Question

Prove :- 4sinxsin(x+pi/3)sin(x+2pi/3) = sin 3x.

Answer 1

I will give you answer after wards

Answer 2

Prove that $4sinxsin(x+\frac{\pi}{3} )sin(x+\frac{2\pi}{3} )=sin3x$

Explanation:  

• we have a trigonometric identity of sine of sum of two angles as , $sin(A+B)=sinAcosB+cosAsinB$
• using this we get,          
•             $sin(x+\frac{\pi}{3} )=sinxcos\frac{\pi}{3} +cosxsin\frac{\pi}{3} \\sin(x+\frac{\pi}{3})=\frac{sinx}{2}+\frac{\sqrt{3} cosx}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -----(a)$  
• similarly we get,      
•             $sin(x+\frac{2\pi}{3} )=sinxcos\frac{2\pi}{3} +cosxsin\frac{2\pi}{3} \\sin(x+\frac{2\pi}{3})=-\frac{sinx}{2}+\frac{\sqrt{3} cosx}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -----(b)$  
• Putting (a) and (b) in the main function we get,
•               $L.H.S=4sinx(\frac{\sqrt{3}cosx +sinx}{2} )(\frac{\sqrt{3}cosx -sinx}{2} ) \\L.H.S=sinx(3cos^2x-sin^2x)\\L.H.S=sinx(3(1-sin^2x)-sin^2x)\\L.H.S=sinx(3-4sin^2x)\\L.H.S=3sinx-4sin^3x$---------(i)
• Applying the sum of angles identity to R.H.S of the main function we get,
• $R.H.S=sin(3x)=sin(2x+x)\\R.H.S=sin2xcosx+cos2xsinx\\R.H.S=(2sinxcosx)cosx+(1-2sin^2x)sinx\\R.H.S=sinx(2cos^2x+1)-2sin^3x\\R.H.S=sinx(2-2sin^2x+1)-2sin^3x\\R.H.S=3sinx-4sin^3x$------(ii)
• From (i) and (ii) we get that RHS and LHS are equal. Hence, proved.