use trigonometric identities to write sin x in terms of cos x. \nchoose the correct expression equal to sin…

use trigonometric identities to write sin x in terms of cos x. \nchoose the correct expression equal to sin x \na. sin x = \\frac{1}{1 - cos x} \nb. sin x = \\pm\\sqrt{1 - cos^{2}x} \nc. sin x = \\pm\\sqrt{1 - cos x} \nd. sin x = (1 - cos x)(1 + cos x)

use trigonometric identities to write sin x in terms of cos x. \nchoose the correct expression equal to sin x \na. sin x = \\frac{1}{1 - cos x} \nb. sin x = \\pm\\sqrt{1 - cos^{2}x} \nc. sin x = \\pm\\sqrt{1 - cos x} \nd. sin x = (1 - cos x)(1 + cos x)

Answer

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}\neq\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)\neq(1 - \cos x)) (except in some special cases).
  • Option D: ((1 - \cos x)(1+\cos x)=1-\cos^{2}x=\sin^{2}x\neq\sin x)

Answer:

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