Another identity proof.

**JennyFlowers** · May 5th 2010, 11:09 AM

One more, guys.

$sin\hspace{5pt}2t - cot\hspace{5pt} t = - cot\hspace{5pt}t\hspace{5pt}cos\hspace{5pt}2t$

On this one, I made a bit of progress, I think.

Starting from the RHS:

$- \frac{cos t}{sin t} * (2 cos^2 t - 1)$

I think that might be the right direction to start in, but I can't see where it goes from here.

Thanks again!

**e^(i*pi)** · May 5th 2010, 11:30 AM

Originally Posted by JennyFlowers

One more, guys.

$sin\hspace{5pt}2t - cot\hspace{5pt} t = - cot\hspace{5pt}t\hspace{5pt}cos\hspace{5pt}2t$

On this one, I made a bit of progress, I think.

Starting from the RHS:

$- \frac{cos t}{sin t} * (2 cos^2 t - 1)$

I think that might be the right direction to start in, but I can't see where it goes from here.

Thanks again!

As you have $\sin(2t)$ on the LHS you are looking to have $2\sin(x)\cos(x)$ in the numerator. Therefore it is easier if you use the equivalent expression for cos(2x) in terms of sin which is $\cos(2x) = 1-2\sin^2(x)$ instead.

Spoiler:

**JennyFlowers** · May 5th 2010, 11:49 AM

Thank you, that did the trick!

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