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

Solution

Given,

\[ 3\sin x+4\cos x=5 \]

Now square both sides:

\[ (3\sin x+4\cos x)^2=5^2 \]

\[ 9\sin^2x+16\cos^2x+24\sin x\cos x=25 \]

Using

\[ \sin^2x+\cos^2x=1 \]

we get

\[ 9\sin^2x+16(1-\sin^2x)+24\sin x\cos x=25 \]

But an easier method is to observe that

\[ 3^2+4^2=5^2 \]

Hence,

\[ 3\sin x+4\cos x \]

attains its maximum value \(5\) only when

\[ \sin x=\frac{3}{5}, \qquad \cos x=\frac{4}{5} \]

Therefore,

\[ 4\sin x-3\cos x = 4\left(\frac{3}{5}\right) – 3\left(\frac{4}{5}\right) \]

\[ = \frac{12}{5}-\frac{12}{5} \]

\[ =0 \]

Final Answer

\[ \boxed{ 4\sin x-3\cos x=0 } \]

Correct Option: (a)