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

Prove the identity
tan^{2}(x)  sin^{2}(x) = tan^{2}(x) sin^{2}(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(105^{o}) = ( 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.
tan^{2}(x)  sin^{2}(x) = sin^{2}(x) / cos^{2}(x)  sin^{2}(x)
= [ sin^{2}(x)  cos^{2}(x) sin^{2}(x) ] / cos^{2}(x)
= sin^{2}(x) [ 1  cos^{2}(x) ] / cos^{2}(x)
= sin^{2}(x) sin^{2}(x) / cos^{2}(x)
= sin^{2}(x) tan^{2}(x) which is equal to the right hand side of the given identity.

Use the identities cos(2x) = 2 cos^{2}(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 cos^{2}(x)  1 ] / [ sin(x) + 2 sin(x) cos(x) ]
= [ cos(x) + 2 cos^{2}(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 sin^{2}(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 sin^{2}(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(105^{o}) = sin(60^{o} + 45^{o})
= sin(60^{o})cos(45^{o}) + cos(60^{o}) sin(45^{o})
= (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 sin^{2}(x) = 17/25
b) Use identity: cos(4x) = 1  2 sin^{2}(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(35^{o}) or AC = 20 sin(35^{o})
Note that the two internal angles B and C of triangle ABC add up to 90^{o} and therefore the third angle of triangle ABC is a right angle.
We can therefore write
tan(32^{o}) = AB / AC
Which gives AB = AC tan(32^{o})
= 20 sin(35^{o})tan(32^{o}) = 7.17 ( rounded to 3 significant digits)


More High School Math (Grades 10, 11 and 12)  Free Questions and Problems With Answers
More Middle School Math (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Author 
email
Home Page