the result on your sketch.
If an expression between the sides of a triangle is homogeneous in a, b and c, we obtain an equivalent expression by replacing a,b and c by sin(A), sin(B) and sin(C).
Example:
In a triangle
b.sin(AC) = 3.c.cos(A+C) <=> sin(B).sin(AC) = 3.sin(C).cos(A+C)
In any triangle ABC we have
a^{2} = b^{2} + c^{2}  2 b c cos(A) b^{2} = c^{2} + a^{2}  2 c a cos(B) c^{2} = a^{2} + b^{2}  2 a b cos(C) 
If the angle A is a right angle, then the proof is obvious.
Now, suppose the angle A is an acute angle.
a^{2} = h^{2} + p^{2} (*) b^{2} = h^{2} + q^{2} = h^{2} + (c  p)^{2} so, h^{2} = b^{2}  (c  p)^{2} (**) From (*) and (**) a^{2} = b^{2}  (c  p)^{2} + p^{2} = b^{2}  (c^{2}  2 p c + p^{2}) + p^{2} = b^{2}  c^{2} + 2 p c = b^{2} + c^{2} + 2 p c  2 c^{2} = b^{2} + c^{2} + 2 c (p  c) = b^{2} + c^{2}  2 c (c  p) = b^{2} + c^{2}  2 c q = b^{2} + c^{2}  2 c b cos(A)Now, suppose the angle A is an obtuse angle.
The proof proceeds in the same manner as above.
Draw a new picture and work this out as an exercise.
This cosine rule can also be proved using the dot product of vectors.
See Proof cosine rule
sin^{2} (pi/3) = sqrt( 1  cos^{2} (pi/3)) = sqrt(3)/2 So, sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2. tan(pi/3) = sqrt(3)
cos^{2}(pi/4)+sin^{2}(pi/4) = 1 => 2cos^{2}(pi/4) = 1 => cos (pi/4) = sqrt(1/2) So, cos (pi/4) = sin(pi/4) = sqrt(1/2) tan(pi/4) = 1
Substitute all the sides in de Cosine Rule to compute the angles.
Example: a=4 b=5 c=7
The Cosine Rule gives
58 = 70 cos(A) 40 = 56 cos(B) 8 = 40 cos(C) A = 34.05° B = 44.41° C = 101.53°Test : A + B + C = ...
Calculate the third angle and then the sides with the Sine Rule.
Example: a=4 A=34° B=45°
The third angle is C =101°
From the Sine Rule
4 sin(45°) b =  = 5.06 sin(34°) 4 sin(101°) c =  = 7.02 sin(34°)Test : draw a sketch of the triangle
Use the Cosine Rule.
Example: b=5 c=7 A=34.05°
From the Cosine Rule
a^{2} = 25 + 49  70 cos(34.05°) => a = 4 The other two formulas of the cosine rule give 40 = 56 cos(B) 8 = 40 cos(C) B = 44.41° C = 101.53°Test : A + B + C = ...
Draw a sketch. There are three cases.
1) no solutions
2) one solution
3) two solutions
From a sketch we see that there are no solutions.
From a sketch we see that there is one solution. We use the Sine Rule.
7 5 c  =  =  sin(60°) sin(B) sin(C)So, sin(B)= 0.6186 and this gives us two supplementary solutions for B.
From a sketch we see that there are two solutions for B. We use the Sine Rule.
4.5 5 c  =  =  sin(60°) sin(B) sin(C)So, sin(B)= 0.96225 and this gives us two supplementary solutions for B.
sin : R > R : x > sin(x)is called, the sine function.
cos : R > R : x > cos(x)is called, the cosine function.
tan : R > R : x > tan(x)is called, the tangent function.
cot : R > R : x > cot(x)is called, the cotangent function.
Example 1
y = sin(4x)
The graph of this function arises from the graph of sin(x) when we compress the
graph of sin(x) towards the yaxis with a factor 4.
From this it follows that the period of sin(4x) is pi/2. The function y = sin(ax) has a period
2.pi/a if a > 0.
Similar rules apply to the other trigonometric functions. Thus the period of tan(x/3) is 3.pi.
Example 2
y = sin(x+5)
The graph of this function comes about by moving the graph of sin(x) five units to the left.
The period does not change.
Example 3
y = tan(x)+5
The graph of this function is obtained by moving the graph of tan(x) five units upwards.
The period does not change.
Example 4
We start with y = tan(x). We compress the graph towards the yaxis with a factor 3. The new function is y = tan(3x). We move the graph two units to the right. The new function is y = tan(3(x2)) . Finally, we move the last graph two units downwards. We obtain y = tan(3x 6)2. The period is pi/3.
Generalization:
The period of A sin(a x + b ) is 2 pi/a
The period of A cos(a x + b ) is 2 pi/a
The period of A tan(a x + b ) is pi/a
The period of A cot(a x + b ) is pi/a
The period of A / sin(a x + b ) is 2 pi/a
The period of A / cos(a x + b ) is 2 pi/a
The period of A / tan(a x + b ) is pi/a
The period of A / cot(a x + b ) is pi/a
If f(x) is a function with period = a g(x) is a function with period = b Then f(x)+g(x) is a function with period = c <=> There are strictly positive and relatively prime integers m and n such that c = m.a = n.bExamples
sin(2x) has pi as period and cos(3x) has 2pi/3 as period .
Now, c = 2.(pi) = 3.(2pi/3). So, 2 pi is a period of sin(2x) + cos(3x)
sin(pi x) has 2 as period and tan(2 pi x/7) has 7/2 as period.
