trig identities help!?

Question

make LHS = RHS
cos(x+y)cosy + sin(x+y)siny = cosx

I expanded both cos(x+y) and sin(x+y) then expanded again, and that's where I'm stuck, was I even suppose to expand again?

Please provide a full solution, thanks!

Answer 1

To solve this you need the sum/difference identities and the Pythagorean trig. identities.

cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
sin^2(y) + cos^2(y) = 1

So, cos(x+y)cos(y) + sin(x+y)sin(y)
= cos(y)[cos(x)cos(y) - sin(x)sin(y)] + sin(y)[sin(x)cos(y) + cos(x)sin(y)]
= cos(x) cos^2(y) - sin(x)sin(y)cos(y) + sin(x)sin(y)cos(y) + cos(x)sin^2(y)
= cos(x)[sin^2(y)+cos^2(y)] = cos(x)

Hope this helps!

Best of luck,

~Angel

Answer 2

LHS
cos y (cosx cosy - sin x sin y)
sin y (sin x cos y + cos x sin y)
Sum
cos x cos ² y + cos x sin ² y
cos x (cos ² y + sin ² y)
cos x

RHS
cos x

LHS = RHS

Answer 3

Use this FORMULA
cos (a - b) = cos a cos b + sin a sin b

cos (x + y) cos y + sin (x + y) sin y = cos x

LHS
= cos (x + y) cos y + sin (x + y) sin y
= cos (x + y - y)
= cos x
= RHS

Answer 4

cos (x+y) cos y + sin(x+y) sin y
= [ cos x cos y - sin x sin y ] cos y + [ sin x cos y + cos x sin y ] sin y
= cos x cos y cos y - sin x sin y cos y + sin x cos y sin y + cos x sin y sin y
= cos x cos² y + cos x sin² y
= cos x [ cos² y + sin² y ]
= cos x [ 1 ]
= cos x
:)