# Double-Angle and Multiple-Angle Identities

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate sin 2α, cos 2α, and tan 2α if sin α = 1/2 and the angle α lies in the 1st quadrant.

### Example 2

Calculate sin 2β, cos 2β, and cot 2β if cos 2β= −5/13 and the angle β lies in the 2nd quadrant.

### Example 3

Find $$\frac{{\cos \alpha }}{{2 - 3\sin \alpha }}$$ if $$\tan \frac{\alpha}{2} = 3.$$

### Example 4

Find $$\frac{{2\sin \beta }}{{4 + 5\cos \beta }}$$ if $$\cot \frac{\beta}{2} = -2.$$

### Example 5

Simplify the expression $\frac{{1 + \sin 2\alpha }}{{{{\left( {\sin \alpha + \cos \alpha } \right)}^2}}}.$

### Example 6

Simplify the expression $\frac{{2\sin \beta - \sin 2\beta }}{{2\sin \beta + \sin 2\beta }}.$

### Example 7

Find $$\cos4\alpha$$ if $$\tan\alpha = 2.$$

### Example 8

Determine $$\sin 4\beta$$ if $$\cot\beta = -3.$$

### Example 9

Verify the identity $\frac{{\sin 3\alpha }}{{\sin \alpha }} - \frac{{\cos 3\alpha }}{{\cos \alpha }} = 2.$

### Example 10

Determine which is greater: $$\tan 2\alpha$$ or $$\tan\alpha$$ if $$0 \lt \alpha \lt \frac{\pi }{4}?$$

### Example 1.

Calculate $$\sin2\alpha,$$ $$\cos2\alpha,$$ and $$\tan2\alpha$$ if $$\sin\alpha = \frac{1}{2}$$ and the angle $$\alpha$$ lies in the $$1\text{st}$$ quadrant.

Solution.

First we determine $$\cos\alpha$$ and $$\tan\alpha.$$ Given that $$\alpha$$ is in the $$1\text{st}$$ quadrant where cosine is positive and using the Pythagorean trigonometric identity, we have

$\cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } = \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} = \sqrt {1 - \frac{1}{4}} = \sqrt {\frac{3}{4}} = \frac{{\sqrt 3 }}{2}.$

The tangent is given by

$\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{1}{{\sqrt 3 }}.$

Now we can find the double angle trigonometric functions:

$\sin 2\alpha = 2\sin \alpha \cos \alpha = 2 \cdot \frac{1}{2} \cdot \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{2};$
$\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2};$
$\tan 2\alpha = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{{2 \cdot \frac{1}{{\sqrt 3 }}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} = \frac{{\frac{2}{{\sqrt 3 }}}}{{1 - \frac{1}{3}}} = \frac{{\frac{2}{{\sqrt 3 }}}}{{\frac{2}{3}}} = \frac{2}{{\sqrt 3 }} \cdot \frac{3}{2} = \frac{3}{{\sqrt 3 }} = \sqrt 3 .$

### Example 2.

Calculate $$\sin2\beta,$$ $$\cos2\beta,$$ and $$\cot2\beta$$ if $$\cos\beta = -\frac{5}{13}$$ and the angle $$\beta$$ lies in the $$2\text{nd}$$ quadrant.

Solution.

The sine is positive in the $$2\text{nd}$$ quadrant. Therefore,

$\sin \beta = \sqrt {1 - {{\cos }^2}\beta } = \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = \sqrt {1 - \frac{{25}}{{169}}} = \sqrt {\frac{{144}}{{169}}} = \frac{{12}}{{13}}.$

Using the double-angle identities, we get

$\sin 2\beta = 2\sin \beta \cos \beta = 2 \cdot \frac{{12}}{{13}} \cdot \left( { - \frac{5}{{13}}} \right) = - \frac{{120}}{{169}};$
$\cos 2\beta = {\cos ^2}\beta - {\sin ^2}\beta = {\left( { - \frac{5}{{13}}} \right)^2} - {\left( {\frac{{12}}{{13}}} \right)^2} = \frac{{25}}{{169}} - \frac{{144}}{{169}} = - \frac{{119}}{{169}};$

By definition,

$\cot 2\beta = \frac{{\cos 2\beta }}{{\sin 2\beta }} = \frac{{ - \frac{{119}}{{169}}}}{{ - \frac{{120}}{{169}}}} = \frac{{119}}{{120}}.$

### Example 3.

Find $$\frac{{\cos \alpha }}{{2 - 3\sin \alpha }}$$ if $$\tan \frac{\alpha}{2} = 3.$$

Solution.

