trigonometric identity

**singleton2787** · May 19th 2008, 07:05 PM

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

**Mathstud28** · May 19th 2008, 07:10 PM

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

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

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

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

rewrite the other one in a more hospitable form

**singleton2787** · May 19th 2008, 07:23 PM

Problem 1 (simplify) 1/sin(-x) * (1-cos^2x)

**Mathstud28** · May 19th 2008, 07:28 PM

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

**singleton2787** · May 19th 2008, 07:35 PM

how does $1-\cos^2(x)=\sin^2(x)$ ?

**Mathstud28** · May 19th 2008, 07:40 PM

Originally Posted by singleton2787

how does $1-\cos^2(x)=\sin^2(x)$ ?

Because $\cos^2(x)+\sin^2(x)=1$

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

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

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

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

Simplifying we get

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

Usually in trig classes this indentity is just taken to be true

**singleton2787** · May 19th 2008, 07:53 PM

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

**Reckoner** · May 19th 2008, 08:18 PM

Originally Posted by singleton2787

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

First, convert everything to sines and cosines using the identities

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

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

Then, combine the fractions in the numerator and denominator, reduce, factor and cancel, and you should be left with $-\csc x$ 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,

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

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

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

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