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Verify the identity

sin^3 0 +cos^3 0 / sin 0+ cos 0 = 1 - sin 0 cos 0

User Lureen
by
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1 Answer

1 vote

Answer:

See below for proof of identity.

Explanation:

Since both terms of
\sin^3\theta+\cos^3\theta are perfect cubes, we can use the formula
a^3+b^3=(a+b)(a^2-ab+b^2) where
a=\sin\theta and
b=\cos\theta:


\displaystyle (\sin^3\theta+\cos^3\theta)/(\sin\theta+\cos\theta)\\\\((\sin\theta+\cos\theta)(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta))/(\sin\theta+\cos\theta)\\ \\((\sin\theta+\cos\theta)(1-\sin\theta\cos\theta))/(\sin\theta+\cos\theta)\\\\1-\sin\theta\cos\theta

Thus, the identity is proven.

User Geekbro
by
6.7k points