# Half Angle Questions With Solutions

 

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

## Review of the Half Angle Formulas

The double angle formulas are given by
A - $\qquad \sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 - \cos \theta} {2}}$

B - $\qquad \cos \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 + \cos \theta} {2}}$

C 1 - $\qquad \tan \left(\dfrac{\theta}{2} \right) = \pm \sqrt { \dfrac{1 - \cos \theta}{1 + \cos \theta} }$,   C 2 - $\quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\sin \theta}{1 + \cos \theta}$,   C 3 - $\quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{1 - \cos \theta}{\sin \theta}$

NOTES
1) Selecting the + or - in these formulas depends on the quadrant where $\dfrac{\theta}{2}$ is located
2) There are three formulas for $\quad \tan \left(\dfrac{\theta}{2} \right)$

## Examples With Solutions

Example 1
Given that $\cos(\theta) = - 0.9$ and $\theta$ is in quadrant 3 ,
a) Find the quadrant of $\dfrac{\theta }{2}$ and the signs of $\sin \left(\dfrac{\theta }{2} \right)$ and $\cos \left(\dfrac{\theta }{2} \right)$
and find the exact and approximate values of
b) $\quad \sin \left(\dfrac{\theta }{2} \right)$
a) $\quad \cos \left(\dfrac{\theta }{2} \right)$

Solution to Example 1
a)
Given that $\theta$ is in quadrant 3 and may be written in inequality form as follows
$\qquad \pi \lt \theta \lt \dfrac{3 \pi}{2}$
Divide all terms in the above inequality by $2$ to obtain
$\qquad \dfrac{\pi}{2} \lt \dfrac{ \theta}{2} \lt \dfrac{3 \pi}{4}$
The above inequality tells us that $\dfrac{ \theta}{2}$ is in quadrant 2 and therefore
$\sin \left(\dfrac{\theta }{2} \right) \gt 0$ and $\cos \left(\dfrac{\theta }{2} \right) \lt 0$

b)
According to the above formulas in A, we have
$\qquad \sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 - \cos \theta} {2}} \qquad (I)$
$\cos(\theta)$ is given.

We need to decide whether to use the $+$ or the $-$ in front of the formula in $(I)$. In part a) we found the sign of $\sin \left(\dfrac{\theta}{2} \right)$ to be positive, hence
$\qquad \sin \left(\dfrac{\theta}{2} \right) = \color{red}{ + } \sqrt {\dfrac{1 - \cos \theta} {2}}$

Substitute $\cos(\theta) = - 0.9$ (given) in the above to obtain
$\qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 + 0.9} {2}}$

Simplify to obtain
$\qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{19}{20}} \approx 0.97467$

c)

According to the above formulas in B, we have
$\qquad \cos \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 + \cos \theta} {2}} \qquad (II)$

In part a) bove we determined that $\cos \left(\dfrac{\theta}{2} \right)$ is negative , hence
$\qquad \cos \left(\dfrac{\theta}{2} \right) = \color{red}{ - } \sqrt {\dfrac{1 + \cos \theta} {2}}$

Substitute $\cos(\theta)$ by its value in the above to obtain
$\qquad \cos \left(\dfrac{\theta}{2} \right) = - \sqrt {\dfrac{1 - 0.9} {2}} = - \dfrac{\sqrt{5}}{10} \approx - 0.22360$

Example 2
Given that $\quad \tan(\theta) = - 4$ and $\theta$ is in quadrant 2 , find the exact and approximate values of $\tan \left(\dfrac{\theta }{2} \right)$ .
Solution to Example 2
According to the formula C2 given above in A, we have
$\qquad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\sin \theta}{1 + \cos \theta} \qquad (I)$

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

Given that $\qquad \tan(\theta) = - 4$, we can write
We use right triangle and use $|\tan(\theta)| = 4$
$\qquad |\tan(\theta)| = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} = 4 = \dfrac{4}{1}$,
and say that
$\text{Opposite Side} = 4$ 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
$\qquad \text{Hypotenuse} = \sqrt{4^2 + 1^2} = \sqrt {17}$

