Class 11MathematicsTrigonometric FunctionsTrigonometrical ratios of sum and difference of two and three angles
EasyProve that $\sin (n+1) x \sin (n+2) x + \cos (n+1) x \cos (n+2) x = \cos x$.
(N/A) We know the trigonometric identity: $\cos (A - B) = \cos A \cos B + \sin A \sin B$.
Let $A = (n+2)x$ and $B = (n+1)x$.
Then the expression becomes $\cos (n+2)x \cos (n+1)x + \sin (n+2)x \sin (n+1)x$.
Using the identity,this is equal to $\cos [(n+2)x - (n+1)x]$.
Simplifying the angle: $(n+2)x - (n+1)x = nx + 2x - nx - x = x$.
Therefore,the expression equals $\cos x$,which is the $R.H.S.$
Let $A = (n+2)x$ and $B = (n+1)x$.
Then the expression becomes $\cos (n+2)x \cos (n+1)x + \sin (n+2)x \sin (n+1)x$.
Using the identity,this is equal to $\cos [(n+2)x - (n+1)x]$.
Simplifying the angle: $(n+2)x - (n+1)x = nx + 2x - nx - x = x$.
Therefore,the expression equals $\cos x$,which is the $R.H.S.$
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