Now, c = 7.(2) = 4.(7/2). So, 14 is a period of sin(pi x) + tan(2 pi x/7)
sin(sqrt(2) x) has pi.sqrt(2) as period and cos(2x) has pi as period .
There are no strictly positive integers m and n such that
m.(pi.sqrt(2)) = n.(pi). So, sin(sqrt(2) x) + cos(2x) has NO period!
sin(x) has 2pi as period and cos(pi x) has 2 as period.
There are no strictly positive integers m and n such that
m.(2pi) = n.(2). So, sin(x) + cos(pi x) has NO period!
Example:
y = 2.arcsin(x1) comes about by moving the graph of arcsin(x) one unit to the right, and then by
multiplying all the images by two. The domain is [0,2] and the range is [pi,pi].
cos(uv) = cos(u).cos(v)+sin(u).sin(v) 
cos(pi/32x) = cos(pi/3)cos(2x) + sin(pi/3)sin(2x) = 0.5 cos(2x) + 0.5 sqrt(3) sin(2x)
cos(u + v) = cos(u).cos(v)sin(u).sin(v) 
cos(x + x/2) + cos(x  x/2) = cos(x)cos(x/2) + sin(x)sin(x/2) + cos(x)cos(x/2)  sin(x)sin(x/2) = 2 cos(x)cos(x/2)
sin(u  v) = sin(u).cos(v)cos(u).sin(v) 
sin(x  pi/4) = sin(x) cos(pi/4)  cos(x) sin(pi/4) = (sin(x)cos(x))/sqrt(2)
sin(u + v) = sin(u).cos(v)+cos(u).sin(v) 
sin(u + v) sin(u).cos(v)+cos(u).sin(v) tan(u+v) =  =  cos(u + v) cos(u).cos(v)sin(u).sin(v)Dividing the dominator and denominator by cos(u).cos(v) we have
tan(u) + tan(v) tan(u+v) =  1  tan(u).tan(v) 
tan(u) + tan(pi/4) tan(u) + 1 1 + tan(u) tan(u+pi/4) =  =  =  1  tan(u).tan(pi/4) 1  tan(u) 1  tan(u)
tan(u)  tan(v) tan(uv) =  1 + tan(u).tan(v) 
sin(2u) = 2sin(u).cos(u) 
sin(x) = 2 sin(x/2).cos(x/2) sin(4x) = 2 sin(2x).cos(2x) = 4 sin(x) cos(x) cos(2x) 12 sin(8x) cos(8x) = 6 sin(16x)
cos(2u) = cos^{2} (u)  sin^{2} (u) 
tan(u) + tan(u) 2 tan(u) tan(2u) =  =  1  tan(u).tan(u) 1 tan(u)tan(u)
2 tan(u) tan(2u) =  1 tan^{2}(u) 
1 cot(2x) =  tan(2x) 1  tan^{2}(x) =  2 tan(x)
1 + cos(2u) = 1+cos^{2} (u)sin^{2} (u) = 2 cos^{2} (u) 1  cos(2u) = 1cos^{2} (u)+sin^{2} (u) = 2 sin^{2} (u)
1 + cos(2u) = 2 cos^{2} (u) 1  cos(2u) = 2 sin^{2} (u) 
Applications:
1 + 2 cos(x) + cos(2x) = 2 cos(x) + ( 1 + cos(2x)) = 2 cos(x) + 2 cos^{2} (x) = 2 cos(x) (1 + cos(x)) = 2 cos(x) 2 cos^{2} (x/2) = 4 cos(x) cos^{2} (x/2)Since 2 pi is the period of (1 + 2 cos(x) + cos(2x)), it follows that the period of cos(x) cos^{2} (x/2) is 2pi.
The period of tan^{2}(4x) is equal to the period of 1+tan^{2}(4x).
The period of 1+tan^{2}(4x) is equal to the period of 1/ cos^{2}(4x).
The period of 1/ cos^{2}(4x) is equal to the period of cos^{2}(4x).
The period of cos^{2}(4x) is equal to the period of 0.5(1+cos(8x)).
The period of 0.5(1+cos(8x)) is equal to the period of cos(8x).
And this period is pi/4.
Solution:
About the three sides we know :
a b c  =  =  7 8 3 Since similar triangles have the same angles, we can use a = 7 , b = 8 and c = 3 as sides of the triangle. From the cosine rule we can write b^{2} + c^{2}  a^{2} cos(A) =  = 1/2 2 b c Now we use the Carnot formulas 1  cos(A) 2 sin^{2}(A/2)  =  = tan^{2}(A/2) = 1/3 1 + cos(A) 2 cos^{2}(A/2)
We know : 1 + cos(2u) = 2 cos^{2} (u)
Now take 2u = pi/6 radians , then u = pi/12 radians.
Now the exact value of cos(pi/6) is sqrt(3)/2. We can use the Carnot formula to calculate cos(pi/12).
cos^{2}(pi/12) = (1 + cos(pi/6))/2 = (1 + sqrt(3)/2)/2 = (2 + sqrt(3))/4 So, cos(pi/12) = (1/2). sqrt(2 + sqrt(3))
cos(2u) = 2 cos^{2}(u) 1 2 =   1 1 + tan^{2} (u) 1  tan^{2}(u) =  1 + tan^{2} (u) We know: 2 tan(u) tan(2u)=  1  tan^{2} (u) Hence, 2 tan(u) sin(2u) =  1 + tan^{2} (u)
Let t = tan(u) , then 1  t^{2} cos(2u) =  1 + t^{2} or 1  tan^{2}(u) cos(2u) =  1 + tan^{2} (u) 2t sin(2u) =  1 + t^{2} or 2 tan(u) sin(2u) =  1 + tan^{2} (u) 2t tan(2u) =  1  t^{2} or 2 tan(u) tan(2u)=  1  tan^{2} (u) 
Application:
The equation of a line d is y = 3 x + 4 .
u = the angle fom the xaxis to this line.
