Problems about trigonometry

• READ THIS FIRST
  
• Identities
  
• Equations
  
• Other problems
  

READ THIS FIRST

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Identities

• 
  

  prove that : cos(t)+sin(t) cos(2t) ------------- = ------------- cos(t)-sin(t) 1- sin(2t)

  
  

   
    cos(2t)
  ---------- =
   1- sin(2t)
  
  
  cos2(t)-sin2(t)
  ---------------    =
  1-2sin(t)cos(t)
  
       (cos(t)-sin(t))(cos(t)+sin(t))
  ------------------------------------------ =
    cos(t)cos(t)-2sin(t)cos(t)+sin(t)sin(t)
  
  (cos(t)-sin(t))(cos(t)+sin(t))
  ------------------------------- =
  (cos(t)-sin(t))(cos(t)-sin(t))
  
  cos(t)+sin(t)
  ------------
  cos(t)-sin(t)
  
  

• 
  

  show that sin(p)-sin(q) p+q -------------- = cot(-----) cos(q)-cos(p) 2

  
  

   
  
  sin(p)-sin(q)
  -------------- =
  cos(q)-cos(p)
  
  2cos((p+q)/2)sin((p-q)/2)
  -------------------------- =
  -2sin((p+q)/2)sin((q-p)/2)
  
  cos((p+q)/2)
  ------------ =
  sin((p+q)/2)
  
        p+q
   cot(-----)
         2
  

• 
  

  Three real numbers a,b,c are successive terms of an arithmetic sequence. Show that sin(a)+sin(b)+sin(c) --------------------- = tan(b) cos(a)+cos(b)+cos(c)

  
  
  We can represent the tree terms by b-v,b,b+v . Then we have to show that

   
  sin(b-v)+sin(b)+sin(b+v)
  ------------------------- = tan(b) <=>
  cos(b-v)+cos(b)+cos(b+v)
  
  
  2sin(b)cos(v)+sin(b)
  --------------------- = tan(b) <=>
  2cos(b)cos(v)+cos(b)
  
  
  sin(b)(2cos(v)+1)
  ----------------- = tan(b) <=>
  cos(b)(2cos(v)+1)
  
  
  sin(b)
  ------- = tan(b)
  cos(b)
  
  

• 
  

  Given: a+b+c=pi Prove that cos(2a)+cos(2b)+cos(2c)+1=-4cos(a)cos(b)cos(c)

  
  

   
  cos(2a)+cos(2b)+cos(2c)+1=
  
  2cos(a+b)cos(a-b)+2cos2 (c) =
  
          but     a+b=pi-c =>cos(a+b)=-cos(c)
  
  -2cos(c)cos(a-b)+2cos2(c) =
  
  -2cos(c).(cos(a-b)-cos(c)) =
  
  -2cos(c).(cos(a-b)-cos(a+b)) =
  
  -2cos(c).2cos(a)cos(b) =
  
  -4cos(a)cos(b)cos(c)
  

• 
  

  Prove that tan(pi/8) = sqrt(2) - 1

  
  
  We know that tan(pi/4) = 1.
  Let tan(pi/8) = x .
  We know that

   
                2 tan(u)
  tan(2u)  = ---------------
              1- tan(u)tan(u)
  

  Make u = pi/8

   
                 2 x
          1  = -------
                1- x.x
  
  <=>     x2 + 2x - 1 = 0
  
  Since tan(pi/8) is positive, we only take the positive root.
  
          x = sqrt(2) - 1
  

• 
  

  a+b+c=pi . Prove that tan(a)+tan(b)+tan(c) =tan(a).tan(b).tan(c)

  
  

   
          b+c = pi - a
  
  =>      tan(b+c) = tan(pi - a)
  
          tan(b) + tan(c)
  =>      ----------------- = -tan(a)
          1 - tan(b).tan(c)
  
  
  =>      tan(b) +tan(c) = -tan(a).(1 - tan(b).tan(c))
  
  =>      tan(a)+tan(b)+tan(c) =tan(a).tan(b).tan(c)
  

• 
  

  Prove that sin(7.pi/12) = (sqrt(6) +sqrt(2))/4

  
  

   
  sin(7.pi/12) = sin (pi/3 + pi/4)
  
  = sin(pi/3).cos(pi/4) + cos(pi/3).sin(pi/4)
  
  = sqrt(3)/2 .sqrt(2)/2 + 1/2 . sqrt(2)/2
  
  = (sqrt(6) +sqrt(2))/4
  

• 
  

  a+b+c=pi. Prove that cos(a).cos(a) + cos(b).cos(b) + cos(c).cos(c) + 2cos(a).cos(b).cos(c) =1

  
  

   
          a+b = pi - c
  
  =>      cos(a+b) = cos(pi - c)
  
  =>      cos(a+b) = -cos(c)
  
  =>      cos(a).cos(b) - sin(a).sin(b) = - cos(c)
  
  =>      cos(a).cos(b) + cos(c) = sin(a).sin(b)
  
  =>      (cos(a).cos(b) + cos(c))2  = (sin(a).sin(b))2
  
  
  =>      (cos(a).cos(b))2 +2.cos(a).cos(b).cos(c) + cos2(c)
  
                          = (1 -  cos2(a)).(1 - cos2(b))
  
  =>      ...
  
