# Thread: Verifying Triogonomic Identies (Double Angle)

1. ## Verifying Triogonomic Identies (Double Angle)

Hi All,

I am stuck on these two problems. Basically, I understand the double angle identities, but not when sin or cos is square (i.e. sin^2(2x), etc)

Here are the problems...

1. -1/2(sec^2(x) = -2sin^2(x)csc^2(2x)

2. 1/2(sin(4x)) = 2sinxcosx - 4sin(sin^3(x))cos(x)

If my notation is screwy let me know...

Any help is appreciated. Thanks!

2. Originally Posted by lziz
Hi All,

I am stuck on these two problems. Basically, I understand the double angle identities, but not when sin or cos is square (i.e. sin^2(2x), etc)

Here are the problems...

1. -1/2(sec^2(x) = -2sin^2(x)csc^2(2x)

2. 1/2(sin(4x)) = 2sinxcosx - 4sin(sin^3(x))cos(x)

If my notation is screwy let me know...

Any help is appreciated. Thanks!
Is Q1. $\displaystyle -\frac{1}{2\sec^2{x}}$ or $\displaystyle -\frac{1}{2}\sec^2{x}$?

Is Q2. $\displaystyle \frac{1}{2\sin{(4x)}}$ or $\displaystyle \frac{1}{2}\sin{(4x)}$?

3. The latter for both questions. Also, aren't they the same thing anyway...?

4. Originally Posted by lziz
The latter for both questions. Also, aren't they the same thing anyway...?
No they aren't.

The latter are the equivalent of

$\displaystyle -\frac{\sec^2{x}}{2}$

and

$\displaystyle \frac{\sin{(4x)}}{2}$.