# Trig identities help

• Nov 9th 2008, 06:41 PM
cokeclassic
Trig identities help
Hi i have question

tanx - cotx / tanx + cotx + 1 = 2sinx

so...
sin/cos - cos/sin / sin/cos + cos/sin

then mutlple fracs and get

sin^2x - cos ^2x/ sinxcosx / sin^2x + cos^2x/ sinxcosx

and then what do i do?
• Nov 9th 2008, 06:58 PM
Chop Suey
Is this your question? $\frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}} + 1 = 2\sin{x}$

I'm going to assume that the left hand side is written correctly, but I'm going to ask you to recheck the right hand side.

After this step:
$\frac{\frac{\sin{x}}{\cos{x}} - \frac{\cos{x}}{\sin{x}}}{\frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}} \cdot \frac{\sin{x}\cos{x}}{\sin{x}\cos{x}}$

We get:
$\frac{\sin^2{x}-\cos^2{x}}{\sin^2{x}+\cos^2{x}} + 1$

Recall that $\sin^2{x}+\cos^2{x} = 1$

Therefore:

$= \sin^2{x}-\cos^2{x}+1 = 2\sin^2{x}$
• Nov 9th 2008, 07:07 PM
cokeclassic
Quote:

Originally Posted by Chop Suey
Is this your question? $\frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}} + 1 = 2\sin{x}$

I'm going to assume that the left hand side is written correctly, but I'm going to ask you to recheck the right hand side.

After this step:
$\frac{\frac{\sin{x}}{\cos{x}} - \frac{\cos{x}}{\sin{x}}}{\frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}} \cdot \frac{\sin{x}\cos{x}}{\sin{x}\cos{x}}$

We get:
$\frac{\sin^2{x}-\cos^2{x}}{\sin^2{x}+\cos^2{x}} + 1$

Recall that $\sin^2{x}+\cos^2{x} = 1$

Therefore:

$= \sin^2{x}-\cos^2{x}+1 = 2\sin^2{x}$

Sorry it is 2sin^2x, but i still dont understand the steps
i see how you got
$\frac{\sin^2{x}-\cos^2{x}}{\sin^2{x}+\cos^2{x}} + 1$
but after that how did you end up with 2sin^2? how do you divide that :S
• Nov 9th 2008, 07:20 PM
Chop Suey
Quote:

Originally Posted by cokeclassic
how do you divide that :S

I didn't. I used the Pythagorus identity $\sin^2{x}+\cos^2{x} = 1$

Then notice that $1-\cos^2{x} = \sin^2{x}$
• Nov 9th 2008, 07:28 PM
cokeclassic
Ok I see.... thanks...

so basically

sin^2 + cos^2/ sin^2 + cos^2 is the same as
sin^2 + cost ^2 since it equal 1
• Nov 9th 2008, 07:32 PM
Chop Suey
$\frac{\sin^2{x} + \cos^2{x}}{\sin^2{x} + \cos^2{x}} = 1$

But I think you might have made a typo and meant instead:
$\frac{\sin^2{x} - \cos^2{x}}{\sin^2{x} + \cos^2{x}}$

In that case, yes, it equals $\sin^2{x} - \cos^2{x}$ since $\sin^2{x} + \cos^2{x} = 1$
• Nov 9th 2008, 07:39 PM
cokeclassic
Er yeah, ty