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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae2 Marks
Question
Show that:
$\sin50^\circ\cos85^\circ=\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}$
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Answer

$\sin50^\circ\cos85^\circ=\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}$
$\text{LHS}=\sin50^\circ\cos85^\circ=\frac{2\sin50^\circ\cos85^\circ}{2}$
$\because\ 2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$\Rightarrow\ \frac{2\sin50^\circ\cos85^\circ}{2}=\frac{1}{2}[\sin(50^\circ+85^\circ)+\sin(50^\circ-85^\circ)]$
$=\ \frac{1}{2}[\sin135^\circ+\sin(-35^\circ)]$
$=\ \frac{1}{2}[\sin(90^\circ+45^\circ)-\sin35^\circ]$$[\because\ \sin(-\theta)=-\sin\theta]$
$=\ \frac{1}{2}[\cos45^\circ-\sin35^\circ]$$[\because\ \sin(90^\circ+\theta)=\cos\theta]$
Now,
$\cos45^\circ=\frac{1}{\sqrt2}$
$=\ \frac{1}{2}\Big[\frac{1}{\sqrt2}-\sin35^\circ\Big]$
$=\ \frac{1-\sqrt2\sin35^\circ}{2\sqrt2}=\text{RHS}$

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