We think you are located in South Africa. Is this correct?

# Double Angle Identities

## 4.3 Double angle identities (EMCGD)

### Derivation of $$\sin2\alpha$$ (EMCGF)

We have shown that $$\sin\left(\alpha +\beta \right)=\sin\alpha\cos \beta +\cos\alpha\sin \beta$$. If we let $$\alpha =\beta$$, then we can write the formula as:

\begin{align*} \sin \left(2\alpha \right) &= \sin\left(\alpha +\alpha \right) \\ & = \sin \alpha\cos \alpha +\cos\alpha\sin \alpha \\ \therefore \sin 2\alpha & = 2\sin\alpha\cos \alpha \end{align*}

### Derivation of $$\cos2\alpha$$ (EMCGG)

Similarly, we know that $$\cos\left(\alpha +\beta \right)=\cos\alpha\cos \beta -\sin\alpha\sin \beta$$. If we let $$\alpha =\beta$$, then we have:

\begin{align*} \cos \left(2\alpha \right) &=\cos\left(\alpha +\alpha \right) \\ & = \cos \alpha\cos \alpha -\sin\alpha\sin \alpha \\ \therefore \cos 2\alpha & = \cos^{2}\alpha -\sin^{2}\alpha \end{align*}

Using the square identity, $$\sin^{2} \alpha + \cos^{2}\alpha = 1$$, we can also derive the following formulae:

\begin{align*} \cos 2\alpha & = \cos^{2}\alpha -\sin^{2}\alpha \\ & = \left( 1 - \sin^{2}\alpha \right) - \sin^{2}\alpha \\ \therefore \cos 2\alpha & = 1 - 2 \sin^{2}\alpha \end{align*}

And

\begin{align*} \cos 2\alpha & = \cos^{2}\alpha -\sin^{2}\alpha \\ & = \cos^{2}\alpha - \left( 1 - \cos^{2}\alpha \right) \\ & = \cos^{2}\alpha - 1 + \cos^{2}\alpha \\ \therefore \cos 2\alpha & = 2 \cos^{2}\alpha - 1 \end{align*}

Double angle formulae

• $$\sin 2\alpha = 2\sin\alpha\cos \alpha$$
• $$\cos 2\alpha = \cos^{2}\alpha -\sin^{2}\alpha$$
• $$\cos 2\alpha = 1 - 2 \sin^{2}\alpha$$
• $$\cos 2\alpha = 2 \cos^{2}\alpha - 1$$

## Worked example 7: Double angle identities

If $$\alpha$$ is an acute angle and $$\sin \alpha = \text{0,6}$$, determine the value of $$\sin 2 \alpha$$ without using a calculator.

### Draw a sketch

We convert $$\text{0,6}$$ to a fraction so that we can use the ratio to represent the sides of a triangle.

\begin{align*} \sin \alpha &= \text{0,6} \\ &= \frac{6}{10} \end{align*}

### Use the double angle formula to determine the value of $$\sin 2 \alpha$$

\begin{align*} \sin 2 \alpha &= 2 \sin \alpha \cos \alpha \\ &= 2 \left( \frac{6}{10} \right) \left( \frac{8}{10} \right) \\ &= \frac{96}{100} \\ &= \text{0,96} \end{align*}

$\sin 2 \alpha = \text{0,96}$

Check the answer using a calculator:

\begin{align*} \sin \alpha &= \text{0,6} \\ \therefore \alpha &\approx \text{36,87} ° \\ 2 \alpha &\approx \text{73,74} ° \\ \therefore \sin \left( \text{73,74} ° \right) &\approx \text{0,96} \end{align*}

## Worked example 8: Double angle identities

Prove that $$\frac{\sin\theta +\sin2\theta }{1 + \cos\theta +\cos2\theta } = \tan\theta$$.

For which values of $$\theta$$ is the identity not valid?

### Consider the given expressions

The right-hand side (RHS) of the identity cannot be simplified, so we simplify the left-hand side (LHS). We also notice that the trigonometric function on the RHS does not have a $$2\theta$$ dependence, therefore we will need to use the double angle formulae to simplify $$\sin2\theta$$ and $$\cos2\theta$$ on the LHS.

### Prove the left-hand side equals the right-hand side

\begin{align*} \text{LHS}& = \frac{\sin\theta + 2\sin\theta \cos\theta }{1+\cos\theta +\left(2{\cos}^{2}\theta -1\right)} \\ & = \frac{\sin\theta \left(1+2\cos\theta \right)}{\cos\theta \left(1+2\cos\theta \right)} \qquad \text{(factorise)} \\ & = \frac{\sin\theta }{\cos\theta } \\ & = \tan \theta \\ & = \text{RHS} \end{align*}

### Identify restricted values of $$\theta$$

We know that $$\tan \theta$$ is undefined for $$\theta = \text{90} ° + k \cdot \text{180} °, k \in \mathbb{Z}$$.

