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How does `(cos^2x - sin^2x) / (cos^2x + sin^2x)` get to `cos (2x)` ?
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High School Teacher
`(cos^2x - sin^2x) /(cos^2x + sin^2x)`
To show that the given expression is equal to cos (2x), apply the Pythagorean identity which states that `cos^2x + sin^2x = 1` .
`= (cos^2x - sin^2x) /1`
`= cos^2x - sin^2x`
Next, apply the double angle identity of cosine which is `cos (2x) = cos^2x - sin^2x` .
Hence, `(cos^2x-sin^2x)/(cos^2x+sin^2x) = cos(2x)` .
Posted by mjripalda on December 13, 2012 at 2:10 AM (Answer #1)
Replacing `sin^2 x + cos^2 x` with 1, the given expression becomes:
cos^2 x - sin^2 x
You may write `cos^2 x` as `cos x*cos x` and `sin^2 x` as `sin x*sin x` such that:
`cos^2 x - sin^2 x = cos x*cos x - sin x*sin x `
You need to use the following trigonometric identity, such that:
`cos alpha*cos beta - sin alpha*sin beta = cos (alpha + beta)`
Reasoning by analogy yields:
`cos x*cos x - sin x*sin x = cos(x + x) = cos (2x)`
Thus, evaluating the equivalent expression of the fraction `(cos^2 x - sin^2 x)/(sin^2 x + cos^2 x)` yields that it may be converted into `cos 2x` .
Posted by sciencesolve on December 13, 2012 at 6:06 AM (Answer #2)
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