Rewrite with only sin x and cos x. sin 3x - cos x

Question

Rewrite with only sin x and cos x.  sin 3x - cos x

jhannybean · Answer

hint: $\large \sin(3x)=\sin(2x+x)=\sin(2x)\cos(x)+\cos(2x)\sin(x)$

anonymous · Answer

Could you please explain further? I have a difficult time grasping trig concepts.

jhannybean · Answer

Well, ok. you have $$\large \sin(3x)-\cos(x)$$ and you can rewrite sin(3x) as the following, $$\large \sin(2x+x) = \sin(2x)\cos(x)+\cos(2x)\sin(x)$$ So $$\large \sin(3x)-\cos(x) = \sin(2x)\cos(x)+\cos(2x)\sin(x)-\cos(x)$$

anonymous · Answer

So would it be 3 sin x cos x - sin^3x - cos x?

jhannybean · Answer

Oh...hm..$$\large \sin(2x)\cos(x)+\cos(2x)\sin(x)-\cos(x)$$$$\large 2\sin(x)\cos(x)\cos(x)+(2\cos^{2}x-1)(\sin(x))-\cos(x)$$$$\large 2\sin(x)\cos^2(x)+2\cos^2(x)\sin(x)-\sin(x)-\cos(x)$$

jhannybean · Answer

Right? I'm going to need a little help here too... haha.

anonymous · Answer

I'm not sure. If it helps, the choices are: 

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

jhannybean · Answer

Bleh.

anonymous · Answer

Yeah, I'm really awful at these things. Thank you so much for your help, though!

jhannybean · Answer

I'll figure it out soon enough, lol.

jhannybean · Answer

bbs. dont leave the problem!

anonymous · Answer

So, out of the choices I listed about, which would it be? Thanks for your help!

zzr0ck3r · Answer

o there are choices lol , one sec

jhannybean · Answer

hahaha. wooow. good job zz

zzr0ck3r · Answer

um im confused, all the answers are not in terms of only cos(x) sin(x)

anonymous · Answer

How are they not?

zzr0ck3r · Answer

sin(3x) in the first one
cos(2x) in the second
sin(3x) in the third
sin(3x) in the fourth

zzr0ck3r · Answer

im confused I think...
is this sin(3x) or sin(x)^3?

anonymous · Answer

Choice A: 2 sin x - sin^3x - cos x

anonymous · Answer

Choice B: 2 sin x cos^2x + sin x - 2 sin^3x - cos x

unklerhaukus · Answer

$$\sin(3x)-\cos(x)\
=\sin(2x)\cos(x)+\cos(2x)\sin(x)-\cos(x)\
=\Big(\sin(x)\cos(x)+\cos(x)\sin(x)\Big)\cos(x)+\Big(1-2\sin^2(x)\Big)\sin(x)-\cos(x)\
=2\sin(x)\cos^2(x)+\sin(x)-2\sin^3(x)-\cos(x)$$

anonymous · Answer

Thank you so much, UnkleRhaukus!

anonymous · Answer

It's okay!!

unklerhaukus · Answer

i have used$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$ like @Jhannybean suggested
and$$\cos(2u)=1-2\sin^2(u)$$

jhannybean · Answer

such a hard question!!