Note that division by zero on the LHS is not allowed, so the identity will also be undefined for:

\begin{align*} 1 + \cos\theta +\cos2\theta & = 0 \\ \cos\theta \left( 1 + 2\cos\theta \right) & = 0 \\ \therefore \cos \theta = 0 &\text{ or } 1 + 2\cos\theta = 0 \\ & \\ \text{For } \cos \theta = 0, \quad \theta &= \text{90} ° + k \cdot \text{180} ° \\ & \\ \text{For } 1 + 2 \cos \theta = 0, \quad \cos \theta &= - \frac{1}{2} \\ \therefore \theta &= \text{120} ° + k \cdot \text{360} ° \\ \text{or } \theta &= \text{240} ° + k \cdot \text{360} ° \end{align*}

for $$k \in \mathbb{Z}$$.

## Double angle identities

Exercise 4.3

Given $$5 \cos \theta = -3$$ and $$\theta < \text{180} °$$. Determine the value of the following, without a calculator:

$$\cos 2 \theta$$

Draw a sketch:

$$\sin \left( \text{180} ° - 2 \theta \right)$$
\begin{align*} \sin \left( \text{180} ° - 2 \theta \right) &= \sin 2 \theta \\ &= 2 \sin \theta \cos \theta \\ &= 2 \left( \frac{4}{5} \right) \left( - \frac{3}{5} \right) \\ &= - \frac{24}{25} \end{align*}
$$\tan 2 \theta$$
\begin{align*} \tan 2 \theta &= \frac{\sin 2 \theta}{\cos 2 \theta} \\ &= - \frac{24}{25} \times \left( - \frac{25}{7} \right) \\ &= \frac{24}{7} \end{align*}

Given $$\cos \text{40} ° = t$$, determine (without a calculator):

$$\cos \text{140} °$$
\begin{align*} \cos \text{140} ° &= \cos \left( \text{180} ° - \text{40} ° \right) \\ &= - \cos \text{40} ° \\ &= - t \end{align*}
$$\sin \text{40} °$$
\begin{align*} \sin \text{40} ° &= \sqrt{1 - \cos^{2} \text{40} ° } \\ &= \sqrt{1 - t^{2}} \end{align*}
$$\sin \text{50} °$$
\begin{align*} \sin \text{50} ° &= \sin \left( \text{90} ° - \text{50} ° \right) \\ &= \cos \text{40} ° \\ &= t \end{align*}
$$\cos \text{80} °$$
\begin{align*} \cos \text{80} ° &= \cos 2 \left( \text{40} ° \right) \\ &= 2 \cos^{2} \text{40} ° - 1 \\ &= 2t^{2} - 1 \end{align*}
$$\cos \text{860} °$$
\begin{align*} \cos \text{860} ° &= \cos \left[2( \text{360} °) + \text{140} ° \right] \\ &= \cos \text{140} ° \\ &= \cos ( \text{180} ° - \text{40} °) \\ &= - \cos \text{40} ° \\ &= -t \end{align*}
$$\cos ( -\text{1 160} °)$$
\begin{align*} \cos (- \text{1 160} °) &= \cos \text{1 160} ° \\ &= \cos \left( 3( \text{360} °) + \text{80} ° \right) \\ &= \cos ( \text{80} °) \\ &= \cos 2( \text{40} °) \\ &= 2 \cos^{2} \text{40} ° - 1 \\ &= 2 t^{2} - 1 \end{align*}
Prove the identity: $$\frac{1}{\sin 2 A} - \frac{1}{\tan 2A} = \tan A$$
\begin{align*} \text{LHS} &= \frac{1}{\sin 2 A} - \frac{1}{\tan 2A} \\ &= \frac{1}{\sin 2 A} - \frac{\cos 2A}{\sin 2A} \\ &= \frac{1 - \left( 1 - 2 \sin^{2}A \right)}{\sin 2 A} \\ &= \frac{2 \sin^{2}A}{2 \sin A \cos A} \\ &= \frac{\sin A}{\cos A} \\ &= \tan A \\ &= \text{RHS} \end{align*}