  
  => cos(a).cos(a) + cos(b).cos(b) + cos(c).cos(c) + 2cos(a).cos(b).cos(c) =1
  

• 
  

  Prove that sin(a) + sin(b) + sin(c) = sin(a + b + c) + 4sin((b+c)/2) sin((c+a)/2) sin((a+b)/2)

  
  

   
                          a + b      a - b
  sin(a) + sin(b) = 2 sin(-----) cos(-----)
                            2          2
  
                                  a + b + 2 c     a + b
  sin(a + b + c) - sin(c) = 2 cos ----------- sin(-----)
                                       2            2
  
  Then
  
          sin(a)+ sin(b) + sin(c) - sin(a + b + c)
  
                  a + b       a - b        a + b + 2 c
          = 2 sin(-----) (cos(-----) - cos ----------- )
                    2           2               2
  
                  a + b         a + c     b + c
          = 2 sin(-----).2  sin(-----)sin(-----)
                    2             2         2
  
                 b + c      c + a      a + b
          = 4sin(-----) sin(-----) sin(-----)
                   2          2          2
  

• 
  

  Prove that sin(2 arctan(x)) = 2x/(1 + x2)

  
  

   
  Let arctan(x) = a => tan(a) = x
  
  sin(2 arctan(x)) = sin(2a) = 2sin(a)cos(a) = 2 tan(a).cos2(a)
  
             2 tan(a)
          = -------------------
            1 + tan2(a)
  
          = 2x/(1 + x2)
  

• 
  

  tan(a) and tan(b) are the roots of x2 + p.x + q = 0 Prove that sin(a+b).sin(a+b) + p.sin(a+b).cos(a+b) + q.cos(a+b).cos(a+b) = q

  
  

   
  tan(a) + tan(b) = -p and tan(a).tan(b) = q
  
  sin(a+b).sin(a+b) + p.sin(a+b).cos(a+b) + q.cos(a+b).cos(a+b)
  
    sin(a+b).sin(a+b) + p.sin(a+b).cos(a+b) + q.cos(a+b).cos(a+b)
  = -------------------------------------------------------------
             sin(a+b).sin(a+b) + cos(a+b).cos(a+b)
  
      divide numerator and denominator by cos(a+b).cos(a+b)
  
    tan(a+b).tan(a+b)+ p.tan(a+b) +q
  = ---------------------------------
    tan(a+b).tan(a+b) + 1
  
  
     tan(a) + tan(b)    tan(a) + tan(b)     tan(a) + tan(b)
     -----------------. --------------- + p ----------------  +q
     1 - tan(a)tan(b)   1 - tan(a)tan(b)    1 - tan(a)tan(b)
  = ----------------------------------------------------------------
     tan(a) + tan(b)    tan(a) + tan(b)
     ---------------- . ---------------- + 1
     1 - tan(a)tan(b)   1 - tan(a)tan(b)
  
      - p     - p         - p
     ------. ------ + p. ------ + q
     1 - q   1 - q       1 - q
  = ----------------------------------
      - p     - p
     ------ .------  + 1
     1 - q   1 - q
  
  = .... a little algebra
  
  = q
  

• 
  

  Prove that for all x > 1 2 arctan(x) + arcsin ( 2x/ (1 + x2) ) is constant

  
  

   
  Let arctan(x) = y. Then, x = tan(y) with y in ]pi/4 , pi/2[
  
    2x        2 tan(y)
  -------- = ----------------- = sin(2y) with 2y in ]pi/2 , pi[
  1 + x2    1 + tan(y).tan(y)
  
  <=>
  
    2x
  -------- = sin(pi - 2y) with (pi - 2y) in ]-pi/2 , 0 [
  1 + x2
  
  <=>
            2x
  arcsin( -------- ) = pi - 2y
          1 + x2
  
  Hence,
                                2x
          2 arctan(x) + arcsin --------  = 2y + pi - 2y = pi
                               1 + x2
  

• 
  

  The angles of a triangle ABC are a,b and c. Prove that The triangle is rectangular if and only if sin(4a) + sin(4b) + sin(4c) = 0

  
  

  Part 1: if the triangle is rectangular then sin(4a)+sin(4b)+sin(4c) = 0.

   
          triangle is rectangular => a or b or c is pi/2
  
  Suppose a = pi/2, then  sin(4a) = 0 and  b = pi/2 -c
  
          => 4 b = 2 pi - 4c  => sin(4b) = sin(-4c)
  
          => sin(4b) + sin(4c) = 0
  
  So      sin(4a) + sin(4b) + sin(4c) = 0
  
  Similar proof for b = pi/2 or c = pi/2
  

  Part 2: if sin(4a)+sin(4b)+sin(4c) = 0 then the triangle is rectangular.

   
          a + b + c = pi
  
  =>      4a + 4b + 4c = 4 pi
  
  =>      4a = 4 pi - (4b + 4c)
  
  =>      sin(4a) = sin( -(4b + 4c)) = - sin(4b + 4c)
  
  =>      sin(4a) = - (sin(4b)cos(4c) + cos(4b)sin(4c))
  
  Thus,
  
     sin(4a)+sin(4b)+sin(4c)
  
          = sin(4b) - sin(4b)cos(4c) + sin(4c) - cos(4b)sin(4c)
  <=>
          0 = sin(4b) (1 - cos(4c)) + sin(4c) (1 - cos(4b))
  <=>
          0 = sin(4b) 2 sin2 (2c) + sin(4c).2sin2 (2b)
  <=>
          0 = 2.sin(2b).cos(2b).2 sin2 (2c) +  2.sin(2c).cos(2c).2 sin2 (2b)
  <=>
          0 = 4.sin(2b).sin(2c).sin(2b + 2c)
  <=>
          sin(2b) = 0  or sin(2c) = 0  or sin(2b + 2c) = 0
  <=>
          b = pi/2     or c = pi/2     or b + c = pi/2
  <=>
          b = pi/2     or c = pi/2     or  a = pi/2
  

