Trigonometry Problems and Questions with Solutions  Grade 12
\( \) \( \)\( \)\( \)
Grade 12 trigonometry problems and questions with answers and solutions are presented.
Solve the following questions

Prove the identity
\( \tan^2(x)  \sin^2(x) = \tan^2(x) \sin^2(x) \)

Prove the identity
\( \dfrac{1 + \cos(x) + \cos(2x)}{\sin(x) + \sin(2x)} = \cot(x) \)

Prove the identity
\( 4 \sin(x) \cos(x) = \dfrac{\sin(4x)}{\cos(2x)} \)

Solve the trigonometric equation given by
\( \sin(x) + \sin(x/2) = 0 \quad \text{for} \quad 0 \le x \le 2 \pi \)

Solve the trigonometric equation given by
\( (2\sin(x)  1)(\tan(x)  1) = 0 \quad \text{for} \quad 0 \le x \le 2 \pi \)

Solve the trigonometric equation given by
\( \cos(2x) \cos(x)  \sin(2x) \sin(x) = 0 \quad \text{for} \quad 0 \le x \le 2 \pi \)

Simplify the trigonometric expression given by
\( \dfrac{\sin(2x)  \cos(x)}{\cos(2x) + \sin(x)  1 } \)

Prove that
\( \sin(105^{\circ}) = \dfrac{\sqrt 6 + \sqrt 2}{4} \)

If \( \sin(x) = \dfrac{2}{5}\) 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

We start with the left hand side of the given identity
Use the identity \( \tan(x) = \dfrac{\sin(x)}{\cos(x)} \) in the left hand side of the given identity.
\( \tan^2(x)  \sin^2(x) = \left(\dfrac{\sin(x)}{\cos(x)}\right)^2  \sin^2(x) \)
\( = \dfrac{\sin^2(x)}{\cos^2(x)}  \sin^2(x) \)
\( = \dfrac{\sin^2(x)  \cos^2(x) \sin^2(x)}{\cos^2(x)} \)
\( = \dfrac{\sin^2(x) \left(1  \cos^2(x)\right)}{\cos^2(x)} \)
\( = \dfrac{ \sin^2(x) \sin^2(x)}{\cos^2(x)} \)
\( = \sin^2(x) \tan^2(x) \) 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.
\( \dfrac{1 + \cos(x) + \cos(2x)}{\sin(x) + \sin(2x)} \)
\( = \dfrac{1 + \cos(x) + 2 \cos^2(x)  1}{\sin(x) + 2 \sin(x) \cos(x) } \)
\( = \dfrac{\cos(x) + 2 \cos^2(x)}{\sin(x) + 2 \sin(x) \cos(x) } \)
\( = \dfrac{\cos(x) (1 + 2 \cos(x))}{\sin(x) (1 + 2 \cos(x)) } \)
\( = \dfrac{\cos(x)}{\sin(x)} \)
\( = \cot(x) \) which is equal to the right hand side of the given identity.

Use the identity \( \sin(2x) = 2 \sin(x) \cos(x) \) to write \( \sin(4x) = 2 \sin(2x) \cos(2x) \) in the right hand side of the given identity.
\( \dfrac{\sin(4x)}{\cos(2x)} = \dfrac{2 \sin(2x) \cos(2x)}{\cos(2x)}\)
\( = 2 \sin(2x) \)
\( = 2 \times 2 \sin(x) \cos(x) \)
\( = 4 \sin(x) \cos(x) \) which is equal to the left hand side of the given identity.

Use the identity \( \sin(2x) = 2 \sin(x) \cos(x) \) to write \( \sin(x) \) as
\( \sin(x) = \sin(2 \times x/2) = 2 \sin(x / 2) \cos(x / 2) \)
and use in the right hand side of the given equation to write it as follows
\( 2 \sin(x / 2) \cos(x / 2) + \sin(x / 2) = 0 \)
Factor \( \sin(x / 2) \)
\( \sin(x/2) ( 2 \cos(x/2) + 1 ) = 0 \)
which gives two equations to solve
\( \sin(x/2) = 0 \) or \( 2 \cos(x/2) + 1 = 0 \)
a) The equation \( \sin(x / 2) = 0 \) has the solutions \( x / 2 = 0 \) or \( x / 2 = \pi \)
Solve for x to obtain the solutions: \( x = 0 \) or \( x = 2 \pi \)
b) The equation \( 2 \cos(x/2) + 1 = 0 \) leads to \( \cos(x/2) = 1/2 \) which has the solutions \( x/2 = 2 \pi/3 \) and \( x/2 = 4 \pi/3 \)
Solve for x to obtain the solutions: \( x = 4 \pi/3 \) and \( x = 8 \pi/3 \)
Note that \( 8 \pi/3 \) is greather than \( 2 \pi \) and is therefore not accepted.
Final solutions for the given equation are: \( \{ 0 , 4 \pi/3 , 2 \pi \} \)

