There's this last trig identity... Myself and a friend on the other end of computer keep getting a final answer of 2 ... simply 2. But the Answers say csc^2x

(sin(x-[pi]/2))/(cos([pi]-x))+(tan(x-3[pi]/2))/(-tan([pi]+x))

Have a go at it! Thanks

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- Nov 27th 2008, 01:11 PMmike_302Trig Identity Help
There's this last trig identity... Myself and a friend on the other end of computer keep getting a final answer of 2 ... simply 2. But the Answers say csc^2x

(sin(x-[pi]/2))/(cos([pi]-x))+(tan(x-3[pi]/2))/(-tan([pi]+x))

Have a go at it! Thanks - Nov 27th 2008, 01:41 PMMath_Helper
Give 5 minutes and I will solve for you

- Nov 27th 2008, 01:49 PMMath_Helper
- Nov 27th 2008, 02:02 PMmike_302
Okay, I knew it. This is where I start hating trig identities.... at the step where you get 1+1/tanx/tanx ..... 1/tanx/tanx should equal 1 , no?

so 1+1 = 2.... That's how I did it anyways. - Nov 27th 2008, 02:06 PMMath_Helper
1/tanx/tanx is not 1

- Nov 27th 2008, 02:07 PMMath_Helper
you dont have 1/tanx * tanx, in this situation is 1

- Nov 27th 2008, 02:10 PMmike_302
I see it as 1/a/a ... a/a is 1 .. 1/1 is 1 . It's not that I don't believe you. I absolutly respect your answer and all, but I just need MY idea disproven in order to understand what is going on.

Thanks for the patience so far :P - Nov 27th 2008, 02:14 PMMath_Helper
- Nov 27th 2008, 02:17 PMmike_302
Oh wow. Haha, thanks very much. I see now. That was silly of me. BEDMAS TO THE RESCUE.... If it was 1/(a/a), then we would have = 1 , correct.

I was starting to think of m/s^2 for acceleration and its relation to m/s/s so I see now.

Thanks again! - Nov 27th 2008, 02:22 PMMath_Helper
Good luck!

- Nov 27th 2008, 02:25 PMmike_302
It was one of those slip up moments.

Anyways, I got a couple steps further into the solution but after that, it's getting clear that (this is the last question on the homework) this stuff might be one class ahead of us because we haven't reviewed any of the other trig ratios other than sin(-x)=-sin(x) (those 3 ) and sin(pi/2-x)=cos(x) (those 3)

Thanks though. I will haveto complete it after tomorrow's lesson