Question

If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =

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Answer 1

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\[3 \sin x + 4 \cos x = 5\]
\[\frac{3}{5}\sin x + \frac{4}{5}\cos x = 1\]
\[\text{ Let }\cos \alpha = \frac{3}{5}\text{ and }\sin \alpha = \frac{4}{5} . \]
\[ \therefore \cos \alpha \sin x + \sin\alpha \cos x = 1\]
\[ \Rightarrow \sin \left( \alpha + x \right) = \sin\frac{\pi}{2}\]
\[ \Rightarrow \alpha + x = \frac{\pi}{2}\]
\[ \Rightarrow x = \frac{\pi}{2} - \alpha . . . . (1)\]
\[\text{ We have to find the value of }4 \sin x - 3 \cos x . \]
\[4 \sin\left( \frac{\pi}{2} - \alpha \right) - 3 \cos\left( \frac{\pi}{2} - \alpha \right) . . . {\text{ From eq }(1)} \]
\[ = 4\cos \alpha - 3\sin\alpha \]
\[ = 4 \times \frac{3}{5} - 3 \times \frac{4}{5} \left( \because \cos \alpha = \frac{3}{5}\text{ and }\sin \alpha = \frac{4}{5} \right)\]
\[ = 0\]