• 
  

  Given: in a triangle ABC with sides a,b and c A - C A + C b cos(-----) - 3 c cos(-----) = 0 2 2 Prove that a = 2 c

  
  

   
  Proof:
  Appealing to the sine rule we can write
  
                A - C            A + C
          b cos(-----) - 3 c cos(-----) = 0
                  2                2
  <=>
                     A - C                 A + C
          sin(B) cos(-----) - 3 sin(C) cos(-----) = 0
                       2                     2
  
  
  Now,    A + B + C = pi
  
           A + C    pi   B
  =>      (-----) = -- - -
             2      2    2
  
  =>          A + C        B
          cos(-----) = sin -
                2          2
  
  Hence
                     A - C                 A + C
          sin(B) cos(-----) - 3 sin(C) cos(-----) = 0
                       2                     2
  
  <=>
                     A - C                 B
          sin(B) cos(-----) - 3 sin(C) sin - = 0
                       2                   2
  <=>
              B     B     A - C                 B
        2 sin - cos - cos(-----) - 3 sin(C) sin - = 0
              2     2       2                   2
  <=>
              B         B     A - C
          sin - .(2 cos - cos(-----) - 3 sin(C))= 0
              2         2       2
                      B
                  sin - = 0 is impossible in a triangle,
                      2
                              B       A + C
                  and     cos - = sin(-----) , hence
                              2         2
  
                A + C     A - C
  <=>     2(sin(-----)cos(-----) = 3 sin(C)
                  2         2
  
  <=>     sin(A) + sin (C) =  3 sin(C)
  
  <=>     sin(A) = 2 sin(C)
  
  <=>     a = 2 c
  

• 
  

  Given: in a triangle ABC a cos(B) + b cos(C) = A tan(A) tan(B/2) cos(B) + b tan(B/2) sin(C) Prove that the triangle is isosceles.

  
  

   
  Proof:
          a cos(B) + b cos(C) = A tan(A) tan(B/2) cos(B) + b tan(B/2) sin(C)
  
  <=>
          a cos(B)(1 - tan(A) tan(B/2)) = b(tan(B/2) sin(C) - cos(C))
  
  <=>     a   tan(B/2) sin(C) - cos(C)
          - = -----------------------------
          b   cos(B)(1 - tan(A) tan(B/2))
  
          sin(A)     tan(B/2) sin(C) - cos(C)
  <=>     ------- = ---------------------------
          sin(B)     cos(B)(1 - tan(A) tan(B/2))
  
  
  <=>     sin(A) cos(B)(1 - tan(A) tan(B/2)) =sin(B)(tan(B/2) sin(C) - cos(C))
  
  multiplying through by cos(B/2)
  
  
  <=>sin(A)cos(B)(cos(B/2)-tan(A)sin(B/2))=sin(B)(sin(B/2)sin(C)-cos(C)cos(B/2))
  
  
        sin(A)cos(B)
  <=>   ------------(cos(A)cos(B/2)-sin(A)sin(B/2)) = -sin(B)cos(C+B/2)
           cos(A)
  
  <=>     tan(A)cos(B)cos(A + B/2) = - sin(B)cos(C+B/2)
  
  
              Now (A + B/2) + (C+B/2) = pi => cos(A + B/2)  = - cos(C+B/2)
  
  
  <=>     tan(A)cos(B)cos(A + B/2) = sin(B)cos(A + B/2)
  
  <=>     tan(A)cos(B)cos(A + B/2) - sin(B)cos(A + B/2) = 0
  
  <=>     cos(A + B/2) = 0  or tan(A)cos(B) - sin(B) = 0
  
  <=>     A + B/2 = pi/2          or      tan(A)cos(B) = sin(B)
  
  <=>     2A + B = pi             or      tan(A) = tan(B)
  
  <=>     2A + B = A + B + C      or      A = B
  
  <=>     A = C
  
  <=>     the triangle is isosceles.
  

• 
  

  Show that sin(7 pi/12) sin(pi/12) = 1/4

  
  

  We use the fact that (7 pi/12 + pi/12) and (7 pi/12 - pi/12) are special angles and we use a formula so that sin(7 pi/12) sin(pi/12) occurs.

   
     cos(7 pi/12) cos(pi/12) + sin(7 pi/12) sin(pi/12)
  
    = cos(pi/2)
  
    = 0                                (1)
  
  and
  
     cos(7 pi/12) cos(pi/12) - sin(7 pi/12) sin(pi/12)
  
    = cos(8 pi/12)
  
    = cos(2 pi/3)
  
    = -1/2                             (2)
  
  We calculate (1) - (2)
  
     2  sin(7 pi/12) sin(pi/12)  = 1/2
  <=>
        sin(7 pi/12) sin(pi/12)  = 1/4
  

Equations

• 
  

  Solve the quadratic equation in x (sin(2a)).x2 - 2.x.(sin(a)+cos(a)) +2 =0

  
  

   
  Discriminant = 4(sin(a)+cos(a))2  - 8.sin(2a)
  
          = 4.( sin2 (a) + 2.sin(a).cos(a) + cos2 (a) - 4 sin(a)cos(a) )
  
          = 4.(sin(a) - cos(a))2
  
  So, the roots are
              sin(a) + cos(a) + sin(a) - cos(a)         1
          x = --------------------------------- =...= ------
                          sin(2a)                     cos(a)
                     1
  and     x = ... = ------
                    sin(a)
  

• 
  

  Solve cos(2x) + sin2(x) = 1/2

  
  

   
  <=> cos2(x) - sin2(x) + sin2(x) = 1/2
  
  <=> cos2(x) = 1/2
  
  <=> cos(x) = 1/sqrt(2)  or cos(x) = -1/sqrt(2)
  
  <=> cos(x) = cos(pi/4)  or cos(x) = cos(pi - pi/4)
  
       ......             or    ......
  

