# Trigonometry Problems and Questions with Solutions - Grade 12

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

 Prove the identity tan2(x) - sin2(x) = tan2(x) sin2(x) Prove the identity (1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x) Prove the identity 4 sin(x) cos(x) = sin(4x) / cos(2x) Solve the trigonometric equation given by sin(x) + sin(x/2) = 0 for 0 ≤ x ≤ 2 pi Solve the trigonometric equation given by (2sin(x) - 1)(tan(x) - 1) = 0 for 0 ≤ x ≤ 2 pi Solve the trigonometric equation given by cos(2x) cos(x) - sin(2x) sin(x) = 0 for 0 ≤ x ≤ 2 pi Solve the trigonometric equation given by ( sin(2x) - cos(x) ) / ( cos(2x) + sin(x) - 1 ) = 0 for 0 ≤ x ≤ 2 pi Prove that sin(105o) = ( sqrt(6) + sqrt(2) ) / 4 If sin(x) = 2/5 and x is an acute angle, find the exact values of a) cos(2x) b) cos(4x) c) sin(2x) d) sin(4x) Find the length of side AB in the figure below. Round your answer to 3 significant digits. . Solutions to the Above Problems 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. 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) 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. 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 to 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 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 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 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) 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 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 identity: 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 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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