Question

$5\sin x+4\cos x=3\Rightarrow 4\sin x-5\cos x=$

• A. 4
• B. $4\sqrt{2}$
• C. $3\sqrt{2}$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$5\sin x+4\cos x=3$
Squaring both sides $25{\sin }^{2}x+16{\cos }^{2}x+2\ast 5\ast 4\sin x\cos x=9\Rightarrow 9{\sin }^{2}x+7+40\sin x\cos x=0$
$\Rightarrow 9{\sin }^{2}x+7=-40\sin x\cos x$
Now, $4\sin x-5\cos x=\sqrt{(4\sin x-5\cos x{)}^2}$
$=\sqrt{16{\sin }^{2}x+25{\cos }^{2}x-40\sin x\cos x}$
$=\sqrt{16{\sin }^{2}x+25{\cos }^{2}x+9{\sin }^{2}x+7}$
$=\sqrt{25{\sin }^{2}x+25{\cos }^{2}x+7}$
$=\sqrt{25+7}=\sqrt{32}=$
$4\sqrt{2}$