# Double-Angle and Multiple-Angle Identities

### Double-Angle Identities

The double-angle identities are a special case of the addition formulas.

To derive the double-angle formula for sine, we use the sine addition formula:

${\sin \left( {\alpha + \beta } \right) }={ \sin \alpha \cos \beta + \cos \alpha \sin \beta .}$

Putting $$\beta = \alpha$$ gives

${\sin \left( {\alpha + \alpha } \right) = \sin 2\alpha }={ \sin \alpha \cos \alpha + \cos \alpha \sin \alpha }={ 2\sin \alpha \cos \alpha ,}$

that is,

Similarly, substituting $$\beta = \alpha$$ in the cosine addition formula

${\cos \left( {\alpha + \beta } \right) }={ \cos \alpha \cos \beta – \sin \alpha \sin \beta }$

yields the double-angle identity for cosine:

${\cos \left( {\alpha + \alpha } \right) = \cos 2\alpha }={ \cos \alpha \cos \alpha – \sin \alpha \sin \alpha }={ {\cos ^2}\alpha – {\sin ^2}\alpha ,}$

or

Using the Pythagorean trigonometric identity, we can get two more variations for the cosine of a double angle:

Since $${\cos ^2}\alpha = 1 – {\sin ^2}\alpha ,$$ we have

${\cos 2\alpha = {\cos ^2}\alpha – {\sin ^2}\alpha }={ 1 – {\sin ^2}\alpha – {\sin ^2}\alpha }={ 1 – 2\,{\sin ^2}\alpha .}$

Hence, the first variation is given by

If we substitute the identity $${\sin ^2}\alpha = 1 – {\cos ^2}\alpha$$ in the original double-angle formula for cosine, we get the second variation:

${\cos 2\alpha = {\cos ^2}\alpha – {\sin ^2}\alpha }={ {\cos ^2}\alpha – \left( {1 – {{\cos }^2}\alpha } \right) }={ 2\,{\cos ^2}\alpha – 1.}$

Thus,

To derive the double-angle formula for tangent, recall the tangent addition formula

${\tan \left( {\alpha + \beta } \right) }={ \frac{{\tan \alpha + \tan \beta }}{{1 – \tan \alpha \tan \beta }}.}$

Let $$\beta = \alpha.$$ Then

${\tan \left( {\alpha + \alpha } \right) = \tan 2\alpha }={ \frac{{\tan \alpha + \tan \alpha }}{{1 – \tan \alpha \tan \alpha }} }={ \frac{{2\tan \alpha }}{{1 – {{\tan }^2}\alpha }},}$

that is,

The double-angle identity for cotangent has the form

${\cot \left( {\alpha + \beta } \right) = \frac{{\cot \alpha \cot \beta – 1}}{{\cot \alpha + \cot \beta }},}\;\; \Rightarrow {\cot 2\alpha = \frac{{\cot \alpha \cot \alpha – 1}}{{\cot \alpha + \cot \alpha }} }={ \frac{{{{\cot }^2}\alpha – 1}}{{2\cot \alpha }}.}$

Hence,

### Double-Angle Identities in Terms of Tangent

The double angle identities can be expressed in terms of the tangent of the single angle. To prove this, note that

${\sin 2\alpha = 2\sin \alpha \cos \alpha }={ \frac{{2\sin \alpha \cos \alpha }}{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}.}$

Assuming $$\cos \alpha \ne 0,$$ we can divide both the numerator and denominator by $$\cos^2\alpha.$$ This yields

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

We got the following identity:

Now take the double-angle formula for cosine:

${\cos 2\alpha }={ {\cos ^2}\alpha – {\sin ^2}\alpha }={ \frac{{{{\cos }^2}\alpha – {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}.}$

Divide the numerator and denominator by $$\cos^2\alpha$$ again, provided $$\cos\alpha \ne 0$$ or $$\alpha \ne \large{\frac{\pi }{2}}\normalsize + \pi n,$$ $$n \in \mathbb{Z}:$$

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

So, we have

Finally, note that $$\cot 2\alpha = \large{\frac{1}{{\tan 2\alpha }}}\normalsize.$$ Therefore,

where $$\alpha \ne \large{\frac{{\pi n}}{2}}\normalsize,$$ $$n \in \mathbb{Z}.$$

### Triple-Angle Identities

Using double-angle identities and addition formulas, we can derive triple-angle identities for trigonometric functions.

