Trigonometry Problems and Questions with Solutions - Grade 12

Grade 12 trigonometry problems and questions with answers and solutions are presented.




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  1. Prove the indentity

    tan2(x) - sin2(x) = tan2(x) sin2(x)

  2. Prove the indentity

    (1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x)

  3. Prove the indentity

    4 sin(x) cos(x) = sin(4x) / cos(2x)

  4. Solve the trigonometric equation given by

    sin(x) + sin(x/2) = 0 for 0 ≤ x ≤ 2 pi

  5. Solve the trigonometric equation given by

    (2sin(x) - 1)(tan(x) - 1) = 0 for 0 ≤ x ≤ 2 pi

  6. Solve the trigonometric equation given by

    cos(2x) cos(x) - sin(2x) sin(x) = 0 for 0 ≤ x ≤ 2 pi

  7. Solve the trigonometric equation given by

    ( sin(2x) - cos(x) ) / ( cos(2x) + sin(x) - 1 ) = 0 for 0 ≤ x ≤ 2 pi

  8. Prove that

    sin(105o) = ( sqrt(6) + sqrt(2) ) / 4

  9. If sin(x) = 2/5 and x is an accute angle, find the exact values of

    a) cos(2x)

    b) cos(4x)

    c) sin(2x)

    d) sin(4x)

  10. Find the length of side AB in the figure below. Round your answer to 3 significant digits.

    trigonometry grade 12 problem 10.


Solutions to the Above Problems


  1. Use the identity tan(x) = sin(x) / cos(x) in the left hand side of the given identity.

    tan2(x) - sin2(x) = sin2(x) / cos2(x) - sin2(x)

    = [ sin2(x) - cos2(x) sin2(x) ] / cos2(x)

    = sin2(x) [ 1 - cos2(x) ] / cos2(x)

    = sin2(x) sin2(x) / cos2(x)

    = sin2(x) tan2(x) which is equal to the right hand side of the given identity.


  2. Use the identities cos(2x) = 2 cos2(x) - 1 and sin(2x) = 2 sin(x) cos(x) in the left hand side of the given identity.

    [ 1 + cos(x) + cos(2x) ] / [ sin(x) + sin(2x) ]

    = [ 1 + cos(x) + 2 cos2(x) - 1 ] / [ sin(x) + 2 sin(x) cos(x) ]

    = [ cos(x) + 2 cos2(x) ] / [ sin(x) + 2 sin(x) cos(x) ]

    = cos(x) [1 + 2 cos(x)] / [ sin(x)( 1 + 2 cos(x) ) ]

    = cot(x)


  3. Use the identity sin(2x) = 2 sin(x) cos(x) to write sin(4x) = 2 sin(2x) cos(2x) in the right hand side of the given identity.

    sin(4x) / cos(2x)

    = 2 sin(2x) cos(2x) / cos(2x)

    = 2 sin(2x)

    = 2 [ 2 sin(x) cos(x)]

    = 4 sin(x) cos(x) which is equal to the right hand side of the given identity.


  4. Use the identity sin(2x) = 2 sin(x) cos(x) to write sin(x) as sin(2 * x/2) = 2 sin(x / 2) cos(x / 2) in the right hand side of the given equation.

    2 sin(x / 2) cos(x / 2) + sin(x / 2) = 0

    sin(x/2) [ 2 cos(x/2) + 1 ] = 0 factor

    which gives

    sin(x/2) = 0 or 2 cos(x/2) + 1 = 0

    sin(x / 2) = 0 leads to x / 2 = 0 or x / 2 = Pi which leads ro x = 0 or x = 2pi

    2 cos(x/2) + 1 = 0 leads to cos(x/2) = -1/2 which leads to x/2 = 2pi/3 and x/2 = 4pi/3 (the second solution leads to x greater than 2pi)

    solutions: x = 0, x = 4pi/3 and x = 2pi


  5. The given equation is already factored

    (2sin(x) - 1)(tan(x) - 1) = 0

    which means

    2sin(x) - 1 = 0 or tan(x) - 1 = 0

    sin(x) = 1/2 or tan(x) = 1 equivalent equations to the above

    solutions: x = pi/6, 5pi/6, x = pi/4 and x = 5pi/4


  6. Note that cos(2x + x) = cos(2x) cos(x) - sin(2x) sin(x) using the formula for cos(A + B). Hence

    cos(2x) cos(x) - sin(2x) sin(x) = 0 is equivalent to

    cos(3x) = 0

    Solve for 3x to obtain: 3x = pi/2, 3x = 3Pi/2, 3x = 5pi/2, 3x = 7pi/2, 3x = 9pi/2 and 11pi/2

    solutions: x = pi/6, pi/2, 5pi/6, 7pi/6, 3pi/2 and 11pi/6


  7. Use the identities sin(2x) = 2 sin(x) cos(x) and cos(2x) = 1 - 2 sin2(x) to rewrite the given equation as follows

    given equation

    ( sin(2x) - cos(x) ) / ( cos(2x) + sin(x) - 1 ) = 0

    ( 2 sin(x) cos(x) - cos(x) ) / ( 1 - 2 sin2(x) + sin(x) - 1) = 0

    cos(x)( 2 sin(x) - 1 ) / [ - sin(x)( 2 sin(x) - 1 ) ] = 0

    Divide numerator and denominator by 2 sin(x) - 1 to simplify; assuming that 2 sin(x) - 1 is not equal to zero.

    - cos(x) / sin(x) = 0

    -cot(x) = 0

    solutions: x = pi/2 and x = 3pi/2

    We now need to verify that both solutions found make neither the denominator nor 2 sin(x) - 1 equal to zero. (do this as an exercise)



  8. Use the identities sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

    sin(105o) = sin(60o + 45o)

    = sin(60o)cos(45o) + cos(60o) sin(45o)

    = (sqrt(3)/2 )(sqrt(2)/2) + (1/2)(sqrt(2)/2)

    = ( sqrt(6) + sqrt(2) ) / 4


  9. If sin(x) = 2/5 then cos(x) = sqrt(1 - (2/5)2) = sqrt(21)/5

    a) Use identity: cos(2x) = 1 - 2 sin2(x) = 17/25

    b) Use identinty: cos(4x) = 1 - 2 sin2(2 x)

    = 1 - 2 [ 2sin(x) cos(x) ]2

    = 457 / 625

    c) sin(2x) = 2 sin(x) cos(x) = 4 sqrt(21)/25

    d) sin(4x) = sin(2(2x)) = 2 cos(2x) sin(2x)

    = 2 (17/25)(4 sqrt(21)/25) = 136 sqrt(21) / 625


  10. Note that triangle DAC is isosceles and therefore if we draw the perpendicular from D to AC, it will cut AC into two halves and bisect angle D. Hence

    (1/2) AC = 10 sin(35o) or AC = 20 sin(35o)

    Note that the two internal angles B and C of triangle ABC add up to 90o and therefore the third angle of triangle ABC is a right angle. We can therefore write

    tan(32o) = AB / AC

    Which gives AB = AC tan(32o)

    = 20 sin(35o)tan(32o) = 7.17 ( rounded to 3 significant digits)


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Updated: 2 April 2013

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