We know that tan(u) = 3.
Line d' is the reflection of the the line y = 3 in d.
The the angle fom the xaxis to the line d' is 2u.
The slope of line d' is tan(2u).
2t 2 tan(u) 2 . 3 tan(2u) =  =  =  =  0.75 1  t^{2} 1  tan^{2}(u) 1  9
cos(u + v) = cos(u).cos(v)sin(u).sin(v) cos(u  v) = cos(u).cos(v)+sin(u).sin(v) sin(u + v) = sin(u).cos(v)+cos(u).sin(v) sin(u  v) = sin(u).cos(v)cos(u).sin(v)and from this, we have
cos(u + v) + cos(u  v) = 2.cos(u).cos(v) cos(u + v)  cos(u  v) = 2.sin(u).sin(v) sin(u + v) + sin(u  v) = 2. sin(u).cos(v) sin(u + v)  sin(u  v) = 2. cos(u).sin(v)Let x = u + v and y = u  v
Now we have
cos(x) + cos(y) = 2 cos((1/2)(x + y)) cos((1/2)(x  y)) cos(x)  cos(y) = 2 sin((1/2)(x + y)) sin((1/2)(x  y)) sin(x) + sin(y) = 2 sin((1/2)(x + y)) cos((1/2)(x  y)) sin(x)  sin(y) = 2 cos((1/2)(x + y)) sin((1/2)(x  y))
Simpson formulas
x + y x  y cos(x) + cos(y) = 2 cos  cos  2 2 x + y x  y cos(x)  cos(y) = 2 sin  sin  2 2 x + y x  y sin(x) + sin(y) = 2 sin  cos  2 2 x + y x  y sin(x)  sin(y) = 2 cos  sin  2 2 
cos(2x)  cos(2y)  cos(2x) + cos(2y) 2 sin(x+y) sin(xy) =  2 cos(x+y) cos(xy) =  tan(x+y) tan(xy) = tan(y+x) tan(yx)
sin(2a).(1 + tan^{2}(a)) = 2 sin(a) cos(a) . (1/cos^{2}(a)) = 2 sin(a) / cos(a) = 2 tan(a)
2 sin(2a) + sin(4a) = 2 sin(2a) + 2 sin(2a).cos(2a) = 2 sin(2a). (1 + cos(2a)) = 4 sin(a) cos(a) .2 cos^{2}(a) = 8 sin(a) cos^{3}(a)
1  tan^{4}(a) = (1  tan^{2}(a)).(1 + tan^{2}(a)) sin^{2}(a) 1 = ( 1   )  cos^{2}(a) cos^{2}(a) cos(2a) =  cos^{4}(a)
cos^{2}(a)  sin^{2}(a)  2 cos(a) + 1 = cos^{2}(a)  2 cos(a) + cos^{2}(a) = 2 cos^{2}(a)  2 cos(a) = 2 cos(a) (cos(a)  1) = 2 cos(a) (1  cos(a)) = 2 cos(a) 2 sin^{2}(a/2) = 4 cos(a) sin^{2}(a/2)
sin(2a) (1 + 2 cos(2a) ) + 2 sin(3a) = sin(2a) + 2 sin(2a) cos(2a) + 2 sin(3a) = sin(2a) + sin(4a) + 2 sin(3a) = 2 sin(3a) cos(a) + 2 sin(3a) = 2 sin(3a) (1+ cos(a)) = 4 sin(3a) cos^{2}(a/2)
cos^{2}(a) 2 cos(a) + cos(2a) + sin^{2}(a) = 1 + cos(2a)  2 cos(a) = 2 cos^{2}(a)  2 cos(a) = 2 cos(a) (cos(a)  1) = 2 cos(a) (1  cos(a)) = 2 cos(a) 2 sin^{2}(a/2) = 4 cos(a) sin^{2}(a/2)
(1  cos(4a))^{2}  (1  cos^{2}(4a)) (1  cos(4a))^{2} =  (1  cos(4a))(1 + cos(4a)) (1  cos(4a)) =  (1 + cos(4a)) 2 sin^{2}(2a) =  2 cos^{2}(2a) = tan^{2}(2a)
cos^{4}(a)  sin^{4}(a) = (cos^{2}(a)  sin^{2}(a)) (cos^{2}(a) + sin^{2}(a)) = (cos^{2}(a)  sin^{2}(a)) = cos(2a)
sin(a) + sin(b) + sin(c)  sin(a+b+c) = 2 sin( (a+b)/2 ) cos( (ab)/2 ) + 2 cos( (a+b+2c)/2 ) sin( (ab)/2 ) = 2 sin( (a+b)/2 ) ( cos( (ab)/2 )  cos( (a+b+2c)/2 ) = 4 sin( (a+b)/2 ) sin( (a+c)/2 ) sin( (bc)/2 ) ) = 4 sin( (a+b)/2 ) sin( (a+c)/2 ) sin( (b+c)/2 ) )
sin^{2}(a)  sin^{2}(b)  sin^{2}(a+b) = sin^{2}(a)  sin^{2}(b)  (sin(a) cos(b) + cos(a) sin(b))^{2} = sin^{2}(a)  sin^{2}(b)  sin^{2}(a) cos^{2}(b)  cos^{2}(a) sin^{2}(b)  2 sin(a) sin(b) cos(a) cos(b) = sin^{2}(a) (1 cos^{2}(b))  sin^{2}(b) (1 + cos^{2}(a))  2 sin(a) sin(b) cos(a) cos(b) = sin^{2}(a) sin^{2}(b)  sin^{2}(b) (1 + cos^{2}(a))  2 sin(a) sin(b) cos(a) cos(b) = sin^{2}(b) ( sin^{2}(a) 1  cos^{2}(a))  2 sin(a) sin(b) cos(a) cos(b) = 2 sin^{2}(b) cos^{2}(a)  2 sin(a) sin(b) cos(a) cos(b) = 2 sin(b) cos(a) ( sin(b) cos(a) sin(a) cos(b)) = 2 cos(a) sin(b) sin(a+b)
sin(2b+2c) (cos(2b) + cos(2c))  sin(2b)  sin(2c) = sin(2b+2c) 2 cos(b+c) cos(bc)  2 sin(b+c)cos(bc) = 4 sin(b+c) cos(b+c) cos(b+c) cos(bc)  2 sin(b+c)cos(bc) = 2 sin(b+c) cos(bc) ( 2 cos^{2}(b+c)  1) = 2 sin(b+c) cos(bc) ( 2 cos^{2}(b+c)  cos^{2}(b+c)  sin^{2}(b+c) ) = 2 sin(b+c) cos(bc) (cos^{2}(b+c)  sin^{2}(b+c) ) = 2 sin(b+c) cos(bc) cos(2b +2c)
