# Double Angle Questions With Solutions

 

The steps to calculate the exact value of trigonometric functions of a double angle using the double angle formulas are presented. Exercises and their solutions are also included.

## Review of Double Angle Formulas

The double angle formulas are given by
A - $\qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta)$
B - $\qquad \cos(2 \theta) = 1 - 2 \sin^2(\theta) = 2 \cos^2(\theta) - 1$
C - $\qquad \tan(2 \theta) = \dfrac{2 \tan \theta}{1 - \tan^2 \theta}$

## Examples With Solutions

Example 1
Given that $\sin(\theta) = 0.4$ and $\theta$ is in quadrant 2 , find the exact and approximate values of
a) $\quad \sin(2 \theta)$
b) $\quad \cos(2 \theta)$.
Solution to Example 1
a)
According to the above formulas in A, we have
$\qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) \qquad (I)$

$\sin(\theta)$ is given, we therefore need to calculate $\cos(\theta)$.
There are many ways to calculate $\cos(\theta)$.
Let us use the identity $\sin^2(\theta) + \cos^2(\theta) = 1$ which gives
$\qquad \cos^2(\theta) = 1 - \sin^2(\theta)$

Solve for $\cos(\theta)$ to obtain
$\qquad \cos(\theta) = \pm \sqrt {1 - \sin^2(\theta) }$

Angle $\theta$ is in quadrant 2 (given) and therefore $\cos(\theta)$ is negative. Hence
$\qquad \cos(\theta) = - \sqrt {1 - 0.4^2 } = - \sqrt {0.84}$

Substitute $\sin(\theta)$ and $\cos(\theta)$ by their values in $(I)$ to obtain
$\qquad \sin(2 \theta) = 2 \cdot 0.4 (- \sqrt {0.84}) = - 0.16 \sqrt {21} \approx -0.73321$

b)
According to the above formulas in B, we have
$\qquad \cos(2 \theta) = 1 - 2 \sin^2(\theta) \qquad (II)$
Substitute $\sin(\theta)$ by its value in $(II)$ to obtain
$\qquad \cos(2 \theta) = 1 - 2 \cdot 0.4^2 = 0.68$

Example 2
Given that $\quad \tan(\theta) = 5$ and $\theta$ is in quadrant 3 , find the exact and approximate values of $\sin(2 \theta)$ .
Solution to Example 2
According to the formulas given above in A, we have
$\qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) \qquad (I)$

We are given $\tan(\theta)$ and the quadrant of $\theta$.
Let us use the definition of $\tan(\theta)$ in a right triangle
$\qquad \tan(\theta) = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$

Given that $\qquad \tan(\theta) = 5$, we can write
$\qquad \tan(\theta) = \dfrac{5}{1} = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$,
and say that
$\text{Opposite Side} = 5$ and $\text{Adjacent Side} = 1$

Calculate the Hypotenuse of the right triangle using the Pythagorean theorem
$\qquad \text{Hypotenuse} = \sqrt { \text{Opposite Side}^2 + \text{Adjacent Side}^2 }$

Substitute the $\text{Opposite Side}$ and $\text{Adjacent Side}$ by their values to obtain
$\qquad \text{Hypotenuse} = \sqrt{5^2 + 1^2} = \sqrt {26}$

We now calculate $\sin(\theta)$ and $\cos(\theta)$ taking into account the fact that both $\sin(\theta)$ and $\cos(\theta)$ are negative in quadrant 3.
$\qquad \sin(\theta) = - \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = - \dfrac{5}{\sqrt {26}}$

$\qquad \cos(\theta) = - \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = - \dfrac{1}{\sqrt {26}}$

Substitute $\sin(\theta)$ and $\cos(\theta)$ by their values in $(I)$, to obtain
$\qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta) = 2 (- \dfrac{5}{\sqrt {26}}) (- \dfrac{1}{\sqrt {26}})$
Simplify
$\qquad \sin(2 \theta) = \dfrac{10}{26} = \dfrac{5}{13} \approx 0.38461$

Example 3
Given that $\cos(\theta) = 0.8$ and $\theta$ is in quadrant 4 , find the exact and approximate values of
a) $\quad \sec(2 \theta)$
b) $\quad \csc(2 \theta)$
c) $\quad \cot(2 \theta)$.
Solution to Example 3
a)
$\sec(2 \theta)$ is given by
$\qquad \sec(2 \theta) = \dfrac{1}{\cos(2 \theta)}$

Use the formula of $\cos(2 \theta)$ given in B above, we obtain
$\qquad \sec(2 \theta) = \dfrac{1}{2 \cos^2(\theta) - 1}$

Substitute $\cos(\theta) = 0.8$ to obtain
$\qquad \sec(2 \theta) = \dfrac{1}{2 \cdot 0.8^2 - 1} = \dfrac{1}{0.28} \approx 3.57142$

b)
$\csc(2 \theta)$ is given by
$\qquad \csc(2 \theta) = \dfrac{1}{\sin(2 \theta)}$

