# 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.