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If $\sin ^{3} \theta+\cos ^{3} \theta=0$, then what is the value of $\theta$ ?
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Verified Answer
The correct answer is:
$\frac{-\pi}{4}$
Since, $\sin ^{3} \theta+\cos ^{3} \theta=0$
$\Rightarrow(\sin \theta+\cos \theta)\left(\sin ^{2} \theta-\sin \theta \cos \theta+\cos ^{2} \theta\right)=0$
$\left(\because a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right)$
$\Rightarrow(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta)=0$
$\Rightarrow(\sin \theta+\cos \theta)\left(1-\frac{\sin 2 \theta}{2}\right)=0$
$\Rightarrow \sin \theta+\cos \theta=0$
or $\sin 2 \theta=2$
(discarded since $\sin ^{2} \theta=2$ is not possible) $\Rightarrow \sin \theta+\cos \theta=0 \Rightarrow \sin \theta=-\cos \theta$
$\Rightarrow \tan \theta=-1 \Rightarrow \theta=-\pi / 4$
$\Rightarrow(\sin \theta+\cos \theta)\left(\sin ^{2} \theta-\sin \theta \cos \theta+\cos ^{2} \theta\right)=0$
$\left(\because a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right)$
$\Rightarrow(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta)=0$
$\Rightarrow(\sin \theta+\cos \theta)\left(1-\frac{\sin 2 \theta}{2}\right)=0$
$\Rightarrow \sin \theta+\cos \theta=0$
or $\sin 2 \theta=2$
(discarded since $\sin ^{2} \theta=2$ is not possible) $\Rightarrow \sin \theta+\cos \theta=0 \Rightarrow \sin \theta=-\cos \theta$
$\Rightarrow \tan \theta=-1 \Rightarrow \theta=-\pi / 4$
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