• 
  

  Solve 2 sin2(3x) + sin2(6x) = 2

  
  

   
     2 sin2(3x) + sin2(6x) = 2
  <=>
     2 sin2(3x) + 4 sin2(3x).cos2(3x) = 2
  <=>
     4 sin2(3x).cos2(3x) - 2 cos2(3x) = 0
  <=>
     cos2(3x) ( 2 sin2(3x) -1) = 0
  <=>
     cos(3x) = 0  or sin(3x) = 1/sqrt(2) or sin(3x) = -1/sqrt(2)
  <=>
      ....   or  .... or ....
  
  

• 
  

  Solve sin3(x) - sin2(x) - sin(x)/4 +1/4 =0

  
  

   
  <=> sin2(x) . (sin(x) -1) - (sin(x) - 1)/4 = 0
  
  <=>  (sin(x) -1) . ( sin2(x) - 1/4) = 0
  
  <=> sin(x) = 1  or sin(x) = 1/2  or sin(x) = -1/2
  
  <=>  ......     or    ......     or    .....
  

• 
  

  Solve 1 + sin(x) + cos(x) = 0

  
  

   
  <=> sin(x) + ( 1+ cos(x))  = 0
  
  <=> 2 sin(x/2).cos(x/2) + 2 cos2(x/2) = 0
  
  <=> 2 cos(x/2) ( sin(x/2) + cos(x/2) ) =0
  
  First case
  
    cos(x/2) = 0
  
  <=> ....
  
  <=> x = pi + 2.k.pi
  
  Second case
  
       sin(x/2) = - cos(x/2)
  
  <=>  sin(x/2) = cos( pi - x/2 )
  
  <=>  sin(x/2) = sin(pi/2 - pi + x/2)
  
  <=>  sin(x/2) = sin(-pi/2 + x/2)
  
  <=>   ....
  

• 
  

  Solve 3sin(x) + cos(x) = 5

  
  
  Since sin(x) and cos(x) have 1 as an absolute maximum, the equation has no solutions.

• 
  

  Solve: 2.tan(x) = sin(4x) - 2.sin(2x)tan(x)cos(2x)tan(x)

  
  

   
  
  <=>      2.tan(x) = sin(4x) - sin(4x).tan2 (x)
  
  <=>      2.tan(x) = sin(4x)(1 - tan2(x))
  
          2.sin(x)    sin(4x).(cos2(x) - sin2(x))
  <=>     --------- = ----------------------------
           cos(x)          cos(x).cos(x)
  
  <=>     2.sin(x). cos(x) = sin(4x).(cos2 (x) - sin2 (x))  and cos(x) not 0
  
  <=>     sin(2x) - 2.sin(2x). cos(2x).cos(2x)  and cos(x) not 0
  
  <=>     sin(2x) .(1 - 2.cos2 (2x)) = 0 and cos(x) not 0
  
  <=>     2.sin(x). cos(x).(1 - 2.cos2 (2x)) = 0 and cos(x) not 0
  
  <=>     sin(x) = 0  or cos(2x) = 1/sqrt(2) or cos(2x) =-1/sqrt(2)
  
  <=>     x = k.pi  or 2x = pi/4 + k(pi/2)
  
  <=>     x = k.pi  or x = pi/8 + k(pi/4)
  

• 
  

  Given : cos(2x) = a (cos(x) - sin(x)) (a is a real parameter) Solve this equation and discuss.

  
  

   
          cos(2x) = a (cos(x) - sin(x))
  
  <=>     cos2 (x) - sin2 (x) = a (cos(x) - sin(x))
  
  <=>     (cos(x) + sin(x))(cos(x) - sin(x)) = a (cos(x) - sin(x))
  
  <=>     (cos(x) - sin(x))(cos(x) + sin(x) -a) = 0
  
  <=>     cos(x) = sin(x)     or   cos(x) + sin(x) -a = 0
  
  <=>    x = pi/4 + k.pi  or      cos(x) + tan(u).sin(x) = a with u = pi/4
  
  <=>    x = pi/4 + k.pi  or cos(u).cos(x)+sin(u).sin(x)=a.cos(u) with u = pi/4
  
  <=>    x = pi/4 + k.pi  or cos(x - u) = a/sqrt(2)         (*)
  
          The last part has solutions if and only if
  
                  a in [-sqrt(2) , sqrt(2) ]
  Case 1 :   a is not in [-sqrt(2) , sqrt(2) ]
          The solutions are x = pi/4 + k.pi
  
  Case 2 :   a in [-sqrt(2) , sqrt(2) ]
          Now it is possible to choose t = arccos(a/sqrt(2))  then (*) gives
  
          x = pi/4 + k.pi   or
          x = pi/4 + t + 2.k.pi or
          x = pi/4 - t + 2.k.pi
  

• 
  

  Solve F = ( 2 sin(2 x) - 1 ) / (cos(2 x) - 3 cos(x) + 2) > 0 with x in [0,2.pi]

  
  

