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Sagot :
Answer:
b.
Step-by-step explanation:
cos2θ = 2 cos^2 θ - 1
2cos^2 θ = 1 + cos2θ
cos^2 θ = (1 + cos2θ)/2
[tex]\bold{ cos^2\theta = \frac{1+cos 2\theta}{2}}[/tex]
The correct answer is an option (b)
What are trigonometric identities?
"These are the equations that involve trigonometry functions and true for all the values of variables given in the equation."
For given question,
We need to find the correct identities for [tex]cos^2\theta[/tex]
We know that,
[tex]\Rightarrow cos 2\theta = cos^2\theta -sin^2\theta\\\\\Rightarrow cos 2\theta = cos^2\theta -(1-cos^2\theta)~~~~......(since~~sin^2\theta+cos^2\theta=1)\\\\\Rightarrow cos 2\theta = 2cos^2\theta -1\\\\\Rightarrow 1+cos 2\theta = 2cos^2\theta \\\\\Rightarrow cos^2\theta = \frac{1+cos 2\theta}{2}[/tex]
Consider,
[tex]\Rightarrow \frac{1-tan^2\theta}{1+tan^2\theta} = \frac{1-\frac{sin^2\theta}{cos^2\theta} }{1+\frac{sin^2\theta}{cos^2\theta} }~~~~~~..........(since~~tan\theta=\frac{sin\theta}{cos\theta})\\\\\Rightarrow \frac{1-tan^2\theta}{1+tan^2\theta} = \frac{\frac{cos^2\theta-sin^2\theta}{cos^2\theta} }{\frac{cos^2\theta+sin^2\theta}{cos^2\theta} }[/tex]
[tex]\Rightarrow \frac{1-tan^2\theta}{1+tan^2\theta} = \frac{{cos^2\theta-sin^2\theta} }{cos^2\theta+sin^2\theta }\\\\\Rightarrow \frac{1-tan^2\theta}{1+tan^2\theta} = \frac{{cos2\theta} }{1}~~~~~~............(cos^2\theta+sin^2\theta=1~and~cos2\theta=cos^2\theta-sin^2\theta)[/tex]
Therefore, [tex]\bold{ cos^2\theta = \frac{1+cos 2\theta}{2}}[/tex]
The correct answer is an option (b)
Learn more about the trigonometric identities here:
https://brainly.com/question/12537661
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