Question

If $3sin\theta +5cos\theta =4$, then find the value of $5sin\theta -3cos\theta$.

• A. $\pm 3\sqrt{2}$
• B. $\pm 5\sqrt{2}$
• C. $\pm \sqrt{2}$
• D. $\pm 7\sqrt{2}$

Answer 1

Answer: (A)

Let $x=5sin\theta -3cos\theta$, then
$x^2={\left(5sin\theta -3cos\theta \right)}^2=25si{n}^2\theta +9co{s}^2\theta -30sin\theta cos\theta$
$=25\left(1-co{s}^2\theta \right)+9\left(1-si{n}^2\theta \right)-30sin\theta cos\theta$
$=25+9-\left(25co{s}^2\theta +9si{n}^2\theta +30sin\theta cos\theta \right)$
$=34-{\left(3sin\theta +5cos\theta \right)}^2$
$=34-4^2=18$
$\Rightarrow x=\pm \sqrt{18}$
$=\pm 3\sqrt{2}$.