The given equation is already factored
\( (2\sin(x)  1)(\tan(x)  1) = 0 \)
which leads to two equations
\( 2\sin(x)  1 = 0 \) or \( \tan(x)  1 = 0 \)
The above equations may be written as
\( \sin(x) = 1/2 \) or \( \tan(x) = 1 \)
The solutions of \( \sin(x) = 1/2 \) are solutions: \( x = \pi/6 \) and\( x = 5 \pi/6 \)
The solutions of \( \tan(x) = 1 \) are: \( x = \pi /4 \) and \( x = 5 \pi/4 \)
The solutions of the given equation within the given interval are: \( \{\pi/6, 5 \pi/6 , \pi /4 , 5 \pi/4 \}\)

Use the formula for \( \cos(A + B) \) to write
\( \cos(2x + x) = \cos(2x) \cos(x)  \sin(2x) \sin(x) \) .
Hence the given equation
\( \cos(2x) \cos(x)  \sin(2x) \sin(x) = 0 \)
may be written as
\( \cos(3x) = 0 \)
Solve the above equation for \( 3x \) to obtain:
\( 3x = \pi/2 \), \( 3x = 3\pi/2 \), \( 3x = 5\pi/2 \), \( 3x = 7\pi/2 \), \( 3x = 9\pi/2 \) and \( 3x = 11\pi/2 \)
Solve the above for x to obtain the solutions: \( \{\pi/6, \pi/2, 5\pi/6, 7\pi/6, 3\pi/2 , 11\pi/6 \} \)

Use the identities \( \sin(2x) = 2 \sin(x) \cos(x) \) and \( \cos(2x) = 1  2 \sin^2(x) \) to rewrite the given expression as follows
\( \dfrac{\sin(2x)  \cos(x)}{\cos(2x) + \sin(x)  1 } = \dfrac{ 2 \sin(x) \cos(x)  \cos(x)}{1  2 \sin^2(x) + \sin(x)  1 } \)
Simplify the right hand side and factor numerator and denominator
\( = \dfrac{\cos(x)( 2 \sin(x) 1) }{ \sin(x)(  2 \sin(x) + 1) } \)
Simplify
\( =  \dfrac{\cos(x)}{ \sin(x)} \)
\( =  \cot(x) \)

\( 105^{\circ} \) may be written as the sum of two special angles as follows:
\( 105^{\circ} = 60^{\circ} + 45^{\circ}\)
Hence
\( \sin(105^{\circ}) = \sin(60^{\circ} + 45^{\circ}) \)
Use the identities \( \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) \)
\( \sin(105^{\circ}) = \sin(60^{\circ})\cos(45^{\circ}) + \cos(60^{\circ}) \sin(45^{\circ}) \)
Use table of special angles
\( = (\sqrt {3} / 2 )(\sqrt {2}/2) + (1/2)(\sqrt {2}/2) \)
\( = \dfrac{ \sqrt {6} + \sqrt {2} } {4} \)

If \( \sin(x) = 2/5 \) then \( \cos(x) = \sqrt {1  \sin^2 x} = \sqrt{1  (2/5)^2} = \sqrt{21}/5 \)
a) Use identity: \( \cos(2x) = 1  2 \sin^2(x) = 17/25 \)
b) Use identity: \( \cos(4x) = 1  2 \sin^2(2 x) \)
\( = 1  2 [ 2\sin(x) \cos(x) ]^2 \)
\( = 457 / 625 \)
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
.
\( (1/2) AC = 10 \sin(35^{\circ}) \)
which gives
\( AC = 20 \sin(35^{\circ}) \)
Note that the two internal angles B and C of triangle ABC add up to \( 90^{\circ} \) and therefore the third angle of triangle ABC is a right angle.
We can therefore write
\( \tan(32^{\circ}) = AB / AC \)
Which gives
\( AB = AC \tan(32^{\circ}) \)
\( = 20 \sin(35^{\circ})\tan(32^{\circ}) = 7.17 \quad \) ( rounded to 3 significant digits)
References and Links
High School Maths (Grades 10, 11 and 12)  Free Questions and Problems With Answers
Middle School Maths (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
Home Page