cos(u + v) = cos(u).cos(v)sin(u).sin(v) cos(u  v) = cos(u).cos(v)+sin(u).sin(v) sin(u + v) = sin(u).cos(v)+cos(u).sin(v) sin(u  v) = sin(u).cos(v)cos(u).sin(v) Thus cos(u + v) + cos(u  v) = 2.cos(u).cos(v) cos(u + v)  cos(u  v) = 2.sin(u).sin(v) sin(u + v) + sin(u  v) = 2. sin(u).cos(v) sin(u + v)  sin(u  v) = 2. cos(u).sin(v) or 2.cos(u).cos(v) = cos(u + v) + cos(u  v) 2.sin(u).sin(v) = cos(u + v)  cos(u  v) 2. sin(u).cos(v) = sin(u + v) + sin(u  v) 2. cos(u).sin(v) = sin(u + v)  sin(u  v)The period of cos(u).cos(v) is equal to the period of cos(u + v) + cos(u  v)
Examples:
The period of cos(2x).sin(x+3) is equal to the period of sin(3x+3)  sin(x3)
and this period is 2 pi.
The period of cos(4x).cos(x/2) is equal to the period of cos(9x/2) + cos(7x/2)
and this period is 4pi
We can transform many equations of trigonometric functions to the form of a general sine function by using previous formulas.
We give some examples of such transformations.
y = 3 sin(x) <=> y = 3 sin(x  pi)
y = sin(2x) <=> y =  sin(2x) <=> y = sin(2x  pi) <=> y = sin 2(x  pi/2)
y = 4 sin(3x) <=> y = 4 sin(3x)
y = 2 cos(3x) <=> y = 2 sin(pi/2  3x) <=> y = 2 sin(3x pi/2) <=> y = 2 sin(3x  3pi/2) <=> y = 2 sin 3(x  pi/2)
y = 2 cos(3x1) <=> y = 2 sin(pi/2 3x + 1) <=> y = 2 sin(3x  pi/2 1) <=> y = 2 sin 3(x  (pi/6 + 1/3))
y = cos(3x+4)  cos(3x4) <=> y = 2 sin(3x) sin(4) <=> y = (2sin(4)) sin(3x)
y = sin(4x3).cos(4x3) <=> y = (1/2) sin(8x6) <=> y = (1/2) sin(8(x(3/4))
a.sin(u)+b.cos(u) can be brought in the form A.sin(uu_{o}).
Then, the transformation to the general sine function is easy.
a.sin(u) + b.cos(u) = a( sin(u) + (b/a) cos(u) ) Take u_{o} such that tan(u_{o}) =  b/a = a( sin(u)  tan(u_{o}) cos(u) ) = (a/cos(u_{o})) . ( sin(u).cos(u_{o})  sin(u_{o}).cos(u) ) Let A = (a/cos(u_{o})) = A . sin(u  u_{o})Example
3 sin(x)  2 cos(x) = 3( sin(x)  (2/3) cos(x) ) Let tan(u_{o}) = 2/3 ; take u_{o} = 0.588 = 3( sin(x)  tan(u_{o}) cos(x) ) = (3/cos(u_{o})) ( sin(x) cos(u_{o})  cos(x) sin(u_{o}) ) = 3.6055 sin( x  0.588)
cos(u) = cos(v) <=> (u = v + k.2pi) or (u = v + k.2pi)
sin(u) = sin(v) <=> (u = v + 2.k.pi) or (u = pi  v + 2.k.pi)
tan(u) = tan(v) <=> (u = v + k.pi) on condition that tan(u) and tan(v) exist
cot(u) = cot(v) <=> (u = v + k.pi) on condition that cot(u) and cot(v) exist
cos(2x) = cos(pi3x) <=> 2x = (pi3x) + 2.k.pi or 2x = (pi3x) + 2.k'.pi <=> 5x = pi + 2.k.pi or x = pi + 2.k'.pi <=> x = pi/5 + 2.k.pi/5 or x = pi  2.k'.piExample 2
tan(xpi/2) = tan(2x) <=> (xpi/2) = 2x + k.pi <=> x = pi/2 + k.pi <=> x = pi/2  k.pi ( for these values tan(xpi/2) and tan(2x)) exist)Example 3
cos(x) = 1/3 <=> cos(x) = cos(1.91) <=> x = 1.91 +2.k.pi or x = 1.91  2.k.piExample 4
sin(2x) = cos(xpi/3) <=> cos(pi/2  2x) = cos(xpi/3) <=> pi/2  2x = x  pi/3 + 2.k.pi or pi/2  2x =  x + pi/3 + 2.k'.pi <=> 3x =  pi/2  pi/3 + 2.k.pi or x = pi/2 + pi/3 + 2.k'.pi <=> x = pi/6 + pi/9 + 2.k.pi/3 or x = pi/2  pi/3  2.k'.pi <=> x = 5pi/18 + 2.k.pi/3 or x = pi/6  2.k'.piExample 5
3 sin(2x) = cos(2x) <=> 3 tan(2x) = 1 <=> tan(2x) = 1/3 <=> tan(2x) = tan(0.32) <=> 2x = 0.32 + k pi <=> x = 0.16 + k pi/2Example 6
tan(2x) . cot( x + pi/2) = 1 <=> tan(2x) = tan( x + pi/2) <=> 2x = x + pi/2 + k.pi <=> x = pi/2 + k.pi (on condition that tan(2x) and cot( x + pi/2) exist)But cot( x + pi/2) does not exist for x = pi/2 + k.pi !!!!!
