Double-Angle and Multiple-Angle Identities
Solved Problems
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,
\[\tan 2\alpha \gt \tan \alpha .\]
