Question

use trigonometric identities to write sin x in terms of cos x.
choose the correct expression equal to sin x
a. sin x = \frac{1}{1 - cos x}
b. sin x = \pm\sqrt{1 - cos^{2}x}
c. sin x = \pm\sqrt{1 - cos x}
d. sin x = (1 - cos x)(1 + cos x)

Explanation:

Step1: Use Pythagorean identity

The Pythagorean identity is \(\sin^{2}x+\cos^{2}x = 1\).

Step2: Solve for \(\sin x\)

Subtract \(\cos^{2}x\) from both sides of the identity \(\sin^{2}x+\cos^{2}x = 1\). We get \(\sin^{2}x=1 - \cos^{2}x\). Then take the square - root of both sides: \(\sin x=\pm\sqrt{1 - \cos^{2}x}\).

Let's check other options:

• Option A: \(\frac{1}{1 - \cos x}\) is not equivalent to \(\sin x\). We know that \(\frac{1}{1-\cos x}\) is related to the cosecant and cotangent functions (\(\csc x+\cot x=\frac{1}{\sin x}+\frac{\cos x}{\sin x}=\frac{1 + \cos x}{\sin x}

eq\sin x\)).

• Option C: \(\pm\sqrt{1-\cos x}\) is not correct. From \(\sin^{2}x=1 - \cos^{2}x=(1 - \cos x)(1+\cos x)

eq(1 - \cos x)\) (except in some special cases).

• Option D: \((1 - \cos x)(1+\cos x)=1-\cos^{2}x=\sin^{2}x

eq\sin x\)

Answer:

B. \(\sin x=\pm\sqrt{1 - \cos^{2}x}\)