sinθ = tanθ cosθ, so
tanθ - sinθ = tanθ (1-cosθ)
squared that is tan^2θ (1-cosθ)^2
now we can add up the left side:
tan^2θ(1-cosθ)^2 + (1-cosθ)^2
= (tan^2θ + 1)(1-cosθ)^2
= sec^2θ (1-cosθ)^2
on the right side, we have
1-secθ = (cosθ-1)/cosθ
squared, that is sec^2θ (cosθ-1)^2
and they are the same
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1-2cos^2θ = -cos2θ
2sinθ cosθ = sin2θ
so, you have -2cot2θ
-2cot2θ = -2(1-tan^2θ)/2tanθ
= -(1-tan^2θ)/tanθ
= -(cotθ - tanθ)
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(sinθ+cosθ)^2 = sin^2θ + 2sinθcosθ + cos^2θ
expand the other one, then add and it drops right out.
Prove the following identities:
1. (tan theta - sin theta)^2 + (1-cos theta)^2 = (1-sec theta) ^2
2. (1-2cos^2 theta) / sin theta cos theta = tan theta - cot theta
3. (sin theta + cos theta ) ^2 + (sin theta - cos theta ) ^2 = 2
Thank you so much! :)
1 answer