We now calculate $\sin(\theta)$ and $\cos(\theta)$ taking into account the fact that $\sin(\theta)$ is positive and $\cos(\theta)$ is negative in quadrant 2.
$\qquad \sin(\theta) = \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = \dfrac{4}{\sqrt {17}}$

$\qquad \cos(\theta) = - \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = - \dfrac{1}{\sqrt {17}}$

Substitute $\sin(\theta)$ and $\cos(\theta)$ by their values in $(I)$, to obtain
$\qquad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\dfrac{4}{\sqrt {17}}}{1 - \dfrac{1}{\sqrt {17}}}$
Simplify
$\qquad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{4}{\sqrt {17} - 1} \approx 1.28077$

Example 3
Given that $\sin(\theta) = - 0.2$ and $\theta$ is in quadrant 4 , find the exact and approximate values of
a) Find the quadrant of angle $\dfrac{\theta}{2}$ and hence the signs of $\sec \left(\dfrac{\theta}{2} \right)$, $\csc \left(\dfrac{\theta}{2} \right)$ and $\cot \left(\dfrac{\theta}{2} \right)$
b) $\quad \sec \left(\dfrac{\theta}{2} \right)$
c) $\quad \csc \left(\dfrac{\theta}{2} \right)$
d) $\quad \cot \left(\dfrac{\theta}{2} \right)$.
Solution to Example 3
a)
$\theta$ in quadrant 4 could be written as an inequality
$\dfrac{3 \pi}{2} \lt \theta \lt 2\pi$
Divide all terms of the above inequality by 2 to obtain
$\dfrac{3 \pi}{4} \lt \dfrac{\theta}{2} \lt \pi$
The above inequality indicates that $\dfrac{\theta}{2}$ is in quadrant 2 and therefore the signs of $\sec \left(\dfrac{\theta}{2} \right)$, $\csc \left(\dfrac{\theta}{2} \right)$ and $\cot \left(\dfrac{\theta}{2} \right)$ are as follows:
$\sec \left(\dfrac{\theta}{2} \right) \lt 0$,

$\csc \left(\dfrac{\theta}{2} \right) \gt 0$

$\cot \left(\dfrac{\theta}{2} \right) \lt 0$

b)
$\sec \left(\dfrac{\theta}{2} \right)$ is given by
$\qquad \sec \left(\dfrac{\theta}{2} \right) = \dfrac{1}{\cos \left(\dfrac{\theta}{2} \right)} \quad (I)$

Use the formula of $\cos \left(\dfrac{\theta}{2} \right)$ given in B above,

$\qquad \cos \left( \dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 + \cos \theta} {2}} \quad (II)$

We now need to find $\cos \theta$ knowing $\sin(\theta) = - 0.2$. Use the identity $\cos \theta = \pm \sqrt{1 - \sin^2 \theta}$. $\theta$ is in quadrant 4 and therefore $\cos \theta$ is positive, hence
$\qquad \cos \theta = \sqrt {1 - (-0.2)^2 } = \sqrt {0.96}$
In part a) above, we found that $\cos \left( \dfrac{\theta}{2} \right)$ is negative, hence from $(II)$, we have
$\qquad \cos \left( \dfrac{\theta}{2} \right) = - \sqrt {\dfrac{1 + \cos \theta} {2}}$

Substitute $\cos \theta = \sqrt {0.96}$ in the above to obtain
$\qquad \cos \left( \dfrac{\theta}{2} \right) = - \sqrt {\dfrac{1 + \sqrt {0.96} } {2}}$

Substitute the above in $(I)$ to obtain
$\qquad \sec \left(\dfrac{\theta}{2} \right) = \dfrac{1}{ -\sqrt {\dfrac{1 + \sqrt {0.96} } {2}} } = - \sqrt {\dfrac{2} {1 + \sqrt {0.96} }} \approx - 1.00508$

c)
$\csc \left(\dfrac{\theta}{2} \right)$ is given by
$\qquad \csc \left(\dfrac{\theta}{2} \right) = \dfrac{1}{\sin \left(\dfrac{\theta}{2} \right)} \quad (III)$