   
  First we investigate  N = 2 sin(2 x) - 1
  
          2 sin(2 x) - 1 > 0
  
  <=>     sin(2 x) > 1/2
  
  <=>     pi/6 + 2.k.pi < 2x < 5.pi/6 + 2.k.pi
  
  <=>     pi/12 + k.pi < x < 5.pi/12 + k.pi
  
  Now we investigate  D = cos(2 x) - 3 cos(x) + 2
  
          cos(2 x) - 3 cos(x) + 2 > 0
  
  <=>     2 cos2 (x) - 3 sin(x) + 1 > 0
  
  <=>     2(cos(x) - 1/2)(cos(x) - 1) > 0
  
                  since (cos(x) - 1) =< 0 we have
  
  <=>     cos(x) < 1/2
  
  <=>     pi/3 + 2.k.pi < x < 5.pi/3 + 2.k.pi
  
  Now we see that
  
            1         4          5          13       17       20
           --pi       --pi       --pi       --pi     --pi     --pi      2pi
  x  0     12         12         12         12       12       12
  ---------------------------------------------------------------------------
  T  -   -   0    +    +    +     0     -    0    +   0    -   -     -   -
  N  0   -   -    -    0    +     +     +    +    +   +    +   0     -   0
  F  |   +   0    -    |    +     0     -    0    +   0    -   |     +   |
  
  From this table we can easily find the solutions of F > 0 in [0, 2.pi].
  

• 
  

  Solve x 1 1 1 cot(-) - cot(x) = ------ + -------- + -------- 2 sin(x) sin(2 x) sin(4 x) with x in [0, 2pi]

  
  

   
              x
          cot(-) - cot(x)
              2
  
                 x
             cos(-)
                 2    cos(x)
          =  ------ - ------
                 x    sin(x)
             sin(-)
                 2
  
                 x       2 x       2 x
             cos(-)   cos (-) - sin (-)
                 2         2         2
          =  ------ - ------------------
                 x           x      x
             sin(-)    2 sin(-) cos(-)
                 2           2      2
  
  
  
          = ...
  
  
                    1
          = -----------------
                    x      x
              2 sin(-) cos(-)
                    2      2
               1
          = --------
             sin(x)
  
  With this result the equation is
  
              1          1
           -------- + -------- = 0
           sin(2 x)   sin(4 x)
  
              1            1
  <=>      -------- = - --------
           sin(2 x)     sin(4 x)
  
  <=>
          sin(2 x) = - sin(4 x)           with sin(4 x) not 0
  <=>
          sin(2 x) = sin(- 4 x)           with sin(4 x) not 0
  <=>
          2x = -4x + 2.k.pi  or 2x = pi + 4x + 2.k.pi  with sin(4 x) not 0
  <=>
          6x = 2.k.pi or -2x = pi + 2.k.pi        with sin(4 x) not 0
  <=>
          x = k.pi/3  or x = -pi/2 - k.pi         with sin(4 x) not 0
  
  The solutions in [0, 2pi] are
  
          pi/3 ; 2pi/3 ; 4pi/3 ; 5pi/3
  
  

• 
  

  Solve the equation ___ V 2 (sin(x) + cos(x)) + 2 cos4 (x - pi/4 ) = 0

  
  

   
  First   sin(x) + cos(x)
  
                         pi
          = sin(x) + sin(-- - x)
                         2
                  pi          pi
          = 2 sin(--) cos(x - ---)
                  4            4
            ___          pi
          =V 2   cos(x - ---)
                          4
  
  With this, the equation becomes
  
                    pi          4     pi
          2 cos(x - ---) + 2 cos (x - --- ) = 0
                     4                 4
  
  <=>
                  pi                   3     pi
          cos(x - ---) = 0  or  1 + cos (x - --- ) = 0
                   4                          4
  
  <=>          3 pi                       pi
          x =  ---- + k.pi   or   cos(x - ---) = -1
                4                          4
  
  <=>          3 pi                   3 pi
          x =  ---- + k.pi   or   x = ---- + 2.k.pi
                4                      4
  
  <=>
               3 pi
          x =  ---- + k.pi
                4
  

• 
  

  Solve the equation cos6(x) + sin6(x) = 5/8

  
  

   
         cos6(x) + sin6(x) = 5/8
  <=>
  
  (sin2(x) + cos2(x))3 - 3 sin4(x).cos2(x) -3  cos4(x).sin2(x) = 5/8
  
  <=>
  
     1 - 3 sin2(x).cos2(x).(cos2(x) + sin2(x)) = 5/8
  
  <=>
  
          -3  sin2(x).cos2(x) = - 3/8
  
  <=>       sin2(x).cos2(x) = 1/8
  
  <=>       4  sin2(x).cos2(x) = 1/2
  
  <=>         sin2( 2 x )  = 1/2
  

  Take first sin (2x) = sin (pi/4)
  The solutions are :

   
    2x = pi/4 + 2 k pi  or  2x = 3 pi/4 + 2 kpi
  
  <=>
  
    x = pi/8 + kpi    or   x = 3 pi/8 + k pi
  

  Take now the case sin (2x) = sin (- pi/4)
  The solutions are :

   
    2x = -pi/4 + 2 k pi  or  2x = 5 pi/4 + 2 kpi
  
  <=>
  
    x = - pi/8 + kpi    or    x = 5 pi/8 + k pi
  

  All these sets of solutions can be gathered in one expression

   
   x = pi/8 + k pi/4
  

• 
  

  Solve the equation sin(x) + sin(2x) + sin(3x) = cos(x) + cos(2x) + cos(3x)

  
  

   
    sin(x) + sin(2x) + sin(3x) = cos(x) + cos(2x) + cos(3x)
  <=>
    sin(x) - cos(x) +  sin(2x) - cos(2x) + sin(3x) - cos(3x) = 0
  
