Question

Prove cos(A+B)cos(A-B) = cos^2(A) - sin^2(B)

Answer 1

@zepdrix @ParthKohli @No.name

Answer 2

@ikram002p

Answer 3

ok , so expand !

Answer 4

cos (A+B)= ?
cos(A-b) = ?

Answer 5

RHS?
which way tho? i could expand two angles to singles or elimnate the product

Answer 6

so as in (cosAcosB-sinAsinB)(cosAcosB+sinAsinB)?

Answer 7

@ikram002p

Answer 8

\(\cos (A+B )=\cos A \cos B -\sin A \sin B\)
\(\cos (A-B )=\cos A \cos B +\sin A \sin B\)

thus
\((\cos (A+B ))(\cos (A-B ) )=(\cos A \cos B -\sin A \sin B)(\cos A \cos B +\sin A \sin B)\)

Answer 9

oh yeah @CrashOnce what u did is correct :)
now FOIL expand

Answer 10

(cosAcosB)^2-(sinAsinB)^2

Answer 11

correct !
since
(a+b)(a-b)=a^2-b^2

Answer 12

but that isnt cos^2A-sin^2B

Answer 13

hehe , looking for other identity :P

Answer 14

Alternatively, use the identity\[2\cos A \cos B = \cos(A - B) - \cos(A + B)\]In this case,\[\cos(A + B)\cos(A -B ) = \frac{1}{2}\left(\cos (2B)-\cos(2A)\right)\]\[= \frac{1}{2}\left(1 - 2\sin^2 B + \cos^2 A - 1 \right)\]\[= \cdots\]

Answer 15

Correction:\[= \frac{1}{2}\left(1 - 2\sin^2 B + 2\cos^2 A - 1 \right) \]

Answer 16

nice !