Question

my final question for the night, prove that 2cos(x+pi/4)cos(x-pi/4)=cos2x

Answer 1

expand cos(x+pi/4) and cos(x-pi/4) using the identity \[\cos(a+b)=\cos a \cos b-\sin a\sin b\]

Answer 2

i got that far i wrote cos2x =cosx.cos45deg-sinx.sin45deg? what next

Answer 3

if 45deg does that mean all 45deg

Answer 4

please help me finish this question

Answer 5

sorry i lost connection

u mean cos(x+pi/4) =cosx.cos45deg-sinx.sin45deg ?

Answer 6

its double angle formula 2cosx+pi/4

Answer 7

make that just 2cs=

Answer 8

@waterineyes

Answer 9

what do u mean ? by 2cs=

Answer 10

2cosx

Answer 11

well let me tell u something :

cos(x+pi/4) =cosx.cos45deg-sinx.sin45deg right?

Answer 12

but it has to cancel from here to get cos2x?

Answer 13

this equation leaves 0

Answer 14

how do u get 0?
cos 45=sin 45 = 1/(sqrt 2)
this makes cos(x+pi/4) = (1/(sqrt 2))*(cos x -sin x)
similarly do it for cos(x-pi/4) ....

Answer 15

i don,'t know how i got it, if you know how to come up with the answer it would be good if you could show it

Answer 16

i wanna do this completely...then u tell us which part is confusing..

\[\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}\]so using the first identity i gave u\[\cos(x+\frac{\pi}{4})=\cos x \cos \frac{\pi}{4}-\sin x \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}(\cos x-\sin x)\]\[\cos(x-\frac{\pi}{4})=\cos x \cos \frac{\pi}{4}+\sin x \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}(\cos x+\sin x)\] so u will come up with\[2\cos(x-\frac{\pi}{4})\cos(x+\frac{\pi}{4})=2\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}(\cos x-\sin x)(\cos x+\sin x)\\ =(\cos x-\sin x)(\cos x+\sin x)=\cos^2 x-\sin^2 x=\cos 2x\]using the fact that\[(a-b)(a+b)=a^2-b^2\]