Question

If $A+B+C=\frac{\pi }{2},$ then the value of $sin2A+sin2B+sin2C$ is

• A. $2sinAsinBsinC$
• B. $2cosAcosBcosC$
• C. $4sinAsinBsinC$
• D. $4cosAcos\text{ B }cosC$

Answer 1

Answer: (D)

Given, $A+B+C=\frac{\pi }{2}$
$sin2A+sin2B+sin2C$
$=(sin2A+sin2B)+sin2C$
$=2\sin \left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right)+\sin 2C$
$=2\sin (A+B)\cos (A-B)+\sin 2C$
$=2\cos C.\cos (A-B)+2\sin C\cos C$
$\left[\begin{array}{rl}& \because A+B+C=\pi /2\\ & A+B=\pi /2-C\end{array}\right]$
$=2\cos C[\cos (A-B)+\sin C]$
$=2\cos C[\cos (A-B)+\cos (A+B)$
$\left[\because C=\frac{\pi }{2}-(A+B)\right]$
$=2\cos C[2\cos A\cos B]$
$=4\cos A\cos B\cos C$