I am stuck with the simple expressioncos2(θ+α)1−cos2(θ−α)=const.where θ is a variable and α is the number satisfyingα=tan−1(43)

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junoon363km · Accepted Answer

Note that 1−cos2(θ−α)=sin2(θ−α)That gives you:cos2(θ+α)1−cos2(θ−α)=const.=cos2(θ+α)sin2(θ−α)For the numerator:cos⁡(θ+α)=cos⁡θcos⁡α−sin⁡θsin⁡α (1)For the denominator:sin⁡(θ−α)=sin⁡θcos⁡α−cos⁡θsin⁡α (2)Note that since tan⁡α=43, we have a 3:4:5 triangle, using the Pythagorean Theorem, and noting that the leg opposite α must be of length 4, and the leg adjacent to α is length 3. This gives us a hypotenuse of length 5. Calculating cos⁡α=adjacenthypotenuse=35. Likewise sin⁡α=oppositehypotenuse=45.So (1) becomes cos2(θ+α)=(35cos⁡θ−45sin⁡θ)2And (2) becomes sin2(θ−α)=(35sin⁡θ−45cos⁡θ)2

I am stuck with the simple expression \frac{\cos^2(\theta+\alpha)}{1-\cos^2(\theta-\alpha)}=\text{const.} where \theta is a

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