Consider sin3x=sin(2x+x)We know that sin(A+B)=sinAcosB+cosAsinBWhere A=2x, B=x∴sin(2x+x)=sin2xcosx+cos2xsinx0ex0exbut sin2x=2sinxcosx 0ex0excos2x=1 2sin^2x0ex0e ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

sin3x=

Question

$\sin 3 x =$

Open in App

Solution

Consider $\sin 3 x = \sin (2 x + x)$
We know that $\sin (A + B) = \sin A \cos B + \cos A \sin B$
Where $A = 2 x, B = x$
$∴ \sin (2 x + x) = \sin 2 x \cos x + \cos 2 x \sin x b u t \sin 2 x = 2 \sin x \cos x \cos 2 x = 1 - 2 \sin^{2} x \Rightarrow \sin (2 x + x) = (2 \sin x \cos x) \cos x + (1 - 2 \sin^{2} x) \sin x = 2 \sin x \cos^{2} x + \sin x - 2 \sin^{3} x = 2 \sin x (1 - \sin^{2} x) + \sin x - 2 \sin^{3} x = 2 \sin x - 2 \sin^{3} x + \sin x - 2 \sin^{3} x = 3 \sin x - 4 \sin^{3} x$
Hence, $\sin 3 x = 3 \sin x - 4 \sin^{3} x$

Suggest Corrections

Join BYJU'S Learning Program

sin3x=

Consider sin3x=sin2x+xWe know that sinA+B=sinAcosB+cosAsinBWhere A=2x,B=x∴sin2x+x=sin2xcosx+cos2xsinxbutsin2x=2sinxcosxcos2x=1-2sin2x⇒sin2x+x=2sinxcosxcosx+1-2sin2xsinx=2sinxcos2x+sinx-2sin3x=2sinx1-sin2x+sinx-2sin3x=2sinx-2sin3x+sinx-2sin3x=3sinx-4sin3xHence, sin3x=3sinx-4sin3x

$\sin 3 x =$