  First we construct a formula for sin(u) - cos(u)
  
    sin(u) - cos(u)
          = sin(u) - sin(pi/2 - u)
          = 2 cos(pi/4) sin(u - pi/4)
          = 2 sqrt(2) sin(u - pi/4)
  
  So,
  
  
    sin(x) - cos(x) +  sin(2x) - cos(2x) + sin(3x) - cos(3x) = 0
  <=>
     sin(x - pi/4) + sin(2x - pi/4) + sin(3x - pi/4) = 0
  <=>
     2 sin(2x - pi/4) cos(x) + sin(2x - pi/4) = 0
  <=>
     sin(2x - pi/4) ( 2 cos(x) +1) = 0
  <=>
     2x - pi/4 = k pi  of  cos(x) = cos(2pi/3)
  <=>
     x = pi/8 + k pi/2   of  x = ± 2 pi/3 + 2 k pi
  

• 
  

  Solve the system / 2 sin(pi - 2x) = 2 | 3 sin(4x) = cot(5 pi/2) \ 6 pi/5 < x < 11 pi/5

  
  

   
    /   2 sin(pi - 2x) = 2
    |   3 sin(4x) = cot(5 pi/2)
    \   6 pi/5 < x < 11 pi/5
  
  <=>
  
    /   sin(2x) = 1
    |   sin(4x) = 0
    \   6 pi/5 < x < 11 pi/5
  
  <=>
  
    /   sin(2x) = 1
    |   2 sin(2x) cos(2x) = 0
    \   6 pi/5 < x < 11 pi/5
  <=>
  
    /   sin(2x) = 1
    |   cos(2x) = 0
    \   6 pi/5 < x < 11 pi/5
  
  <=>
  
    /  2x = pi/2 + 2 k pi
    \  6 pi/5 < x < 11 pi/5
  
  <=>
    /  x = pi/4 + k pi
    \ 6 pi/5 < x < 11 pi/5
  
  <=>
      x = 5 pi/4
  

• 
  

  Solve sin(x + pi/5) sin(x - pi/5) = cos2(x)

  
  

   
     sin(x + pi/5) sin(x - pi/5) = cos2(x)
  <=>
    (sin(x) cos(pi/5) + cos(x) sin(pi/5)).(sin(x) cos(pi/5) - cos(x) sin(pi/5)) = cos2(x)
  <=>
    (sin2(x) cos2(pi/5) - cos2(x) sin2(pi/5)) = cos2(x)
  <=>
    (1 - cos2(x)) cos2(pi/5) - cos2(x) (1 - cos2(pi/5)) = cos2(x)
  <=>
     cos2(pi/5) - cos2(x) = cos2(x)
  <=>
     2 cos2(x) =  cos2(pi/5)
  <=>
     cos(x) = ± 0.57
  <=>
       x = ± 0.96 + k pi
  

Other problems

• 
  

  cot(x) = 1 + 1/m and cot(y) = 2m + 1 Find tan(x + y)

  
  

   
      tan(x + y)
  
        tan(x) + tan (y)
    = -------------------
       1 - tan(x) tan (y)
  
        m/(m+1) + 1/(2m+1)
    = --------------------------
        1 -  m / ((m+1)(2m+1))
  
       m(2m+1) + m + 1
    = ----------------------
       (m+1)(2m+1)  - m
  
    = ...
  
    = 1
  

• 
  

  Calculate cos(3u) in terms of cos(u)

  
  

   
  cos(2u+u) =
  
  cos(2u)cos(u)-sin(2u)sin(u) =
  
  (cos2(u)-sin2(u))cos(u)-2sin(u)cos(u)sin(u) =
  
  cos3(u) - sin2(u)cos(u)-2sin2(u)cos(u) =
  
  cos3(u) -3sin2(u)cos(u)
  
  cos3(u) -3cos(u)+3cos3(u)
  
  4cos3(u) -3cos(u)
  
  In the same way
  sin(3u) =
  
  -4sin3(u) + 3sin(u)
  

• 
  

  A plane is approaching your home, and you assume that it is traveling at approximately 550 miles per hour. If the angle of elevation of the plane is 16 degrees at one time and one minute later the angle is 57 degrees, approximate the altitude.

  
  
  We assume that the altitude is constant.
  

  

  The red line is the plane. First the plane is at point Q and one minute later the plane is at point R. Point H is my home.

  The distance |QR| is 550/60 = 55/6 miles.
  Let D = the distance |PH|.

   
  In triangle QPH,
  
        tan(16°) = A/D   => D = A.cot(16°)
  
  In triangle RSH
  
        tan(57°) = A/(D - 55/6)
  
                          A
        tan(57°) = --------------------
                    A.cot(16°) - 55/6
  
  
        (A.cot(16°) - 55/6 ).tan(57°) = A
  
        A.(cot(16°).tan(57°) - 1 ) = (55/6) .tan(57°)
  
        A = 3.23 miles.
  

  The altitude is 3.23 miles.

• 
  

  Given: 2. sqrt(x) y = arcsin --------------- 1 + x Calculate: 1) domain 2) range of the function 3) asymptotes

  
  

   
    The image y exists if and only if  2.sqrt(x)/(1+x) is in  [-1,1].
  
  <=>
  
         4x
     ----------   is in  [0,1]
      (1 + x)2
  
  <=>
         4x
     ----------  =< 1    and  ( x>0 or x=0)
      (1 + x)2
  <=>
  
     4x  =< (1 + x)2   and  ( x>0 or x=0)
  
  <=>
  
     (1 + x)2 - 4x >= 0   and  ( x>0 or x=0)
  
  <=>
  
      (1 - x)2  >= 0   en  ( x>0 or x=0)
  
  <=>   ( x>0 or x=0)
  

  The domain is the set of all non-negative real numbers.

