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If $\sin \beta$ is the harmonic mean of $\sin \alpha$ and $\cos \alpha$, and $\sin \theta$ is the arithmetic mean of $\sin \alpha$ and $\cos \alpha$, then which of the following is/are correct? $[2018-I I]$
1) $\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \sin \beta=\sin 2 \alpha$
2) $\sqrt{2} \sin \theta=\cos \left(\alpha-\frac{\pi}{4}\right)$
Select the correct answer using the code given below:
Options:
1) $\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \sin \beta=\sin 2 \alpha$
2) $\sqrt{2} \sin \theta=\cos \left(\alpha-\frac{\pi}{4}\right)$
Select the correct answer using the code given below:
Solution:
2077 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2
From question, $\sin \alpha, \sin \beta$ and $\cos \alpha$ are in $H . P$
then, $\sin \beta=\frac{2 \sin \alpha \cos \alpha}{\sin \alpha+\cos \alpha}$
and $\sin \alpha, \sin \theta$ and $\cos \alpha$ are in A.P.
then, $2 \sin \theta=\sin \alpha+\cos \alpha$
Statement 1: $\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \cdot \sin \beta$
$=(\sin \alpha+\cos \alpha) \cdot \frac{2 \sin \alpha \cos \alpha}{\sin \alpha+\cos \alpha}$
$=2 \sin \alpha \cos \alpha=\sin 2 \alpha$
Statement $2: \cos \left(\alpha-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\sin \alpha+\cos \alpha)$
$=\frac{2 \sin \theta}{\sqrt{2}}=\sqrt{2} \sin \theta$
then, $\sin \beta=\frac{2 \sin \alpha \cos \alpha}{\sin \alpha+\cos \alpha}$
and $\sin \alpha, \sin \theta$ and $\cos \alpha$ are in A.P.
then, $2 \sin \theta=\sin \alpha+\cos \alpha$
Statement 1: $\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \cdot \sin \beta$
$=(\sin \alpha+\cos \alpha) \cdot \frac{2 \sin \alpha \cos \alpha}{\sin \alpha+\cos \alpha}$
$=2 \sin \alpha \cos \alpha=\sin 2 \alpha$
Statement $2: \cos \left(\alpha-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\sin \alpha+\cos \alpha)$
$=\frac{2 \sin \theta}{\sqrt{2}}=\sqrt{2} \sin \theta$
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