How to confirm this is true?

Question

I can only use:

reciprocal identities

quotient identities

Pythagorean identities

sum and difference formulas

double angle formulas

power reducing formulas

half angle formulas

product to sum formulas

sum to product formulas

Using the above, how do I prove: cos (x+y) * cos (x-y) = cos²(x) - sin²(y) (only manipulate one side)

After choosing the right side to work with, the farthest I got was = (cos x+sin y)(cos x- sin y)

Any help is appreciated!!

Answer 1

A fun modification of Arthur's

Starting from

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

(cosxcosy-sinxsiny)*(cosxcosy+sinxsiny)=

cos^2xcos^2y-sin^2xsin^2y

Note how this is now somewhat similar to the right side. We need to add an additional cos^2xsin^2y to get cos^2x; we need to subtract an additional cos^2xsin^2y to get -sin^2y; note that these are the same thing.

Thus, we just need to add and subtract cos^2xsin^2y

so we just needed to write =

cos^2xcos^2y-sin^2xsin^2y+cos^2xsin^2y-cos^2xsin^2y =

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

cos^2x - sin^2y

Answer 2

prove the first equation !

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

(cosxcosy-sinxsiny)*(cosxcosy+sinxsiny)=

cos^2xcos^2y-sin^2xsin^2y=

sin^2+cos^2=1 is a trig identity which you use

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

cos^2x-sin^2ycos^2x-sin^2y+sin^2ycos^2x=

the second and fourth terms cancel and you get...

cos^2x-sin^2y which is the right side of the original equation

Answer 3

(cos (x + y))(cos (x - y))

[cos x cos y - sin x sin y][cos x cos y + sin x sin y] Sum Formula

cos² x cos² y - sin² x sin² y Differnce of squares

cos² x (1 - sin² y) - (1 - cos² x)(sin² y) Pythagorean Identity

cos² x - sin²y cos² x - (sin² y - sin² y cos² x)

cos² x - sin²y cos² x - sin² y + sin² y cos² x

cos² x - sin² y

Q.E.D