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Old November 1st, 2009, 05:21 PM
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Default 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!
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  #2  
Old November 1st, 2009, 07:13 PM
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Originally Posted by lziz View Post
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. -\frac{1}{2\sec^2{x}} or -\frac{1}{2}\sec^2{x}?


Is Q2. \frac{1}{2\sin{(4x)}} or \frac{1}{2}\sin{(4x)}?
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Old November 1st, 2009, 07:54 PM
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The latter for both questions. Also, aren't they the same thing anyway...?
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Old November 1st, 2009, 08:04 PM
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The latter for both questions. Also, aren't they the same thing anyway...?
No they aren't.

The latter are the equivalent of

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

and

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