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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds3 Marks
Question
Prove that: $\sin^2(\frac{\pi}{8}+\frac{\text{x}}{2})-\sin^2(\frac{\pi}{8}-\frac{\text{x}}{2})=\frac{1}{\sqrt{2}}\sin\text{x}$
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Answer

$\text{LHS}=\sin^2\Big(\frac{\pi}{8}+\frac{\text{x}}{2}\Big)-\sin^2\Big(\frac{\pi}{8}-\frac{\text{x}}{2}\Big)$ $=\Big[\sin\Big(\frac{\pi}{8}+\frac{\text{x}}{2}\Big)+\sin\Big(\frac{\pi}{8}-\frac{\text{x}}{\pi}\Big)\Big]\Big[\frac{\pi}{8}+\frac{\text{x}}{2}\Big)-\sin\Big(\frac{\pi}{8}-\frac{\text{x}}{2}\Big)\Big]$ $=\Big[\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{8},\sin\frac{\text{x}}{2}-\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{8}\sin\frac{\text{x}}{2}\Big]$ $=\Big[\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{\text{x}},\sin\frac{\text{x}}{2}-\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{8},\sin\frac{\text{x}}{2}\Big]$ $=\Big(2\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}\Big)\Big(2\cos\frac{\pi}{2},\sin\frac{\text{x}}{2}\Big)$ $=2\sin\frac{\pi}{8},\cos\frac{\pi}{2}2\sin\frac{\text{x}}{2},\cos\frac{\text{x}}{2}$ $=\sin2.\frac{\pi}{8},\sin2.\frac{\text{x}}{2}$ $=\sin\frac{\pi}{4},\sin\text{x}$ $=\frac{1}{\sqrt{2}}\sin\text{x}=\text{RHS}$

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