  ------------

  arcsin(u) has pi/2 as absolute maximum for u = 1.

  2.sqrt(x)/(1+x) = 1 for x = 1.

  So, the image y is maximum for x = 1 en that maximum image is pi/2.

  An x value, in the domain, is non-negative. So, 2.sqrt(x)/(1+x) has 0 as minimum and this happens for x=0.

  The arcsin-function is rising. So, the minimum value of y happens for x=0. And then, the image is 0.

  Conclusion : the range is [0,pi/2] .

  --------------

  Since the range is [0,pi/2], there are no vertical and no horizontal asymptotes.

  If x -> infty
  

   
   Say  u = sqrt(x) ; then:   x -> infty <=> u -> infty
  
        2. sqrt(x)           2 u
   lim -----------  =  lim ----------- = 0
         1 + x              1 + u2
  

  So the x-axis is the only horizontal asymptote

• 
  

  Find the period of 3.sin2 (3x/4) - cos2 (2x/3)

  
  

   
  2.sin2(3x/4) = 1 - cos(3x/2) and this part has period 4.pi/3
  
  
  So 3.sin2(3x/4) has period 4.pi/3              (1)
  
  
  2. cos2(x/3) = 1 + cos(2x/3) and this part has period 3.pi
  
  
  So  cos2(x/3) has period 3.pi                  (2)
  
  
  From (1) and (2) we have
  
  
  the period of  3.sin2 (3x/4) - cos2 (2x/3)   is 12.pi
  
  

  Show that sin(2 arctan(x)) = 2x/(1 + x2)

  
  

   
  Let a = arctan(x)  then tan(a) = x
  
  sin(2 arctan(x)) = sin(2a) = 2sin(a)cos(a) = 2 tan(a).cos(a).cos(a)
  
             2 tan(a)
          = -------------------
            1 + tan2(a)
  
          = 2x/(1 + x2)
  

• 
  

  Calculate the range of the function f(x) = arctan(x2+ 1).

  
  
  The range of (x²+ 1) is [1, infty[ .
  Let the arctan function act on that range.
  Then, the set of all images is [pi/4, pi/2 [ . The range of f(x) = [pi/4, pi/2 [ .
• 
  

  In the figure below the angles in A and B are 60 degrees. Find x.

  
  

  

  Let O = the center of the left circle.
  Call l the perpendicular to AB through O. l intersects AB in S.
  In the triangle AOS, the angle A = 30 degrees. |OS| = 10.
  tan(30°)= 10/|AS|. From this we calculate |AS|.

  x = |AB| - 2.|AS| - 2.10 = 35.4

• 
  

  Given : sin4x + cos4x + sin2x cos2x = m cos(4x) Find: sin2(2x) as a function of m

  
  

   
  
      sin4x + cos4x + sin2x cos2x = m cos(4x)
  <=>
      (sin2x + cos2x)2 =  m cos(4x)  + sin2x cos2x
  <=>
        1 =  m cos(4x) + (1/4) sin2(2x)
  
                      but  1- cos(4x) = 2 sin2(2x)
  <=>
        1 =  m (1 - 2 sin2(2x)) + (1/4) sin2(2x)
  <=>
        1 - m = -2m  sin2(2x)) + (1/4) sin2(2x)
  <=>
       4( 1 - m) = (1 - 8m) sin2(2x)
  <=>
        4( 1 - m)
       ------------ = sin2(2x)
         (1 - 8m)
  

• 
  

  Consider the quadratic equation 2(2 cos(t) + 1) x2 - 4 x + (2 cos(t) -1) = 0 With 0 < t < pi/2 Find the t-values such that the quadratic equation has different real roots.

  
  

  The quadratic equation has different real roots if and only if the discriminant is strictly positive.

   
         D >0
  <=>
        16 - 8 (2 cos(t) + 1) (2 cos(t) -1)  > 0
  <=>
        3 - 4 cos2(t)  > 0
  <=>
         4 cos2(t) -3  < 0
  <=>
          cos2(t) < 3/4
  
                    and  since  0 < t < pi/2
  
  <=>     pi/6 < t < pi/2
  
  

• 
  

  Show that cosec(x) = cot(x/2) - cot(x) Show that cosec(2x) + cosec(4x) + cosec(8x) = sin(7x)/ ( sin(x) sin(8x) ) Solve the equation: cosec(2x) + cosec(4x) + cosec(8x) = cosec(8x) cosec(x)

  
  

  Part 1:

   
     cosec(x) = cot(x/2) - cot(x)
  <=>
          1          cos(x)
     ----------- + ---------  = cot(x/2)
       sin(x)        sin(x)
  <=>
       1 + cos(x)
     ------------- = cot(x/2)
        sin(x)
  <=>
        2 cos2(x/2)
       --------------------- =  cot(x/2)
        2 sin(x/2) cos(x/2)
  <=>
          cos2(x/2)
        ------------------- =  cot(x/2)
          sin(x/2)
  

  Part 2: We rely on the result of Part 1

   
  cosec(2x) + cosec(4x)  + cosec(8x)
  
       = cot(x) - cot(2x) + cot(2x) - cot(4x) + cot(4x) - cot(8x)
  
