## Negative Angle Identities

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
eliotmason
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Joined: Tue Oct 29, 2013 5:36 am
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### Negative Angle Identities

So I don't know if I completely understand this Identity. I am working on my homework and this identity is really bothering me and I can't figure it out:

Write the expression in terms of sine and cosine, and simplify so that no quotients appear in the final expression and all expressions are of $\theta$ only.

$\frac{1-\cos ^{2}(-\theta) }{1+\tan^{2}(-\theta ) }$

*So I know that the negative identity for cos is : $\cos (-\theta )=\cos (\theta )$

- so that will make the top become:

$\frac{1-\cos ^{2}(\theta )}{1+\tan ^{2}(-\theta )}$

then using the pythagorian identity on top it becomes....

$\frac{\sin^{2}(\Theta )}{1+\tan^{2}(-\Theta ) }$

now is where I get lost.... the answer is supposed to be: ${\color{Red}\sin^{2}\theta \cos ^{2}\theta }$

*I dont understand how I use the Negative-Angle Identity to make the bottom to be used with the Pythagorean Identities to get to the answer. Thank you in advance for your time and help.

buddy
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Joined: Sun Feb 22, 2009 10:05 pm
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### Re: Negative Angle Identities

They've got a list of trig idents here. It shows tan(-@) = -tan(@). Since its squared, you get 1+tan^2(@). This is sec^2(@) which is 1/cos^2(@). That matches what they got.

eliotmason
Posts: 3
Joined: Tue Oct 29, 2013 5:36 am
Contact:

### Re: Negative Angle Identities

They've got a list of trig idents . It shows tan(-@) = -tan(@). Since its squared, you get 1+tan^2(@). This is sec^2(@) which is 1/cos^2(@). That matches what they got.
That makes perfect sense! Thank you so much for your help!