${\sin 3\alpha = \sin \left( {2\alpha + \alpha } \right) }={ \sin 2\alpha \cos \alpha + \cos 2\alpha \sin \alpha }={ 2\sin \alpha \cos \alpha \cos \alpha }+{ \left( {{{\cos }^2}\alpha – {{\sin }^2}\alpha } \right)\sin \alpha }={ 2\sin \alpha \,{\cos ^2}\alpha }+{ \left( {1 – 2{{\sin }^2}\alpha } \right)\sin \alpha }={ 2\sin \alpha \left( {1 – {{\sin }^2}\alpha } \right) }+{ \left( {1 – 2\,{{\sin }^2}\alpha } \right)\sin \alpha }={ 2\sin \alpha – 2\,{\sin ^3}\alpha }+{ \sin \alpha – 2\,{\sin ^3}\alpha }={ 3\sin \alpha – 4\,{\sin ^3}\alpha .}$

Thus,

The triple-angle formula for cosine can be proved in a similar manner:

${\cos 3\alpha = \cos \left( {2\alpha + \alpha } \right) }={ \cos 2\alpha \cos \alpha – \sin 2\alpha \sin \alpha }={ \left( {{{\cos }^2}\alpha – {{\sin }^2}\alpha } \right)\cos \alpha }-{ 2\sin \alpha \cos \alpha \sin \alpha }={ \left( {2\,{{\cos }^2}\alpha – 1} \right)\cos \alpha }-{ 2\cos \alpha\,{\sin ^2}\alpha }={ \left( {2\,{{\cos }^2}\alpha – 1} \right)\cos \alpha }-{ 2\cos \alpha \left( {1 – {{\cos }^2}\alpha } \right) }={ 2\,{\cos ^3}\alpha – \cos \alpha }-{ 2\cos \alpha + 2\,{\cos ^3}\alpha }={ 4\,{\cos ^3}\alpha – 3\cos \alpha .}$

Hence,

The triple-angle identity for tangent is given by

$\require{cancel}{\tan 3\alpha = \tan \left( {2\alpha + \alpha } \right) }={ \frac{{\tan 2\alpha + \tan \alpha }}{{1 – \tan 2\alpha \tan \alpha }} }={ \frac{{\frac{{2\tan \alpha }}{{1 – {{\tan }^2}\alpha }} + \tan \alpha }}{{1 – \frac{{2\tan \alpha }}{{1 – {{\tan }^2}\alpha }} \cdot \tan \alpha }} }={ \frac{{\frac{{2\tan \alpha + \tan \alpha \left( {1 – {{\tan }^2}\alpha } \right)}}{\cancel{1 – {{\tan }^2}\alpha }}}}{{\frac{{1 – {{\tan }^2}\alpha – 2\,{{\tan }^2}\alpha }}{\cancel{1 – {{\tan }^2}\alpha }}}} }={ \frac{{2\tan \alpha + \tan \alpha – {{\tan }^3}\alpha }}{{1 – {{\tan }^2}\alpha – 2\,{{\tan }^2}\alpha }} }={ \frac{{3\tan \alpha – {{\tan }^3}\alpha }}{{1 – 3\,{{\tan }^2}\alpha }},}$

that is,

Let’s also consider the triple-angle formula for cotangent:

${\cot 3\alpha = \cot \left( {2\alpha + \alpha } \right) }={ \frac{{\cot 2\alpha \cot \alpha – 1}}{{\cot 2\alpha + \cot \alpha }} }={ \frac{{\frac{{{{\cot }^2}\alpha – 1}}{{2\cot \alpha }} \cdot \cot \alpha – 1}}{{\frac{{{{\cot }^2}\alpha – 1}}{{2\cot \alpha }} + \cot \alpha }} }={ \frac{{\frac{{{{\cot }^3}\alpha – \cot \alpha – 2\cot \alpha }}{\cancel{2\cot \alpha }}}}{{\frac{{{{\cot }^2}\alpha – 1 + 2\,{{\cot }^2}\alpha }}{\cancel{2\cot \alpha }}}} }={ \frac{{{{\cot }^3}\alpha – \cot \alpha – 2\cot \alpha }}{{{{\cot }^2}\alpha – 1 + 2\,{{\cot }^2}\alpha }} }={ \frac{{{{\cot }^3}\alpha – 3\cot \alpha }}{{3\,{{\cot }^2}\alpha – 1}},}$

or

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

Find $$\large{\frac{{\cos \alpha }}{{2 – 3\sin \alpha }}}\normalsize$$ if $$\tan \large{\frac{\alpha}{2}}\normalsize = 3.$$

### Example 4

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

### 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 $$0 \lt \alpha \lt \large{\frac{\pi }{4}}\normalsize?$$

### Example 1.

Calculate $$\sin2\alpha,$$ $$\cos2\alpha,$$ and $$\tan2\alpha$$ if $$\sin\alpha = \large{\frac{1}{2}}\normalsize$$ 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 $$\cos\beta = -\large{\frac{5}{13}}\normalsize$$ 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 $$\large{\frac{{\cos \alpha }}{{2 – 3\sin \alpha }}}\normalsize$$ if $$\tan \large{\frac{\alpha}{2}}\normalsize = 3.$$

Solution.

We use the double-angle identities:

${\cos \alpha = \frac{{1 – {{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}},\;\;}\kern0pt{\sin \alpha = \frac{{2\tan \frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}.}$

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

Solution.

If $$\cot \large{\frac{\beta }{2}}\normalsize = – 2,$$ 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 $${\tan \large{\frac{\beta }{2}}\normalsize}$$ 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 $$\cos 4\alpha$$ 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 $$0 \lt \alpha \lt \large{\frac{\pi }{4}}\normalsize?$$

Solution.

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

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