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Mariana [72]
3 years ago
15

Please prove the following identity

Mathematics
1 answer:
Sedaia [141]3 years ago
4 0

Let's remember the factorizations

a^3 + b^3 = (a+b)(a^2-ab+b^2)

a^3- b^3=(a-b)(a^2+ab+b^2)

Now

\dfrac{\sin ^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} + \dfrac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta}

=\dfrac{(\sin \theta + \cos \theta)(\sin ^2 \theta - \sin \theta\cos \theta + \cos^2 \theta)}{\sin \theta + \cos \theta} + \dfrac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta - \cos \theta}

=\sin ^2 \theta - \sin \theta\cos \theta + \cos^2 \theta + \sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta

=\sin ^2 \theta + \cos^2 \theta + \sin^2 \theta+ \cos^2 \theta

=2 \quad\checkmark

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