# trigonometric identity

• May 19th 2008, 07:05 PM
singleton2787
trigonometric 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 PM
Mathstud28
Quote:

Originally Posted by singleton2787
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

well the right side is indescernible bu

$\displaystyle \frac{1}{\sin(-x)}=\frac{-1}{\sin(x)}=\csc(x)$

and $\displaystyle 1-\cos^2(x)=\sin^2(x)$

So $\displaystyle \sin^2(x)\cdot\csc(x)=\sin(x)$

rewrite the other one in a more hospitable form
• May 19th 2008, 07:23 PM
singleton2787
problem re-written
Problem 1 (simplify) 1/sin(-x) * (1-cos^2x)
• May 19th 2008, 07:28 PM
Mathstud28
Quote:

Originally Posted by singleton2787
Problem 1 (simplify) 1/sin(-x) * (1-cos^2x)

I am very sorry I was unclear...the above work was the solution to that problem...I meant the other one...once again sorry
• May 19th 2008, 07:35 PM
singleton2787
i dont understand..
how does $\displaystyle 1-\cos^2(x)=\sin^2(x)$?
• May 19th 2008, 07:40 PM
Mathstud28
Quote:

Originally Posted by singleton2787
how does $\displaystyle 1-\cos^2(x)=\sin^2(x)$?

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 PM
singleton2787
second problem
thanks for your help can you explain this one? $\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}$
• May 19th 2008, 08:18 PM
Reckoner
Quote:

Originally Posted by singleton2787
thanks for your help can you explain this one? $\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}$

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.