Find Sin3x, x in QII, given that Cscx = 5/3

[Timcago]

Cscx = (5/3), x terminates in quadrant II

Heres what I got so far:

Sinx=(3/5), Cosx=(4/5)

I know the forumla for sin2x, but sin3x???????

I was thinking that maybe i could add the sinx with the sin2x to get what sin3x equals, but that cannot be right, because by that logic i could do sinx + sinx to get sin2x which doesnt work, because adding the sinx plus sinx does give the same value as using the sin2x formula "2sinxcosx".

How do you do the sin3x?

 

[daon]

Re: Find Sin3x

Cscx=(5/3), x terminates in quadrant II

Heres what i got so far

Sinx=(3/5), Cosx=(4/5)

I know the forumla for sin2x, but sin3x???????

I was thinking that maybe i could add the sinx with the sin2x to get what sin3x equals, but that cannot be right, because by that logic i could do sinx + sinx to get sin2x which doesnt work, because adding the sinx plus sinx does give the same value as using the sin2x formula "2sinxcosx".

How do you do the sin3x?

Sin3x = sin(2(3x/2)) = 2sin(3x/2)cos(3x/2)

or...

Sin3x = Sin(x + 2x) = SinxCos2x + CosxSin2x

 

[Timcago]

I will use the second one, thank you for your help.