Help me solve this trig identity please (SOLVED)

[sandman369]

Prove: [2cot(?/2 - x)] / [1 - tan[sup:lo8139y3]2[/sup:lo8139y3]x] = tan(2x)

I tried:

Left Side
= 2 / [tan(?/2 - x)][1 - tan[sup:lo8139y3]2[/sup:lo8139y3]x]
= 2 / ([tan(?/2) - tanx] / [1 + tan(?/2)tanx])(1 - tan[sup:lo8139y3]2[/sup:lo8139y3]x) ---> because tan(A - B) = (tanA - tanB) / (1 + tanAtanB)

but then I don't know what to do... tan(?/2) is undefined, and I don't know how to get rid of any tan(?/2) or denominators of cos(?/2).

 

[Loren]

Re: Help me solve this trig identity please

cot(?/2 - x) = tan x

Then convert everything to sine and cosine. You will find that you soon end up with (sin 2x)/(cos 2x), so back to tan you go.

 

[sandman369]

Re: Help me solve this trig identity please

I don't understand how you got to that step??

 

[Loren]

Re: Help me solve this trig identity please

\(\displaystyle \frac{2\cot (\frac{\pi}{2}-x)}{1-\tan^2x}=\tan 2x\)

Here's your left side first step.

\(\displaystyle \frac{2\tan x}{1-\tan^2x}=?\)

From here, change to sines and cosines and go from there. If you get stuck, show us your steps so we can see if you are on the right track.

 

[sandman369]

Re: Help me solve this trig identity please

Loren, from your first step it looks like you're saying that \(\displaystyle cot(\frac{\pi}{2} - x)\) is equal to \(\displaystyle tanx\) ? Is this a standard identity, or am I missing something? Very confused sorry

edit: Nevermind, after looking through the identities on this site, I found that it is indeed the case... funny, our teacher never even mentioned that property throughout the entire course, yet decides to put this question on our EXAM. Nice. I will give it a go now.


(final)edit: Ok I went through and solved it using that special identity and got it. Then I thought, hmm maybe you can do it without knowing that identity, otherwise I will have to complain to the teacher lol ... So I went through and solved it by converting cot to cos/sin in the first step, and used the sum/difference identities to weed out the cos(pi/2) which gives 0, and it all worked out to the same conclusion.

So in final conclusion, on the exam I stupidly didn't even think to change cot into cos/sin, I kept using 1/tan and getting stuck, but now I figured it out (too late to get the marks though, doh!)

Thanks Loren!

 

[Deleted member 4993]

sin(?/2 - ?) = cos(?) ? provable through expansion of sin(A+B) = sin(A)*cos(B) + sin(B)*cos(A)

and

cos(?/2 - ?) = sin(?) ? provable through expansion of cos(A+B) = cos(A)*cos(B) - sin(B)*sin(A)

from there you can prove:

cot(?/2-?) = [cos(?/2 - ?)]/[sin(?/2 - ?) ] = sin(?)/cos(?) = tan(?)

These are corrollary to the fundamental theorem of angle addition - and all of these cannot be discussed in the class. You are expected to prove these on your own.

However, if similar problems have not been discussed in the class - this will not be a good problem for test (excellent problem for homework though).

As a "thumb-rule" - whenever you are stuck in a tan-cot-sec-cosec mess, try to convert it to a sin-cos mess (sometimes messier) but most of the time you can fight through that.