Evaluating Trigonometric Expressions
In order to evaluate a trigonometric expression, we often need to simplify it first using algebra and trigonometric identities.
To calculate trig functions, it is convenient to express all angles in terms of reference angles.
Solved Problems
Example 1.
Calculate \(\sin 2\alpha \) if
\[\sin \alpha - \cos \alpha = p.\]
Solution.
Using the the Pythagorean trigonometric identity and double-angle formula, we have
\[{p^2} = {\left( {\sin \alpha - \cos \alpha } \right)^2} = {\sin ^2}\alpha - 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = 1 - \sin 2\alpha .\]
Therefore
\[\sin 2\alpha = 1 - {p^2}.\]
Example 2.
Evaluate the trigonometric expression:
\[\sin 40^\circ + 2\sin 20^\circ - \sqrt 3 \cos 40^\circ .\]
Solution.
We denote the expression by \(A\) and represent it in the form:
\[A = 2\left( {\frac{1}{2}\sin 40^\circ - \frac{{\sqrt 3 }}{2}\cos 40^\circ } \right) + 2\sin 20^\circ .\]
Notice that \(\frac{1}{2} = \cos 60^\circ \) and \(\frac{{\sqrt 3 }}{2} = \sin 60^\circ .\) Then using the sine subtraction formula, we have
\[A = 2\left( {\sin 40^\circ \cos 60^\circ - \cos 40^\circ \sin 60^\circ } \right) + 2\sin 20^\circ = 2\sin \left( {40^\circ - 60^\circ } \right) + 2\sin 20^\circ = 2\sin \left( { - 20^\circ } \right) + 2\sin 20^\circ .\]
The sine function is odd, that is, \(\sin \left( { - 20^\circ } \right) = - \sin 20^\circ .\) Therefore
\[A = \cancel{{ - 2\sin 20^\circ }} + \cancel{{2\sin 20^\circ }} = 0.\]
Example 3.
Calculate \({\sin ^4}\alpha + {\cos ^4}\alpha \) if
\[\sin \alpha - \cos \alpha = \frac{1}{2}.\]
Solution.
We first determine \(\sin 2\alpha\) using the Pythagorean trig identity and double-angle formula for sine:
\[\sin \alpha - \cos \alpha = \frac{1}{2}, \Rightarrow {\left( {\sin \alpha - \cos \alpha } \right)^2} = \frac{1}{4}, \Rightarrow {\sin ^2}\alpha - 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = \frac{1}{4}, \Rightarrow 1 - \sin 2\alpha = \frac{1}{4}, \Rightarrow \sin 2\alpha = \frac{3}{4}.\]
Notice that
\[1 = {1^2} = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} = {\sin ^4}\alpha + 2\,{\sin ^2}\alpha \,{\cos ^2}\alpha + {\cos ^4}\alpha .\]
Hence,
\[{\sin ^4}\alpha + {\cos ^4}\alpha = 1 - 2\,{\sin ^2}\alpha \,{\cos ^2}\alpha = 1 - \frac{1}{2} \cdot 4\,{\sin ^2}\alpha \,{\cos ^2}\alpha = 1 - \frac{1}{2}{\left( {2\sin \alpha \cos \alpha } \right)^2} = 1 - \frac{1}{2}{\sin ^2}2\alpha = 1 - \frac{1}{2}{\left( {\frac{3}{4}} \right)^2} = 1 - \frac{1}{2} \cdot \frac{9}{{16}} = 1 - \frac{9}{{32}} = \frac{{23}}{{32}}.\]
Example 4.
Evaluate the trigonometric expression:
\[1 + 2\cos 70^\circ - \frac{{\sin 105^\circ }}{{\sin 35^\circ }}.\]
Solution.
We represent \(\sin 105^\circ\) as
\[\sin 105^\circ = \sin \left( {70^\circ + 35^\circ } \right) = \sin 70^\circ \cos 35^\circ + \cos 70^\circ \sin 35^\circ .\]
Then
\[A = 1 + 2\cos 70^\circ - \frac{{\sin 105^\circ }}{{\sin 35^\circ }} = 1 + 2\cos 70^\circ - \frac{{\sin 70^\circ \cos 35^\circ }}{{\sin 35^\circ }} - \frac{{\cos 70^\circ \cancel{{\sin 35^\circ }}}}{{\cancel{{\sin 35^\circ }}}} = 1 + \cos 70^\circ - \frac{{\sin 70^\circ \cos 35^\circ }}{{\sin 35^\circ }}.\]
Using the double-angle formula for sine and cosine, we get
\[A = 1 + {\cos ^2}\,35^\circ - {\sin ^2}\,35^\circ - \frac{{2\cancel{{\sin 35^\circ }}{{\cos }^2}\,35^\circ }}{{\cancel{{\sin 35}}^\circ }} = 1 - {\sin ^2}\,35^\circ - {\cos ^2}\,35^\circ .\]
Finally recall the Pythagorean trigonometric identity:
\[A = 1 - {\sin ^2}\,35^\circ - {\cos ^2}\,35^\circ = 1 - \underbrace {\left( {{{\sin }^2}\,35^\circ + {{\cos }^2}\,35^\circ } \right)}_{1} = 1 - 1 = 0.\]
Example 5.
Evaluate the trigonometric expression:
\[{\sin ^4}15^\circ + {\cos ^4}15^\circ. \]
Solution.
