Thread: Another identity proof.

One more, guys.

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

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

Starting from the RHS:

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

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

Thanks again!

Originally Posted by JennyFlowers

One more, guys.

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

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

Starting from the RHS:

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

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

Spoiler:

$\displaystyle -\frac{\cos(t)}{\sin(t)} (1-2\sin^2(t)) = -\frac{\cos(t)}{\sin(t)} + \frac{2\sin^2(t)\cos(t)}{\sin(t)}$

That all cancels down to $\displaystyle 2\sin(t)\cos(t) - \cot(t) = \sin(2t) - \cot(t)$

Thank you, that did the trick!