2sin^2(x) - 6sin(x) / sin(x) -3

I'm so confused as to how to simply the top part where its 2sin^2(x) - 6sin(x)

both terms have 2sinx in common, so factor it out.

it has nothing to do with trig identities afaik. just pretend sine is some other function f

[2*f(x)*f(x) - 6*f(x)] / [f(x) - 3]

This post was edited by carteblanche on Dec 10 2013 08:49pm

what he said

2sinx[ (sinx - 3)] / sinx - 3

beccomes

2sinx

o0o0o0o why didn't I see this ;( thank you guys!

What about if there is a constant on the end? I got stumped on this one...

Simplify
tan^2(x) + 5tan(x) -14 / tan^2(x) -8tan(x) +12

factor it dude. use u substitution if it helps you since these really have nothing to do with trig identities

let u = tan(x)

u^2 + 5u - 14
___________
u^2 - 8u + 12

now factor, simplify, and substitute back tan(x) for u

This post was edited by carteblanche on Dec 10 2013 11:57pm

perhaps if you would use proper notation it would be easier, ie [2*sin^2(x)-6*sin(x)] / [sin(x)-3]

let's try it with your "tan" question, first use the hint of 'cartblanche' and replace tan(x) with a (can restore after):

(a^2+5a-14) / (a^2-8a+12)

the first part looks suspiciously like (a+7) (a-2) doesn't it? and the second part like (a-6) (a-2) which leaves us with (a+7) / (a-6) or after restoring:

[tan(x)+7] / [tan(x)-6]

Thank you! I can't believe I'm struggling so hard with such simple things, I think I just need to take a step back and just look at the equation.

I did what you said and got [tan(x) -2][tan(x) +7] / [tan(x) -6][tan(x) -2]

Canceled out the tanx-2s and got...

tanx +7 / tanx -6

Seems right thank you thank you again

EDIT: Just refreshed and saw brmv's post, thank you! Nice to see I got it right

Jsp's homework help community is actually great

This post was edited by Jwoww on Dec 11 2013 12:11am