Use the formula of $\sin \left(\dfrac{\theta}{2} \right)$ given in A above and the signs of $\sin \left( \dfrac{\theta}{2} \right)$ was determined in part a), hence

$\qquad \sin \left( \dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 - \sqrt {0.96}} {2}} \quad (IV)$
Substitute the above in $(III)$ to obtain
$\qquad \csc \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{2}{1 - \sqrt {0.96}}} \approx 9.94936$

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

Substitute $\csc \left(\dfrac{\theta}{2} \right)$ and $\sec \left(\dfrac{\theta}{2} \right)$ by their exact values calculated above in parts a) and b)

$\qquad \cot \left(\dfrac{\theta}{2} \right) = \dfrac{ \sqrt {\dfrac{2}{1 - \sqrt {0.96}}} }{ - \sqrt {\dfrac{2} {1 + \sqrt {0.96} }} } = - \sqrt{\dfrac{1+\sqrt{0.96}}{1-\sqrt{0.96}}} \approx - 9.89897$

## Exercises With Solutions

1. Given $\tan \theta = 4$ and $\theta$ is in quadrant 3, find the exact and approximate values of $\sin \left(\dfrac{\theta}{2} \right)$.
2. Given $\csc \theta = - 5$ and $\theta$ is in quadrant 4, find the exact and approximate values of $\tan \left(\dfrac{\theta}{2} \right)$.

## Solutions to the Above Exercises

1. Use the formula given in A above to write
$\qquad \sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt {\dfrac{1 - \cos \theta} {2}}$

We need to find the quadrant of $\dfrac{\theta}{2}$ to select the + or - in the formula and we also need to determine $\cos \theta$.
$\theta$ is in quadrant 3 (given) may be written using an inequality as
$\qquad \pi \lt \theta \lt \dfrac{3 \pi}{2}$

Divide all terms in the above inequality by $2$ to obtain
$\qquad \dfrac{\pi}{2} \lt \dfrac{ \theta}{2} \lt \dfrac{3 \pi}{4}$

The above means that $\dfrac{ \theta}{2}$ is in quadrant 2 and therefore $\sin \left(\dfrac{\theta}{2} \right)$ is positive and given by
$\qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 - \cos \theta} {2}} \quad (I)$

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

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

We now determine $\cos \theta$
$\qquad |\cos \theta| = \dfrac{\text{ Adjacent Side }}{\text{ Hypotenuse }} = \dfrac{1}{ \sqrt{17} }$

Since $\theta$ is in quadrant 3, $\cos \theta$ is negatine and therefore
$\cos \theta = - \dfrac{1}{ \sqrt{17}}$

Substitute the value of $\cos \theta$ in $(I)$ to obtain
$\qquad \sin \left(\dfrac{\theta}{2} \right) = \sqrt {\dfrac{1 + \dfrac{1}{ \sqrt{17}}} {2}} \approx 0.78820$

2. Given $\csc \theta = - 5$, we write
$\csc \theta = \dfrac{1}{\sin \theta} = - 5$, hence
$\sin \theta = - \dfrac{1}{5}$
Use the identity $\cos^2 \theta = 1 - \sin^2 \theta$ to claculate $\cos \theta$
$\cos \theta = \pm \sqrt {1 - \sin^2 \theta}$
Since $\theta$ is in quadrant 4 (given) , $\cos \theta$ is positive and therefore
$\cos \theta = \sqrt {1 - \left( - \dfrac{1}{5} \right)^2 } = \dfrac{2\sqrt{6}}{5}$
We now use formula C2 given above to write
$\quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{\sin \theta}{1 + \cos \theta}$
Substitute $\sin \theta$ and $\cos \theta$ by their values to obtain
$\quad \tan \left(\dfrac{\theta}{2} \right) = \dfrac{- \dfrac{1}{5}}{1 + \dfrac{2\sqrt{6}}{5}}$
Which may be simplified to
$\quad \tan \left(\dfrac{\theta}{2} \right) = -\dfrac{1}{5+2\sqrt{6}} \approx -0.10102$