We use the double-angle identities:

$\cos \alpha = \frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}},\;\;\sin \alpha = \frac{{2\tan \frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}.$

Substitute these formulas into the original expression and calculate the answer:

$\frac{{\cos \alpha }}{{2 - 3\sin \alpha }} = \frac{{\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}}}{{2 - 3 \cdot \frac{{2\tan \frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}}} = \frac{{\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{\cancel{1 + {{\tan }^2}\frac{\alpha }{2}}}}}{{\frac{{2\left( {1 + {{\tan }^2}\frac{\alpha }{2}} \right) - 6\tan \frac{\alpha }{2}}}{\cancel{1 + {{\tan }^2}\frac{\alpha }{2}}}}} = \frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{2 + 2\,{{\tan }^2}\frac{\alpha }{2} - 6\tan \frac{\alpha }{2}}} = \frac{{1 - {3^2}}}{{2 + 2 \cdot {3^2} - 6 \cdot 3}} = \frac{{1 - 9}}{{2 + \cancel{18} - \cancel{18}}} = \frac{{ - 8}}{2} = - 4.$

### Example 4.

Find $$\frac{{2\sin \beta }}{{4 + 5\cos \beta }}$$ if $$\cot \frac{\beta}{2} = -2.$$

Solution.

If $$\cot \frac{\beta }{2} = - 2,$$ then

$\tan \frac{\beta }{2} = \frac{1}{{\cot \frac{\beta }{2}}} = - \frac{1}{2}.$

Replacing $$\sin\beta$$ and $$\cos\beta$$ with their expressions in terms of $${\tan \frac{\beta }{2}}$$ yields:

$\frac{{2\sin \beta }}{{4 + 5\cos \beta }} = \frac{{2 \cdot \frac{{2\tan \frac{\beta }{2}}}{{1 + {{\tan }^2}\frac{\beta }{2}}}}}{{4 + 5 \cdot \frac{{1 - {{\tan }^2}\frac{\beta }{2}}}{{1 + {{\tan }^2}\frac{\beta }{2}}}}} = \frac{{\frac{{4\tan \frac{\beta }{2}}}{\cancel{1 + {{\tan }^2}\frac{\beta }{2}}}}}{{\frac{{4\left( {1 + {{\tan }^2}\frac{\beta }{2}} \right) + 5\left( {1 - {{\tan }^2}\frac{\beta }{2}} \right)}}{\cancel{1 + {{\tan }^2}\frac{\beta }{2}}}}} = \frac{{4\tan \frac{\beta }{2}}}{{4 + 4\,{{\tan }^2}\frac{\beta }{2} + 5 - 5\,{{\tan }^2}\frac{\beta }{2}}} = \frac{{4\tan \frac{\beta }{2}}}{{9 - {{\tan }^2}\frac{\beta }{2}}} = \frac{{4 \cdot \left( { - \frac{1}{2}} \right)}}{{9 - {{\left( { - \frac{1}{2}} \right)}^2}}} = \frac{{ - 2}}{{\frac{{35}}{4}}} = - \frac{8}{{35}}.$

### Example 5.

Simplify the expression $\frac{{1 + \sin 2\alpha }}{{{{\left( {\sin \alpha + \cos \alpha } \right)}^2}}}.$

Solution.

Using the double-angle identity for sine and Pythagorean identity, we have

$\frac{{1 + \sin 2\alpha }}{{{{\left( {\sin \alpha + \cos \alpha } \right)}^2}}} = \frac{{1 + \sin 2\alpha }}{{{{\sin }^2}\alpha + 2\sin \alpha \cos \alpha + {{\cos }^2}\alpha }} = \frac{{1 + \sin 2\alpha }}{{1 + 2\sin \alpha \cos \alpha }} = \frac{\cancel{1 + \sin 2\alpha }}{\cancel{1 + \sin 2\alpha }} = 1.$

### Example 6.

Simplify the expression $\frac{{2\sin \beta - \sin 2\beta }}{{2\sin \beta + \sin 2\beta }}.$

Solution.

Using the double-angle formula for sine and cosine, we get

$\frac{{2\sin \beta - \sin 2\beta }}{{2\sin \beta + \sin 2\beta }} = \frac{{2\sin \beta - 2\sin \beta \cos \beta }}{{2\sin \beta + 2\sin \beta \cos \beta }} = \frac{{\cancel{2\sin \beta} \left( {1 - \cos \beta } \right)}}{{\cancel{2\sin \beta} \left( {1 + \cos \beta } \right)}} = \frac{{1 - \cos \beta }}{{1 + \cos \beta }} = \frac{{1 - {{\cos }^2}\frac{\beta }{2} + {{\sin }^2}\frac{\beta }{2}}}{{1 + {{\cos }^2}\frac{\beta }{2} - {{\sin }^2}\frac{\beta }{2}}} = \frac{{{{\sin }^2}\frac{\beta }{2} + {{\sin }^2}\frac{\beta }{2}}}{{{{\cos }^2}\frac{\beta }{2} + {{\cos }^2}\frac{\beta }{2}}} = \frac{{\cancel{2}{{\sin }^2}\frac{\beta }{2}}}{{\cancel{2}{{\cos }^2}\frac{\beta }{2}}} = {\tan ^2}\frac{\beta }{2}.$

### Example 7.

Find $$\cos4\alpha$$ if $$\tan\alpha = 2.$$

Solution.

