Question

simplify the expression.

\\( \sqrt { \frac { 1 - \cos 157 ^ { \circ } } { 1 + \cos 157 ^ { \circ } } } \\)

\\( \sqrt { \frac { 1 - \cos 157 ^ { \circ } } { 1 + \cos 157 ^ { \circ } } } = \square ^ { \circ } \\)
(simplify your answer. use integers or decimals for any numbers in the expression. do not include the degree symbol in your answer.)

Explanation:

Step1: Use the trigonometric identity

We know the identity \(\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{1+\cos\alpha}}\). Here \(\alpha = 157^{\circ}\), and \(\frac{\alpha}{2}=\frac{157^{\circ}}{2}=78.5^{\circ}\). Since \(0^{\circ}<78.5^{\circ}<90^{\circ}\), \(\tan\frac{157^{\circ}}{2}>0\).

Step2: Simplify the expression

So \(\sqrt{\frac{1-\cos157^{\circ}}{1 + \cos157^{\circ}}}=\tan\frac{157^{\circ}}{2}\).
Using a calculator, \(\tan78.5^{\circ}\approx5\) (rounded to the nearest integer).

Answer:

\(5\)