By the Pythagorean trig identity:
\[{\sin ^2}15^\circ + {\cos ^2}15^\circ = 1.\]
Squaring both sides gives
\[1 = {1^2} = {\left( {{{\sin }^2}15^\circ + {{\cos }^2}15^\circ } \right)^2} = {\sin ^4}15^\circ + 2\,{\sin ^2}15^\circ {\cos ^2}15^\circ + {\cos ^4}15^\circ .\]
Now we use the double-angle formula for sine:
\[{\sin ^4}15^\circ + {\cos ^4}15^\circ = 1 - 2{\sin ^2}15^\circ {\cos ^2}15^\circ = 1 - \frac{1}{2}{\left( {2\sin 15^\circ \cos 15^\circ } \right)^2} = 1 - \frac{1}{2}{\sin ^2}30^\circ = 1 - \frac{1}{2}{\left( {\frac{1}{2}} \right)^2} = 1 - \frac{1}{8} = \frac{7}{8}.\]
Example 6.
Evaluate the trigonometric expression:
\[\cos \frac{{3\pi }}{5}\cos \frac{{6\pi }}{5}.\]
Solution.
We divide and multiply this expression by \(2\sin \frac{{3\pi }}{5}.\) Then using the double-angle formula, we get
\[A = \cos \frac{{3\pi }}{5}\cos \frac{{6\pi }}{5} = \frac{{\overbrace {2\sin \frac{{3\pi }}{5}\cos \frac{{3\pi }}{5}}^{\sin \frac{{6\pi }}{5}}\cos \frac{{6\pi }}{5}}}{{2\sin \frac{{3\pi }}{5}}} = \frac{{\sin \frac{{6\pi }}{5}\cos \frac{{6\pi }}{5}}}{{2\sin \frac{{3\pi }}{5}}} = \frac{{2\sin \frac{{6\pi }}{5}\cos \frac{{6\pi }}{5}}}{{4\sin \frac{{3\pi }}{5}}} = \frac{{\sin \frac{{12\pi }}{5}}}{{4\sin \frac{{3\pi }}{5}}}.\]
It is obvious that
\[\sin \frac{{12\pi }}{5} = \sin \left( {\frac{{2\pi }}{5} + 2\pi } \right) = \sin \frac{{2\pi }}{5}.\]
Apply the cofunction identity:
\[A = \frac{{\sin \frac{{2\pi }}{5}}}{{4\sin \frac{{3\pi }}{5}}} = \frac{{\sin \left( {\pi - \frac{{3\pi }}{5}} \right)}}{{4\sin \frac{{3\pi }}{5}}} = \frac{{\cancel{{\sin \frac{{3\pi }}{5}}}}}{{4\cancel{{\sin \frac{{3\pi }}{5}}}}} = \frac{1}{4}.\]
Example 7.
Calculate \({\tan ^3}\alpha - \frac{1}{{{{\tan }^3}\alpha }}\) if
\[\tan \alpha - \frac{1}{{\tan \alpha }} = 1.\]
Solution.
Let's factor the difference of cubes:
\[A = {\tan ^3}\alpha - \frac{1}{{{{\tan }^3}\alpha }} = \left( {\tan \alpha - \frac{1}{{\tan \alpha }}} \right)\left( {{{\tan }^2}\alpha + \tan \alpha \cdot \frac{1}{{\tan \alpha }} + \frac{1}{{{{\tan }^2}\alpha }}} \right) = 1 \cdot \left( {{{\tan }^2}\alpha + 1 + \frac{1}{{{{\tan }^2}\alpha }}} \right) = {\tan ^2}\alpha + 1 + \frac{1}{{{{\tan }^2}\alpha }}.\]
Find the sum \({\tan ^2}\alpha + \frac{1}{{{{\tan }^2}\alpha }}:\)
\[1 = {1^2} = {\left( {\tan \alpha - \frac{1}{{\tan \alpha }}} \right)^2} = {\tan ^2}\alpha - 2\tan \alpha \cdot \frac{1}{{\tan \alpha }} + \frac{1}{{{{\tan }^2}\alpha }} = {\tan ^2}\alpha - 2 + \frac{1}{{{{\tan }^2}\alpha }}.\]
Hence,
\[{\tan ^2}\alpha + \frac{1}{{{{\tan }^2}\alpha }} = 3.\]
Substitute this to find the value of the original expression:
\[A = {\tan ^2}\alpha + 1 + \frac{1}{{{{\tan }^2}\alpha }} = 3 + 1 = 4.\]
Example 8.
Evaluate:
\[\sqrt 3 \left( {\tan 255^\circ - \tan 195^\circ } \right).\]
Solution.
We use the cofunction identities:
\[\tan 255^\circ = \tan \left( {270^\circ - 15^\circ } \right) = \cot 15^\circ ;\]
\[\tan 195^\circ = \tan \left( {180^\circ + 15^\circ } \right) = \tan 15^\circ .\]
By the double-angle formula, we get:
\[\sqrt 3 \left( {\tan 255^\circ - \tan 195^\circ } \right) = \sqrt 3 \left( {\cot 15^\circ - \tan 15^\circ } \right) = \sqrt 3 \left( {\frac{1}{{\tan 15^\circ }} - \tan 15^\circ } \right) = \frac{{\sqrt 3 \left( {1 - {{\tan }^2}15^\circ } \right)}}{{\tan 15^\circ }} = \frac{{2\sqrt 3 }}{{\frac{{2\tan 15^\circ }}{{1 - {{\tan }^2}15^\circ }}}} = \frac{{2\sqrt 3 }}{{\tan 30^\circ }} = \frac{{2\sqrt 3 }}{{\frac{1}{{\sqrt 3 }}}} = 2\sqrt 3 \cdot \sqrt 3 = 6.\]
See more problems on Page 2.
