Question

verify the following identity


sin2t-cot t=-cot t*cos2t

Answer 1

just a quick question... is it


1.
sin(2t)

or

2.

\[\sin^2(t)\]

Answer 2

@campbell_st
Sin(2t)

Answer 3

option 1

Answer 4

ok... start with the left hand side

\[\sin(2t) = 2\sin(t) \cos(t)\]

and

\[\cot(t) = \frac{\cos(t)}{\sin(t)}\]

so you have

\[2\sin(t)\cos(t) - \frac{\cos(t)}{\sin(t)}\]

you noe need a common demoninator so multiply the 1st term by sin(t)

and you get

\[\frac{2\sin^2(t) \cos(t) - \cos(t)}{\sin(t)} = \frac{\cos(t)(2\sin(t) - 1)}{\sin(t)}\]

which is

\[\cot(t) \times (2\sin^2(t) - 1)\]

now look at the different forms of

\[\cos(2t) \]

and I think you'll find there is one where

\[\cos(2t) = 1 - 2\sin^2(t) \]

so reversing the order means you'll get

-cos(2t)

and hopefully its obvious from here