Use the formula of $\sin(2 \theta)$ given in A above, we obtain
$\qquad \csc(2 \theta) = \dfrac{1}{2 \cos(\theta) \sin(\theta) } \qquad (I)$

We need to find $\sin(\theta)$ given $\cos(\theta) = 0.8$. The identity $\sin^2(\theta) + \cos^2(\theta) = 1$ gives
$\qquad \sin^2(\theta) = 1 - \cos^2(\theta)$

Solve for $\sin(\theta)$ to obtain
$\qquad \sin(\theta) = \pm \sqrt {1 - \cos^2(\theta) }$

Angle $\theta$ is in quadrant 4 (given) and therefore $\sin(\theta)$ is negative. Hence
$\qquad \sin(\theta) = - \sqrt {1 - 0.8^2} = - 0.6$

Substitute $\cos(\theta)$ and $\sin(\theta)$ by their values in $(I)$ above, we obtain
$\qquad \csc(2 \theta) = \dfrac{1}{2 \cdot 0.8 \cdot (-0.6)} = - \dfrac{1}{0.96} \approx -1.04166$

c)
We use the identity
$\qquad \cot(2 \theta) = \dfrac{\cos(2 \theta)}{\sin(2 \theta)} = \dfrac{\csc(2 \theta)}{\sec(2 \theta)}$

Substitute $\csc(2 \theta)$ and $\sec(2 \theta)$ by their exact values calculated in parts a) and b)
$\qquad \cot(2 \theta) = \dfrac{- \dfrac{1}{0.96}}{\dfrac{1}{0.28} } = -\dfrac{7}{24} \approx -0.29166$

## Exercises With Solutions

1. Given $\cos \theta = - 0.2$ and $\theta$ is in quadrant 2, find the exact and approximate values of $\tan (2 \theta)$.
2. Given $\cot \theta = 4$ and $\theta$ is in quadrant 3, find the exact and approximate values of $\sin (2 \theta)$.
3. Given $\sec \theta = -4$ and $\theta$ is in quadrant 3, find the exact and approximate values of $\sec (2 \theta)$.

## Solutions to the Above Exercises

1. Formula given in C above is written as
$\qquad \tan(2 \theta) = \dfrac{2 \tan \theta}{1 - \tan^2 \theta} \quad (I)$
$\sin \theta = \sqrt{1-(-0.2)^2} = \sqrt{ 0.96}$ .
$\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{\sqrt{0.96}}{-0.2} = - \sqrt{24}$
Substitute $\tan \theta = - \sqrt{24}$ in $(I)$
$\qquad \tan(2 \theta) = \dfrac{2 (- \sqrt{24})}{-23} = \dfrac{4\sqrt{6}}{23} \approx 0.42599$

2. Let us use the definition of $\cot(\theta)$ in a right triangle
$\qquad \cot(\theta) = \dfrac{\text{Adjacent Side}}{\text{ Opposite Side}}$

Given that $\qquad \cot(\theta) = 4$, we can write
$\qquad \cot(\theta) = \dfrac{4}{1} = \dfrac{\text{Adjacent Side}}{\text{Opposite Side}}$,
and write that
$\text{Adjacent Side} = 4$ and $\text{Opposite Side} = 1$

Calculate the Hypotenuse of the right triangle using the Pythagorean theorem
$\qquad \text{Hypotenuse} = \sqrt { \text{Adjacent Side }^2 + \text{Opposite Side}^2 } = \sqrt{4^2+1^2} = \sqrt{17}$

In quadrant $3$, $\sin (\theta)$ and $\cos (\theta)$ are negative and using the definition of $\sin (\theta)$ and $\cos(\theta)$, we have

$\sin (\theta) = - \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = - \dfrac{1}{\sqrt{17}}$

$\sin (\theta) = - \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = - \dfrac{4}{\sqrt{17}}$
Substitute in the formula $\qquad \sin(2 \theta) = 2 \sin(\theta) \cos(\theta)$ in A above, to obtain

$\qquad \sin(2 \theta) = 2 (- \dfrac{1}{\sqrt{17}}) (- \dfrac{4}{\sqrt{17}}) = \dfrac{8}{17} \approx 0.47058$

3. Given $\sec \theta = -4$ and $\sec \theta = \dfrac{1}{\cos \theta}$
we have
$\dfrac{1}{\cos \theta} = - 4$

Solve for $\cos \theta$
$\cos \theta = - \dfrac{1}{4}$

We are asked to calculate $\sec (2 \theta)$ which is given by
$\sec (2 \theta) = \dfrac{1}{\cos (2 \theta)}$

Use formula B, given above, to write
$\sec (2 \theta) = \dfrac{1}{ 2 \cos^2(\theta) - 1 }$

Substitute $\cos \theta$ by its value found above,
$\sec (2 \theta) = \dfrac{1}{ 2 \left(- \dfrac{1}{4} \right)^2 - 1 } = - \dfrac{8}{7} \approx -1.14285$