# 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**

We'll write the formula for cos 3B =4(cos B)^3 - 3cos B and cos 2B = 2(cos B)^2 - 1

4(cos B)^3 - 3cos B + cos B = 2[2(cos B)^2 - 1]cosB

We'll combine like terms from the left side:

4(cos B)^3 - 2cos B = 2[2(cos B)^2 - 1]cosB

We'll remove the brackets from the right side:

**4(cos B)^3 - 2cos B = 4(cos B)^3 - 2cos B q.e.d.**