trigonometry math

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show that cos3B+cosB=2(cos2B)cosb.

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We have to show that cos 3B + cos B = 2(cos 2B)* cos B

We use the relations: cos 2B = 2*(cos B)^2 - 1 and cos 3B = 4(cos B)^3 - 3cos B

We start with the left hand side

cos 3B + cos B

=> 4(cos B)^3 - 3cos B + cos B

=> 4(cos B)^3 - 2*cos B

=> 2* cos B ( 2* ( cos b)^2 - 1)

=> 2 * cos B * cos 2B

which is the right hand side.

The required relation cos 3B + cos B = 2(cos 2B)* cos B is proved

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