So, tan(2x) . cot( x + pi/2) = 1 has no solutions !
The expression " on condition that ...." is not redundant!
2sin^{2} (2x)+sin(2x)1=0 <=> (let t = sin(2x) ) 2t^{2} + t  1 = 0 <=> t = 0.5 or t = 1 <=> sin(2x) = 0.5 or sin(2x) = 1 <=> sin(2x) = sin(pi/6) or sin(2x) = sin(pi/2) <=> 2x = pi/6 +2.k.pi or 2x = pi  pi/6 +2.k.pi or 2x = pi/2 +2.k.pi or 2x = pi + pi/2 +2.k.pi <=> x = pi/12 + k.pi or x = 5pi/12 + k.pi or x = pi/4 + k.pi or x = 3pi/4 + k.piSometimes it is convenient to view these solutions on the unit circle.
Example 2
cos 10x + 7 = 8 cos 5x <=> cos 10x  8 cos 5x + 7 =0 <=> 1 + cos 10x  8 cos 5x + 6 =0 <=> 2 cos^{2} 5x  8 cos 5x + 6 =0 <=> cos^{2} 5x  4 cos 5x + 3 = 0 Let t = cos 5x t^{2}  4t + 3 = 0 <=> t = 3 or t = 1 <=> cos 5x = 1 <=> cos 5x = cos 0 <=> 5x = 2kpi <=> x = 2kpi / 5Examples
tan^{2} (3x)+tan(3x)=0 sin^{2} (x)(sin(x)+1)0.25(sin(x)+1) = 0 cos(2x)+sin^{2} (x) = 0.5 tan(2x)cot(2x) = 1Check your results by plotting graphs.
3.sin(2x)2.sin(x) = 0 <=> 6sin(x)cos(x)2.sin(x) = 0 <=> 2.sin(x).(3cos()1) = 0 <=> sin(x) = 0 or cos(x) = 1/3 <=> x = k.pi or x = 1.23 + 2.k.pi or x = 1.23 + 2.k'.piExamples
tan(x)tan(4x)+tan^{2} (x) = 0 sin(7x)sin(x) = sin(3x) cos(4x) + cos(2x) + cos(x) = 0 sin(5x)+sin(3x) = cos(2x)cos(6x)Check your results by plotting graphs.
a.sin(u) + b.cos(u) = a( sin(u) + (b/a) cos(u) ) Take u_{o} such that tan(u_{o}) =  b/a = a( sin(u)  tan(u_{o}) cos(u) ) = (a/cos(u_{o})) . ( sin(u).cos(u_{o})  sin(u_{o}).cos(u) ) Let A = (a/cos(u_{o})) = A . sin(u  u_{o}) = A . cos(pi/2  u + u_{o}) = A . cos(u  u_{o}')Example
3 sin(x)  2 cos(x) = 3( sin(x)  (2/3) cos(x) ) Let tan(u_{o}) = 2/3 ; take u_{o} = 0.588 = 3( sin(x)  tan(u_{o}) cos(x) ) = (3/cos(u_{o})) ( sin(x) cos(u_{o})  cos(x) sin(u_{o}) ) = 3.6055 sin( x  0.588) or = 3.6055 cos( x  2.1598)Plot the graph of 3 sin(x)  2 cos(x) and the graph of 3.6055 sin( x  0.588)
With this method we can solve the equation
a.sin(u)+b.cos(u) = c
Example
3.sin(2x)+4.cos(2x) = 2 <=> sin(2x) + 4/3 .cos(2x) = 2/3 Let tan(t) = 4/3 <=> sin(2x) + tan(t) .cos(2x) = 2/3 <=> sin(2x)cos(t)+cos(2x)sin(t) = 2/3.cos(t) <=> sin(2x+t) = 2/3.cos(t) since 2/3.cos(t) = 0.4 <=> sin(2x+0.927) = sin(0.39) <=> 2x + 0.927 = 0.39 +2.k.pi or 2x + 0.927 = pi  0.39 +2.k'.pi <=> ....
Example
3 sin(2x) + 4 cos(2x) = 2 Let tan(x) = t <=> 2 t 1  t^{2} 3  + 4  = 2 1 + t^{2} 1 + t^{2} <=> 6 t + 4  4 t^{2} = 2 + 2 t^{2} <=> 6 t^{2}  6 t  2 = 0 <=> 3 t^{2}  3 t 1 = 0 <=> t = 1.26 or t = 0.26 <=> tan(x) = 1.26 or tan(x) = 0.26 <=> x = 0.9 + k pi or x = 0.26 + k pi
Now, we have in view the equations which are homogeneous in sin(u) and cos(u).
