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$sin(11x)+sin(9x) = (sin(10x)cos(x)+cos(10x)sin(x))+(sin(10x)cos(x)−cos(10x)sin(x)) = 2sin(10x)cos(x) $

I dont understand how you go from the left hand side $sin(11x) + sin(9x)$ to the right hand side.

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Using $$\displaystyle \bullet \; \sin(A+B) = \sin A\cos B+\cos A\sin B$$

And $$\displaystyle \bullet \; \sin(A-B) = \sin A\cos B-\cos A\sin B$$

So $$\sin(11x) = \sin (10x+x) =\sin10x \cos x+\cos 10 x\sin x.......(1)$$

and $$\sin (9x) = \sin (10x-x) =\sin10x \cos x-\cos 10 x\sin x.........(2)$$

Now Add these two equation, We get

$$\sin (11x)+\sin (9x) = 2\sin 10x\cos x$$

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$$\sin { 11x } =\sin { \left( 10x+x \right) =\sin { 10x\cos { x } +\cos { 10x\sin { x } } } } \\ $$ $$\sin { 9x } =\sin { \left( 10x-x \right) =\sin { 10x\cos { x } -\cos { 10x\sin { x } } } } \\ $$

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