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Q)

Sin3x.cosx.cosx.cosx+cos3x.sinx.sinx.sinx

Sin3x.cosx.cosx.cosx+cos3x.sinx.sinx.sinx

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A)
Cos³ x . Sin 3x + Sin³x .  Cos3x

 

Now, We know,

Cos 3 θ = 4 Cos³θ - 3 Cos θ.

∴ 4 Cos³θ  = Cos 3 θ +  3 Cos θ.

∴ Cos³θ  = ( Cos 3 θ +  3 Cos θ)/4

 

∴ Cos³ x . Sin 3x =  (Cos 3 x +  3 Cos x)/4 . Sin 3x

    

Also,  

Sin 3θ = 3 Sin θ - 4 Sin³θ

∴ 4 Sin³θ = 3 Sin θ - Sin 3θ

∴  Sin³θ = (3 Sin θ - Sin 3θ)/4

 

∴  Sin³x . Cos 3x  = (3 Sin x - Sin 3x)/4 . Cos 3x

 

 

∴ Cos³ x . Sin 3x + Sin³x . 3 Cos x  =  (Cos 3 x +  3 Cos x)/4 . Sin 3x + (3 Sin x - Sin 3x)/4 . Cos 3x

= 1/4[(Cos 3 x +  3 Cos x)Sin 3x + ( Sin 3x - Sin 3x) . Cos3x

= 1/4[Sin 3x.Cos 3x + Sin 3x. 3Cos x + 3Sin x.Cos 3 x  - Sin 3x  Cos 3x].

 

= 3/4[Sin 3x Cos x + Cos 3x Sinx]

(∵  SinA CosB + CosA SinB = Sin(A+B) )

 

= 3/4[Sin 4x]
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