First we express the function $$\cos 4\alpha$$ in terms of $$\cos 2\alpha:$$

$\cos 4\alpha = 2\,{\cos ^2}2\alpha - 1.$

Now we apply the double-angle identity for cosine in terms of tangent:

$\cos 2\alpha = \frac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}.$

Hence,

$\cos 4\alpha = 2\,{\cos ^2}2\alpha - 1 = 2 \cdot {\left( {\frac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}} \right)^2} - 1 = 2 \cdot {\left( {\frac{{1 - {2^2}}}{{1 + {2^2}}}} \right)^2} - 1 = 2 \cdot {\left( { - \frac{3}{5}} \right)^2} - 1 = 2 \cdot \frac{9}{{25}} - 1 = \frac{{18 - 25}}{{25}} = - \frac{7}{{25}}.$

### Example 8.

Determine $$\sin 4\beta$$ if $$\cot\beta = -3.$$

Solution.

It is clear that

$\tan \beta = \frac{1}{{\cot \beta }} = - \frac{1}{3}.$

Using the double-angle identity for sine, we can write

$\sin 4\beta = 2\sin 2\beta \cos 2\beta .$

Next, we express the double-angle functions in terms of $$\tan\beta:$$

$\sin 4\beta = 2\sin 2\beta \cos 2\beta = 2 \cdot \frac{{2\tan \beta }}{{1 + {{\tan }^2}\beta }} \cdot \frac{{1 - {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }} = 2 \cdot \frac{{2 \cdot \left( { - \frac{1}{3}} \right)}}{{1 + {{\left( { - \frac{1}{3}} \right)}^2}}} \cdot \frac{{1 - {{\left( { - \frac{1}{3}} \right)}^2}}}{{1 + {{\left( { - \frac{1}{3}} \right)}^2}}} = 2 \cdot \frac{{ - \frac{2}{3}}}{{1 + \frac{1}{9}}} \cdot \frac{{1 - \frac{1}{9}}}{{1 + \frac{1}{9}}} = 2 \cdot \frac{{ - \frac{2}{3}}}{{\frac{{10}}{9}}} \cdot \frac{{\frac{8}{9}}}{{\frac{{10}}{9}}} = 2 \cdot \left( { - \frac{6}{{10}}} \right) \cdot \frac{8}{{10}} = - \frac{{96}}{{100}} = - \frac{{24}}{{25}}.$

### Example 9.

Verify the identity $\frac{{\sin 3\alpha }}{{\sin \alpha }} - \frac{{\cos 3\alpha }}{{\cos \alpha }} = 2.$

Solution.

Recall the triple-angle formulas:

$\sin 3\alpha = 3\sin \alpha - 4{\sin ^3}\alpha ,$
$\cos 3\alpha = 4{\cos ^3}\alpha - 3\cos \alpha .$

Then the left-hand side of the expression is given by

$\frac{{\sin 3\alpha }}{{\sin \alpha }} - \frac{{\cos 3\alpha }}{{\cos \alpha }} = \frac{{3\sin \alpha - 4\,{{\sin }^3}\alpha }}{{\sin \alpha }} - \frac{{4\,{{\cos }^3}\alpha - 3\cos \alpha }}{{\cos \alpha }} = \frac{{\cancel{\sin \alpha} \left( {3 - 4\,{{\sin }^2}\alpha } \right)}}{{\cancel{\sin \alpha} }} - \frac{{\cancel{\cos \alpha} \left( {4\,{{\cos }^2}\alpha - 3} \right)}}{{\cancel{\cos \alpha} }} = 3 - 4\,{\sin ^2}\alpha - 4\,{\cos ^2}\alpha + 3 = 6 - 4\left( {\underbrace {{{\sin }^2}\alpha + {{\cos }^2}\alpha }_1} \right) = 6 - 4 = 2.$

### Example 10.

Determine which is greater: $$\tan 2\alpha$$ or $$\tan\alpha$$ if $$0 \lt \alpha \lt \frac{\pi }{4}?$$

Solution.

Note that the tangent function is increasing. Therefore, if $$0 \lt \alpha \lt \frac{\pi }{4},$$ then $$0 \lt \tan \alpha \lt 1.$$

It follows from here:

$0 \lt {\tan ^2}\alpha \lt 1,\;\; \Rightarrow 0 \lt 1 - {\tan ^2}\alpha \lt 1.$

From the other side,

$\tan \alpha \gt 0,\;\; \Rightarrow 2\tan \alpha \gt \tan \alpha ,\;\; \Rightarrow \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \gt \frac{{\tan \alpha }}{{1 - {{\tan }^2}\alpha }}.$

Since $${0 \lt 1 - {\tan ^2}\alpha \lt 1,}$$ we can write

$\frac{{\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \gt \tan \alpha .$

We have got the double inequality

$\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \gt \frac{{\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \gt \tan \alpha ,$

which mean that

$\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \gt \tan \alpha .$

The left-hand side of the inequality is equal to $$\tan 2\alpha.$$ Therefore,

$\tan 2\alpha \gt \tan \alpha .$