Procedure
2.cos^{3} (x)+2.sin^{2} (x)cos(x) = 5.sin(x)cos^{2} (x) <=> cos(x).(2.cos^{2} (x)+2.sin^{2} (x)  5.sin(x)cos(x)) = 0 <=> The simple part cos(x) = 0 gives us x = pi/2 + k.pi In the second part, we divide both sides by cos^{2} (x). Then we have 2.tan^{2} (x)  5.tan(x) +2 = 0 Let t = tan(x) <=> 2.t^{2}  5 t + 2 = 0 <=> t = 0.5 or t = 2 <=> tan(x) = 0.5 or tan(x) = 2 <=> x = 0.464 +k.pi or x = 1.107 +k.pi
1/sin(x) + 1/cos(x) = sqrt(3) 
1/sin(x) + 1/cos(x) = sqrt(3) sin(x) + cos(x) <=>  = sqrt(3) sin(x) cos(x) <=> sin(x) + cos(x) = sqrt(3) sin(x) cos(x) (1) If we square both sides of the equation (1), only the product sin(x)cos(x) occurs. Then, we can find the value of sin(x)cos(x) and with this we'll simplify the equation (1).
From (1) it follows ( sin(x) + cos(x))^{2} = 3 (sin(x) cos(x))^{2} <=> 1 + 2 sin(x) cos(x) = 3 (sin(x) cos(x))^{2} Let y = sin(x) cos(x) <=> 3 y^{2}  2 y 1 = 0 <=> y = 1 or 1/3 <=> sin(x) cos(x) = 1 or sin(x) cos(x) = 1/3 If sin(x) cos(x) = 1 then sin(2x) = 2 and this is impossible. Conclusion: From (1) it follows that sin(x) cos(x) = 1/3 (2)  Now, we use this result. We bring (2) in (1) sin(x) + cos(x) =  sqrt(3)/3 (3) If (1) is true, then (2) is true and thereby (3) is true. Now, we show that the reverse is true. We start with (3). sin(x) + cos(x) =  sqrt(3)/3 => (sin(x) + cos(x)^{2} = 1/3 => 1 + 2 sin(x) cos(x) = 1/3 => sin(x) cos(x) = 1/3 and this is (2) So, If (3) is true , then (2) is true and thereby (1) is true. Conclusion:A second method
(1) and (3) are equivalent equations. We'll solve (3) now.  sin(x) + cos(x) =  sqrt(3)/3 <=> cos(pi/2 x) + cos(x) =  sqrt(3)/3 <=> 2 cos(pi/4) cos (pi/4x) =  sqrt(3)/3 <=> sqrt(2) cos (pi/4x) =  sqrt(3)/3 <=> cos (pi/4x) = 1/ sqrt(6) Let a = arccos(1/sqrt(6)) <=> cos (pi/4x) = cos(a) <=> pi/4  x = a + 2 k pi or pi/4  x = a + 2 k pi <=> x = pi/4  a + 2 k pi or x = pi/4 + a + 2 k pi The solutions of the given equation are the values pi/4  a + 2 k pi en pi/4 + a + 2 k pi with a = arccos(1/sqrt(6))
Many equations can be solved in different ways. We'll show an alternative way to solve the equation
1/sin(x) + 1/cos(x) = sqrt(3)The period of the function on the lefthand side is 2 pi. If we have the solutions to the equation in [0, 2pi], then we know all solutions.
First, we calculate the solutions in [0, 2pi]. Since the righthand side of the equation is positive, solutions are only possible when the lefthand side is positive too.
By plotting the function 1/sin(x) + 1/cos(x), we see that the image is positive in the intervals (0,pi/2) ; (3pi/4, pi) and (3pi/2, 7pi/4).
The solutions can only occur in these intervals. If we restrict the values of x to these intervals, then both sides of the equation are positive and we can write
1/sin(x) + 1/cos(x) = sqrt(3) <=> (1/sin(x) + 1/cos(x))^{2} = 3 sin(x) + cos(x) <=> ()^{2} = 3 sin(x) cos(x) <=> 1 + 2 sin(x) cos(x) = 3 sin^{2}(x) cos^{2}(x) Let y = sin(x) cos(x) <=> 3 y^{2}  2 y  1 = 0 <=> y = 1 or y = 1/3 Case y = 1 sin(x) cos(x) = 1 <=> 2 sin(x) cos(x) = 2 <=> sin(2x) = 2 In this case, there are no solutions Case y = 1/3 2 sin(x) cos(x) = 2/3 <=> sin(2x) = 2/3 let b = arcsin(2/3) ; b = 0.7297 <=> sin(2x) = sin(b) <=> 2x = b + 2 k pi or 2x = (pib) + 2kpi <=> x = b/2 + k pi or x = pi/2  b/2 + k pi Now, we will take only the xvalues located in the intervals (0,pi/2) ; (3pi/4, pi) en (3pi/2, 7pi/4). There are 2 solutions : x = b/2 + pi = 2.7767 and x = pi/2  b/2 + pi = 3pi/2  b/2 = 5.077 All solutions to the equation 1/sin(x) + 1/cos(x) = sqrt(3) are b/2 + pi + 2 k pi en 3pi/2  b/2 + 2 k pi
'=<' means equal or less than
'>=' means equal or greater than
We draw the solutions for (x/2) on the unit circle.
Now, we see that:
sin(x/2) > 1/2 <=> pi/6 + 2 k pi < x/2 < 5pi/6 + 2 k pi <=> pi/3 +4 k pi < x < 5pi/3 + 4 k piFor each k, we have an open interval with solutions.
