# Trigonometry Problems and Questions with Solutions - Grade 12

 

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

## Solve the following questions

1. Prove the identity
$$\tan^2(x) - \sin^2(x) = \tan^2(x) \sin^2(x)$$

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

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

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

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

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

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

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

9. 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) $$\sin(4x)$$

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

## Solutions to the Above Problems

1. 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.

2. 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.

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

4. 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: $$\{ 0 , 4 \pi/3 , 2 \pi \}$$

5. The given equation is already factored
$$(2\sin(x) - 1)(\tan(x) - 1) = 0$$
$$2\sin(x) - 1 = 0$$   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: $$\{\pi/6, 5 \pi/6 , \pi /4 , 5 \pi/4 \}$$

6. 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: $$\{\pi/6, \pi/2, 5\pi/6, 7\pi/6, 3\pi/2 , 11\pi/6 \}$$

7. 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
$$= - \dfrac{\cos(x)}{ \sin(x)}$$
$$= - \cot(x)$$

8. $$105^{\circ}$$ 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
$$= (\sqrt {3} / 2 )(\sqrt {2}/2) + (1/2)(\sqrt {2}/2)$$
$$= \dfrac{ \sqrt {6} + \sqrt {2} } {4}$$

9. 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) $$\sin(4x) = \sin(2(2x)) = 2 \cos(2x) \sin(2x)$$
$$= 2 (17/25)(4 \sqrt{21}/25) = 136 \sqrt{21} / 625$$

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 .
$$(1/2) AC = 10 \sin(35^{\circ})$$
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)