Been 20 years since I've identified trig identities. We are stuck on a particular type and would appreciate some explanation. 1/sin(-x) * (1-cos^2x) that's cos(squared x) and 1+1/cos(-x) / -sinx-tanx

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- May 19th 2008, 07:05 PMsingleton2787trigonometric identity
Been 20 years since I've identified trig identities. We are stuck on a particular type and would appreciate some explanation. 1/sin(-x) * (1-cos^2x) that's cos(squared x) and 1+1/cos(-x) / -sinx-tanx

- May 19th 2008, 07:10 PMMathstud28
- May 19th 2008, 07:23 PMsingleton2787problem re-written
Problem 1 (simplify) 1/sin(-x) * (1-cos^2x)

- May 19th 2008, 07:28 PMMathstud28
- May 19th 2008, 07:35 PMsingleton2787i dont understand..
how does $\displaystyle 1-\cos^2(x)=\sin^2(x)$?

- May 19th 2008, 07:40 PMMathstud28
Because $\displaystyle \cos^2(x)+\sin^2(x)=1$

This can be derived multiple ways...the most intuitive being

On the unit circle which has a radius of one any of the points on the circle have coordinates

$\displaystyle (\cos(x),\sin(x))$

So the distance from the origin to any point on the circle which I stated earlier is 1

So setting up the distance equation we have

$\displaystyle \sqrt{(0-\cos(x))^2+(0-\sin(x))^2}=1^2$

Simplifying we get

$\displaystyle \sqrt{\cos^2(x)+\sin^2(x)}=1^2\Rightarrow{\cos^2(x )+\sin^2(x)=1}$

Usually in trig classes this indentity is just taken to be true - May 19th 2008, 07:53 PMsingleton2787second problem
thanks for your help can you explain this one? $\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}$

- May 19th 2008, 08:18 PMReckoner
First, convert everything to sines and cosines using the identities

$\displaystyle \sec x = \frac1{\cos x}$

$\displaystyle \tan x = \frac{\sin x}{\cos x}$

Then, combine the fractions in the numerator and denominator, reduce, factor and cancel, and you should be left with $\displaystyle -\csc x$ if you do it right. When working this one, it may be helpful to observe that, since sine is an odd function and cosine is even,

$\displaystyle \sin\left(-x\right) = -\sin x$

$\displaystyle \cos\left(-x\right) = \cos x$

If you have difficulty with the simplification, let us know.