Restrictions:

\begin{align*} \sin 2 A &\ne 0 \\ \therefore 2A &\ne \text{0} ° + k \cdot \text{180} ° \\ \therefore A &\ne \text{0} ° + k \cdot \text{90} °, \quad k \in \mathbb{Z} \\ \text{And } \tan 2 A &\ne 0 \\ \therefore 2A &\ne \text{90} ° + k \cdot \text{180} ° \\ \therefore A &\ne \text{45} ° + k \cdot \text{90} °, \quad k \in \mathbb{Z} \\ \text{And for } \tan A &: \\ A &\ne \text{90} ° + k \cdot \text{180} °, \quad k \in \mathbb{Z} \end{align*}
Hence, solve the equation $$\frac{1}{\sin 2 A} - \frac{1}{\tan 2A} = \text{0,75}$$ for $$\text{0} ° < A < \text{360} °$$.
\begin{align*} \frac{1}{\sin 2 A} - \frac{1}{\tan 2A} &= \text{0,75} \\ \therefore \tan A &= \text{0,75} \\ \therefore A &= \text{36,87} ° \\ \text{ or } A &= \text{180} ° + \text{36,87} °\\ &= \text{216,87} ° \end{align*}

Without using a calculator, find the value of the following:

$$\sin \text{22,5} °$$
\begin{align*} 2 \times \text{22,5} ° &= \text{45} ° \\ \cos 2 \theta &= 1 - 2 \sin^{2} \theta \\ \cos \text{45} ° &= 1 - 2 \sin^{2} \left( \text{22,5} ° \right) \\ \frac{1}{\sqrt{2}} &= 1 - 2 \sin^{2} \left( \text{22,5} ° \right) \\ \frac{1}{\sqrt{2}} - 1 &= - 2 \sin^{2} \left( \text{22,5} ° \right) \\ \frac{1 - \sqrt{2}}{\sqrt{2}} &= - 2 \sin^{2} \left( \text{22,5} ° \right) \\ \frac{1 - \sqrt{2}}{-2\sqrt{2}} &= \sin^{2} \left( \text{22,5} ° \right) \\ \frac{ \sqrt{2} - 1}{2\sqrt{2}} &= \sin^{2} \left( \text{22,5} ° \right) \\ \therefore \sqrt{ \frac{ \sqrt{2} - 1}{2\sqrt{2}} } &= \sin \text{22,5} ° \\ \sqrt{ \frac{ \sqrt{2} - 1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}} &= \sin \text{22,5} ° \\ \sqrt{ \frac{ 2 - \sqrt{2} }{4} } &= \sin \text{22,5} ° \end{align*}

Check the answer using a calculator.

$$\cos \text{67,5} °$$
\begin{align*} \cos \text{67,5} ° &= \cos \left( \text{90} ° - \text{22,5} ° \right) \\ &= \sin \left( \text{22,5} ° \right) \\ &= \sqrt{ \frac{ 2 - \sqrt{2} }{4} } \end{align*}

Check the answer using a calculator.

Prove the identity: $$\tan 2 x + \frac{1}{\cos 2 x} = \frac{\sin x + \cos x }{\cos x - \sin x }$$
\begin{align*} \text{LHS} &= \tan 2x + \frac{1}{\cos 2x} \\ &= \frac{\sin 2x}{\cos 2x} + \frac{1}{\cos 2x} \\ &= \frac{\sin 2x + 1}{\cos 2x} \\ &= \frac{2 \sin x \cos x + \cos^{2}x + \sin^{2}x}{\cos^{2}x - \sin^{2}x} \\ &= \frac{ \left( \cos x + \sin x \right)^{2}}{\left( \cos x - \sin x \right)\left( \cos x + \sin x \right) } \\ &= \frac{ \cos x + \sin x }{ \cos x - \sin x } \\ &= \text{RHS} \end{align*}

Restrictions:

\begin{align*} \cos 2 x &\ne 0 \\ \therefore 2x &\ne \text{90} ° + k \cdot \text{180} ° \\ \therefore x &\ne \text{45} ° + k \cdot \text{90} °, \quad k \in \mathbb{Z} \\ \text{And } \cos x &\ne \sin x \\ \therefore x &\ne \text{45} ° + k \cdot \text{180} °, \quad k \in \mathbb{Z} \\ \text{And for } & \tan 2 x \\ \therefore 2x &\ne \text{90} ° + k \cdot \text{180} ° \\ \therefore x &\ne \text{45} ° + k \cdot \text{90} °, \quad k \in \mathbb{Z} \end{align*}
Explain why the identity is undefined for $$x = \text{45} °$$

Consider the denominator on the LHS:

\begin{align*} \cos 2x &= \cos 2 \left( \text{45} ° \right) \\ &= \cos \text{90} ° \\ &= 0 \end{align*}

Consider the denominator on the RHS:

\begin{align*} \cos \text{45} ° &= \sin \text{45} ° \\ \therefore \cos \text{45} ° - \sin \text{45} ° &= 0 \end{align*}

Therefore, the identity will be undefined because division by zero is not permitted.