Question

Rewrite with only sin x and cos x.

sin 3x - cos x

Answer 1

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

Answer 2

These are my options:
2 sin x - sin3x - cos x

2 sin x cos2x + sin x - 2 sin3x - cos x

2 sin3x cos4x + 1

3 sin x cos x - sin3x - cos x

Answer 3

my method would be breaking 3x in to 2x + x

Answer 4

Actually my options are:
2 sin x - sin^3x - cos x

2 sin x cos^2x + sin x - 2 sin^3x - cos x

2 sin3x cos^4x + 1

3 sin x cos x - sin^3x - cos x

Answer 5

yea - i got that wrong - not my day today

Answer 6

Can someone actually explain then? Please and thank you :)

Answer 7

i'll think i'll come back tomorrow lol

Answer 8

sin 3x - cos x

for now, im going to ignore the - cos x

Answer 9

sin 3x = sin ( 2x +x)

sin(A+B)=sin A cos B + cos A sin B
in this case, A = 2x , B = x

Answer 10

sin ( 2x + x) = sin 2x cos x + cos 2x sin x

Answer 11

then double angle identities

Answer 12

That's what I have but then what do we do with the - cos x

Answer 13

Oh okay

Answer 14

throw the -cos x onto the entire thing?

Answer 15

Do we use the sine double angle identity?

Answer 16

yes because we have a sin 2x

we also have a cos 2x that we need to get rid of so we need to use the cosine double angle identity

Answer 17

Use \(\cos 2x = 1-2\sin^2x\) and it comes out nicely from where you've started...

Answer 18

So it would come out to 2 sin^3x cos^4x + 1?

Answer 19

@whpalmer4

Answer 20

@completeidiot

Answer 21

\[\sin 3x - \cos x = \sin(2x+x) -\cos x= \sin 2x\cos x + \cos 2x\sin x - \cos x\]\[\sin 2x\cos x + \cos 2x\sin x - \cos x= (2\sin x \cos x)cos x+\cos 2x\sin x - \cos x\]\[=(2\sin x \cos x)\cos x+(1-2\sin^2x)\sin x - \cos x\]\[= 2\sin x\cos^2x + \sin x - 2\sin^3x - \cos x\]

Answer 22

I like that the \(\cos 2x\) identity gives you 3 choices, so you can head in 3 different directions if needed. I dislike that the \(\cos 2x\) identity gives 3 you choices, because you need to know which direction to go in :-)

Answer 23

(I tried the other 2 first!)