V = { U (pi/3 +4 k pi , 5 pi/3 + 4 k pi)  k in Z } k
We draw the solutions for (2x) on the unit circle.
Now, we see that:
tan(2x) < 1/3 <=> pi/2 + k pi < 2x < 0.32 + k pi <=> pi/4 + k pi/2 < x < 0.16 + k pi/2For each k, we have an open interval with solutions.
V = { U (pi/4 + k pi/2 , 0.16 + k pi/2)  k in Z } k
This is a variation on the previous example.
The figure is the same as the previous one, but now it gives the
solutions for (2x + pi/5).
Now we have :
tan(2x + pi/5 ) < 1/3 <=> pi/2 + k pi < 2x + pi/5 < 0.32 + k pi <=> pi/2 pi/5 + k pi < 2x < 0.32  pi/5 + k pi <=> 7 pi/20 + k pi/2 < x < 0.16 pi/10 + k pi/2The solution set V is:
V = { U (7 pi/20 + k pi/2 , 0.16  pi/10 + k pi/2)  k in Z } k
Let t = sin(x) .
2 t^{2}  3 t + 1 < 0 <=> 2 (t  1)(t  1/2) < 0 A sign study of the left hand side gives <=> 1/2 =< t =< 1 <=> 1/2 =< sin(x) =< 1 Draw the solutions for x on the unit circle. We see that: <=> pi/6 + 2 k pi =< x =< 5pi/6 + 2 k piThe solution set is
V = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi]  k in Z } k
One can also factor the left hand side directly and investigate the sign in a periodinterval.
2 sin^{2}(x)  3 sin(x) + 1 <=> 2 (sin(x)  1)(sin(x)  1/2)We take a simple periodinterval [0,2pi). We investigate the sign of each factor.
x 0 pi/6 pi/2 5pi/6 pi 2pi  sin(x)1       0                sin(x)1/2   0 + + + + + + 0            product + + 0    0   0 + + + + + + + + + + The solutions in the periodinterval are
The solution set is :
V = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi]  k in Z } k
We investigate first the inequality in the periodinterval [0,2pi).
The values 0 ; pi/2 ; pi ; 3pi/2 are no solutions . We investigate the other values of x in each quadrant.
cosec(x) < sec(x) <=> 1/ sin(x) < 1 / cos(x) now we have sin(x).cos(x) > 0 <=> cos(x) < sin(x)The solution set is ( pi/4, pi/2 ).
Now we have cos(x) < 0 and sin(x) > 0. There are no solutions.
cosec(x) < sec(x) <=> 1/ sin(x) < 1 / cos(x) now we have sin(x).cos(x) > 0 <=> cos(x) < sin(x)The solution set is ( pi , 5 pi/4 )
The solution set is ( 3 pi/2 , 2pi)
(pi/4 + 2 k pi , pi/2 + 2 k pi ) (pi + 2 k pi , 5pi/4 + 2 k pi ) (3pi/2 + 2 k pi , 2 pi + 2 k pi ) with k in Z
cot(x) + 1  > 0 sin (x) cot(x) + 1 <=>  > 0 and sin (x) not 0 sin (x) cos(x) + sin(x) <=>  > 0 and sin (x) not 0 sin^{2}(x) <=> cos(x) + sin(x) > 0 and sin (x) not 0 <=> sin(x) + sin(pi/2 x) > 0 and sin (x) not 0 with Simpson's formulas <=> 2 sin(pi/4) cos( x  pi/4) > 0 <=> cos( x  pi/4) > 0 and sin (x) not 0Using the unit circle we see that
<=> pi/2 + 2k pi < x  pi/4 < pi/2 + 2k pi and sin (x) not 0 <=> pi/4 + 2k pi < x < 3pi/4 + 2k pi and sin (x) not 0The solution set V is the union of the open intervals
V = { U (pi/4 + 2k pi , 3pi/4 + 2k pi )  k in Z } \ { k pi  k in Z } k
Example 1
arcsin(2x) = pi/4 + arcsin(x) <=> / arcsin(2x) = a  arcsin(x) = b (1) \ a = pi/4 + b / sin(a) = 2x =>  sin(b) = x \ a = pi/4 + b => / sin(pi/4 + b) = 2x \ sin(b) = x sum formulas => / cos(b) + sin(b) = 2x.sqrt(2) \ sin(b) = x => / cos(b) = 2x.sqrt(2)  x \ sin(b) = x => (2x.sqrt(2)  x)^{2} + x^{2} = 1 => .... => x = +0.4798 or x = 0.4798We test these values against the initial equation. The only solution is 0.4798.
Example 2
arctan(x+1) = 3.arctan(x1) <=> / arctan(x+1) = a  arctan(x1) = b \ a = 3 b => / tan(a) = x + 1  tan(b) = x  1 \ a = 3 b => / tan(3b) = x+ 1 \ tan(b) = x  1 3 tan(b)  tan^{3}(b) but tan(3b) =  1  3.tan^{2}(b) 3(x1)  (x1)^{3} => x+1 =  1  3 (x1)^{2} => (x+1) (1  3 (x1)^{2}) = 3(x1)  (x1)^{3} => ... => x = 0 or x = sqrt(2) or x = sqrt(2)We test these values against the initial equation. The only solution is sqrt(2).
