Can you solve this (CosA CosB – SinA SinB)CosB + (SinA CosB + CosASinB)SinB =CosA using identities?

Answer 1

Yes.

Working just with the left side:

First, multiply through the parentheses. You then have:

cos A cos^2 B - sin A sin B cos B
+ sin A sin B cos B + cos A sin^2 B

notice the middle two terms cancel each other out, so you're left with:

cos A cos^2 B + cos A sin^2 B

factor out the cos A

cos A (cos^2 B + sin^2 B)

remember the identity cos^2 B + sin^2 B = 1,
so you have:

cos A (1) = cos A

Answer 2

Sure. LHS = cos(A+B)cosB + sin(A+B)sinB = cos ((A+B)-B) = cosA

Answer 3

(cosA cosB-sinA sinB)cosB+(sinAcosB+cosA sin B)sinB
=cos(A+B)cosB+sin(A+B)sinB
=cos(A+B-B)
=cosA
because cosA cosB-sinAsinB=cos(A+B)
sinAcosB+cosAsinB=sin(A+B)
cosAcosB+sinAsinB=sin(A-B)

Answer 4

yes

Using

Cos(A+B) = Cos(A)Cos(B) - Sin(A)Sin(B)
and
Sin(A+B) = Sin(A)Cos(B) + Cos(A)Sin(B)

Cos(A+B)Cos(B) + Sin(A+B)Sin(B) = Cos((A+B)-B) = Cos(A)

Cos(A) = Cos(A)

(oops, got the sign backward the first time around)

The tricky ones are the Cos(A+B) and Cos(A-B) identities, where the signs are reversed:

Cos(A+B) = Cos(A)Cos(B) MINUS Sin(A)Sin(B)
Cos(A-B) = Cos(A)Cos(B) PLUS Sin(A)Sin(B)

In the case of Sines, it is the order that could be tricky:

Sin(A+B) = Sin(A)Cos(B) + Cos(A)Sin(B)
Sin(A-B) = Sin(A)Cos(B) - Cos(A)Sin(B)

Cos is CosCos SinSin, change the sign
Sin is SinCos CosSin, same sign