Question

Rewrite with only sin x and cos x.

cos 3x

Answer 1

cos(3x) = cos(x+x+x) = cos(x+2x)
= cos(x)*cos(2x) - sin(x)*sin(2x)
= cos(x)*(cos(x)*cos(x) - sin(x)*sin(x)) - sin(x)*(sin(x)*cos(x)+cos(x)*sin(x))
=cos^3(x) - sin^2(x)*cos(x) - sin^2(x)*cos(x) - sin^2(x)*cos(x)
= cos^3(x) - 3*sin^2(x)*cos(x)

Answer 2

um the answer choices are
cos x - 4 cos x sin2x

-sin3x + 2 sin x cos x

-sin2x + 2 sin x cos x

2 sin2x cos x - 2 sin x cos x

Answer 3

i assumed right

Answer 4

cos x - 4 cos x sin^2x

-sin^3x + 2 sin x cos x

-sin^2x + 2 sin x cos x

2 sin^2x cos x - 2 sin x cos x

Answer 5

@DUDE..IM..A..DUCK

Then how do I do it?
btw don't just give the answer

Answer 6

mind if i help?

Answer 7

sure

Answer 8

\[\cos(3x)=\cos(x+2x)\]\[\cos(x+2x) = (cosx)(\cos2x) - (sinx)(\sin2x)\]\[=(cosx)(\cos^{2}x-\sin^{2}x) - sinx(2sinxcosx)\]\[=cosx(\cos^{2}x-\sin^{2}x-2\sin^{2}x)\]

Answer 9

do you follow me up till here?

Answer 10

so far yes

Answer 11

since\[\cos^{2}x = 1-\sin^{2}x\]\[cosx(1-\sin^{2}x-\sin^{2}x-2\sin^{2}x)\]\[cosx(1-4\sin^{2}x)\]\[cosx-4cosxsin^{2}x\]

Answer 12

and you are done...

Answer 13

ask me any questions if you dont get something

Answer 14

wow thanks
but how did you get this or do this part?
=(cosx)(cos^2x−sin^2x)−sinx(2sinxcosx)
=cosx(cos^2x−sin^2x−2sin^2x)

Answer 15

I am just taking out cosx as the factor

Answer 16

i will do that part in more detail... please wait

Answer 17

ok no problem

Answer 18

\[cosx(\cos^{2}x-\sin^{2}x)-sinx(2sinxcosx)\]\[cosx(\cos^{2}x-\sin^{2}x) - 2\sin^{2}xcosx\]\[cosx(\cos^{2}x-\sin^{2}x-2\sin^{2}x)\]

Answer 19

do you see now?

Answer 20

okay but real quick what happend to the xcosx?

Answer 21

there is no xcosx... I think you are mistaking (sinx)(cosx), normally writtenas sinxcosx with sin xcosx...

Answer 22

oh okay got it! haha thanks

Answer 23

haha... have fun and good luck with similar questions...

Answer 24

thanks