       = cot(x) -  cot(8x)
  
          cos(x)      cos(8x)
       = -------  -  --------
          sin(x)      sin(8x)
  
          cos(x) sin(8x) -  sin(x)  cos(8x)
      = --------------------------------------------
                sin(x)  sin(8x)
  
           sin(7x)
      = ----------------
          sin(x)  sin(8x)
  

  Part 3: We rely on the result of Part 2

   
           sin(7x)
        ---------------- = cosec(8x)  cosec(x)
          sin(x)  sin(8x)
  
  <=>
         sin(7x) = 1
  <=>
         7x = pi/2 + 2.k.pi
  <=>
         x = pi/14 + 2 k pi/7
  

• 
  

  Show that: If 5 sin(a) = sin(a+2b) then 2 tan(a+b) = 3 tan(b)

  
  

   
    sin(a + 2b) = sin( (a+b) + b) = sin(a+b) cos(b) + cos(a+b) sin(b)
    5 sin(a) = 5 sin((a+b) - b) = 5 sin(a+b) cos(b) - 5 cos(a+b) sin(b)
  So,
  
       5 sin(a) = sin(a+2b)
  
  =>   5 sin(a+b) cos(b) - 5 cos(a+b) sin(b)  =  sin(a+b) cos(b) + cos(a+b) sin(b)
  
                  dividing each term by cos(a+b). cos(b)
  
  =>   5 tan(a+b) - 5 tan(b) = tan(a+b) + tan(b)
  
  =>    2 tan(a+b) = 3 tan(b)
  

• 
  

  The positive numbers a, b, c and d satisfy the conditions: / a + b + c + d = 2 pi | tan(b) - 2 tan(a) = -1 | tan(c) + tan(a) = 0 \ tan(d) + 3 tan(a) = 2 Find tan(a)

  
  

   
  Let x=  tan(a) ; then
  
  / (a + b) + (c + d) = 2 pi   (1)
  | tan(b)  = 2x - 1
  | tan(c)  =  -x
  \ tan(d)  = 2 - 3x
  
  From  (1)
  
     tan(a+b) = - tan(c+d)
  <=>
      tan(a) + tan(b)            tan(c) + tan(d)
     -------------------  = -  --------------------
       1 - tan(a) tan(b)         1 -  tan(c) tan(d)
  <=>
     (tan(a) + tan(b)).(1 -  tan(c) tan(d)) = -(1 - tan(a) tan(b)).(tan(c) + tan(d))
  <=>
      ...
  <=>
       -x3 + x2 - x + 1 = 0
  <=>
       (x - 1)(x2+1) = 0
  <=>
        tan(a) = 1
  

• 
  

  The roots of the quadratic equation x2 + p x + q = 0 are tan(a) and tan(b). These roots are different from 1 an -1. Find a quadratic equation with roots tan(2a) and tan(2b). The coefficients of the new quadratic equation are functions of p and q.

  
  

   
  Let  t= tan(a) and  t'=tan(b)
  Then  t + t' = -p and  t t'= q
  
              2 t
  tan(2a) = --------
            1 - t2
  
              2 t'
  tan(2b) = ---------
            1 - t'2
  
                          t(1-t'2)+ t'(1-t2)
  tan(2a) + tan(2b) = 2  ----------------------
                            (1-t2) (1-t'2)
  
             t + t' - t t' (t'+t)
        = 2 ----------------------------------
              1 - (t+t')2 +2 t t' + t2 t'2
  
               -p + qp
        = 2 -----------------------
             1 - p2 + 2q + q2
  
           2p (q-1)
        = --------------
          (1-q)2 - p2
  
  
                               4q
  tan(2a) tan(2b) = ... =  --------------
                           (1-q)2 - p2
  
  The new quadratic equation is
  
      ((1-q)2 - p2) x2 + 2p(1-q) x + 4q = 0
  

• 
  

  Find the domain of the function x + 1 y = arccos -------- x - 1

  
  

   
       x is in the domain of the function
  <=>
        x + 1                      x + 1
      -------  > or = -1   and   -------- < of = 1
        x - 1                      x - 1
  <=>
         x + 1                      x + 1
       ------- + 1 > or = 0   and  ------- -1 < of = 0
         x - 1                      x - 1
  <=>
         2 x                        2
       --------  > or = 0   and  ------- < 0
         x - 1                    x - 1
  
  sign investigation
  
       x       0     1          x     1
      ------------------- and  ----------------
            +  0  -  | +           -  |   +
                XXXXXX                XXXXXXXXX
  

  Conclusion: The domain of the function = ] -infty , 0 ]

• 
  

  The angles of a triangle are a-b, a and a+b. cos(a-b).cos(a).cos(a+b) = -1/8. Find the angles of the triangle.

  
  

  The sum of the angles is 180 degrees. So 3a = 180 degrees and a is 60 degrees.

   
     cos(a) = 1/2 and sin(a) = sqrt(3)/2
   
  
     cos(a - b) cos(a + b)
  
  =  (cos(a) cos(b) + sin(a) sin(b)) (cos(a) cos(b) - sin(a) sin(b))
  
  =  (cos2(a) cos2(b) - sin2(a) sin2(b))
  
  =  0.25 cos2(b) - 0.75 (1 - cos2(b))
  
  =  cos2(b) - 0.75
  
  So,    0.5 (cos2(b) - 0.75) = -1/8
  
    <=>  cos2(b) = 1/2
  

  1. First case : cos(b) = -sqrt(2)/2

     Now b = 135 degrees. This is impossible because a - b must be positive.

  2. Second case : cos(b) = sqrt(2)/2

     b = 45 degrees.
     The angles of the triangle are 15 degrees ; 60 degrees ; 105 degrees .