# How do you prove #sin 3 theta = 3 sin theta - 4 sin^3 theta#?

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#### Explanation

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see explanation

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Express the left hand side as

#sin3theta=sin(theta+2theta)# now expand the right side of this equation using

#color(blue)"Addition formula"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))#

#rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)#

#color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))# The right hand side is expressed only in terms of

#sintheta's# so we use

#cos2theta=1-2sin^2theta........(1)#

#color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)# Replace

#cos2theta" and " sin2theta# by the expansions (1) and (2)

into (A)

#sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)# and expanding brackets gives.

#sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)#

#color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))# Replace

#cos^2theta=1-sin^2theta" into (B)"#

#rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)# and expanding 2nd bracket gives.

#sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta# Finally, collecting like terms.

#sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"#

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