Example 3
arctan(x) + arctan(2x) = pi/4 <=> / arctan(x) = a  arctan(2x) = b \ a + b = pi/4 => / x = tan(a)  2x = tan(b) \ a + b = pi/4 => / x = tan(a) \ 2x = tan(pi/4a) 1  tan(a) but tan(pi/4a) =  since tan(pi/4) = 1 1 + tan(a) 1  x => 2x =  1 + x => ... => x = (3+sqrt(17))/4 or x = (3sqrt(17))/4We test these values against the initial equation. The only solution is (3+sqrt(17))/4.
Example 4
arctan( (x+1)/(x+2) )  arctan ( (x1)/(x2) ) = arccos( 3/sqrt(13) ) / arctan( (x+1)/(x+2) ) = a <=>  arctan( (x1)/(x2) ) = b  arccos( 3/sqrt(13) ) = c \ a  b = c / tan(a) = (x+1)/(x+2) =>  tan(b) = (x1)/(x2)  cos(c) = 3/sqrt(13) \ a  b = c tan(a)  tan(b) but tan(ab) =  and after calculation one finds 1 + tan(a) tan(b) 2 x / tan(c) =  =>  2 x^{2}  5  \ cos(c) = 3/sqrt(13) From the last equation it follows 1 + tan^{2}(c) = 1/cos^{2}(c) = 13/9 => tan^{2}(c) = 4/9 There are now two cases First case : 2 x / tan(c) =  =>  2 x^{2}  5  \ tan(c) = 2/3 => .... => x = 1 or x = 5/2 Second case: 2 x / tan(c) =  =>  2 x^{2}  5  \ tan(c) =  2/3 => .... => x = 1 or x = 5/2We test these values against the initial equation. The only solutions are 1 and 5/2.
Example 1
________  2 \ 1  p Show that cot(arcsin(p)) =  p 
let b = arcsin(p) ,then sin(b) = p with b in [pi/2 , pi/2]. So, cos(b) = sqrt( 1  p^{2}) and ________  2 \ 1  p cot(arcsin(p)) = cot(b) =  pExample 2
Prove that the following equation has no solutions for x > 0
cos(arctan(x)) + x sin(arctan(x)) = x 
Let u = arctan(x) ; Since x > 0 is u in (0, pi/2) and tan(u) = x 1 + tan^{2}(u) = 1 + x^{2} <=> 1/ cos^{2}(u) = 1 + x^{2} <=> cos(u) = 1/ sqrt(1+x^{2})                 sin(u) = tan(u) . cos(u) <=> sin(u) = x / sqrt(1+x^{2})                 cos(arctan(x)) + x sin(arctan(x)) = x <=> cos(u) + x sin(u) = x <=> 1/ sqrt(1+x^{2}) + x^{2} / sqrt(1+x^{2}) = x <=> 1 + x^{2} = x . sqrt(1+x^{2}) <=> (1 + x^{2})^{2} = x^{2} . (1+x^{2}) <=> 1 + x^{2} = x^{2} and this equation has no solutions.
Example 3
Find the domain of arccos(arcsin(x)) 
x belongs to the domain of arccos(arcsin(x)) <=>  1 =< arcsin(x) =< 1 and since the sine function increases in [1,1] <=> sin(1) =< x =< sin(1)Conclusion : The domain of arccos(arcsin(x)) is [ sin(1), sin(1) ]
Example 4
Find the domain of arcsin(arccos(x)) 
x belongs to the domain of arcsin(arccos(x)) <=>  1 =< arccos(x) =< 1 <=> 0 =< arccos(x) =< 1 and since the cosine function decreases in [0,1] <=> cos(0) >= x >= cos(1)Conclusion : The domain of arcsin(arccos(x)) is [cos(1),1]
Example 5
f(x) = arccos(a. arcsin(x)) with a > 0
Find the values of a such that the domain D of f(x) is as large as possible. Calculate the value of a such that the domain D = [0.5 ; 0.5] 
x belongs to the domain of arccos(a. arcsin(x)) <=>  1 =< a . arcsin(x) =< 1 <=> 1/a =< arcsin(x) =< 1/a (*) First case: 1/a =< pi/2 Now (*) is equivalent with sin(1/a) =< x =< sin(1/a) (**) In this case the domain D is maximum if and only if sin(1/a)is maximum. Then a = 2/pi. Second case: 1/a > pi/2 Now (*) is equivalent with pi/2 =< arcsin(x) =< pi/2 In this case the domain D is always [1,1]. Conclusion: If a =< 2/pi , then the domain D is as large as possible. From (**) it follows that domain D is [1/2,1/2] <=> sin(1/a) = 1/2 <=> a = 6/piRemark : you can illustrate and explore many of the previous steps with a function plotter.
Example 6
Prove that the following expression holds for all the positive integers n.
1 arctan  = arctan(1/n)  arctan(1/(n+1)) n^{2} + n + 1 
1 1 1 0 <  <  <  =< 1 n^{2} + n + 1 n+1 n and therefore 1 1 1 0 < arctan  < arctan  < arctan  =< pi/4 n^{2} + n + 1 n+1 n This means that the expressions 1 arctan  and arctan(1/n)  arctan(1/(n+1)) n^{2} + n + 1 are in (0, pi/4 ] for all n. Therefore, the expression 1 arctan  = arctan(1/n)  arctan(1/(n+1)) n^{2} + n + 1 is equivalent with 1  = tan ( arctan(1/n)  arctan(1/(n+1)) ) n^{2} + n + 1 We simplify the right side tan ( arctan(1/n)  arctan(1/(n+1)) ) tan (arctan(1/n))  tan (arctan(1/(n+1))) =  1 + tan (arctan(1/n)) . tan (arctan(1/(n+1))) 1/n  1/(n+1) =  1 + (1/n)(1/(n+1)) 1 =  n^{2} + n + 1Remark : you can illustrate and explore many of the previous steps with a function plotter.