# Verify the Trigonometric Identity

1. Nov 21, 2016

### FritoTaco

1. The problem statement, all variables and given/known data

Problem 1
: $csc(tan^{-1}\dfrac{x}{2})=\sqrt{\dfrac{x^{2}+4}{x}}$

Problem 2: $\sqrt{\dfrac{1-sinx}{1+sinx}}=\dfrac{|cosx|}{1+sinx}$

2. Relevant equations

Quotient Identities
$tan\theta=\dfrac{sin\theta}{cos\theta}$

$cos\theta=\dfrac{cos\theta}{sin\theta}$

Pythagorean Identites
$sin^{2}\theta+cos^{2}=1$

$cot^{2}\theta+1=csc^{2}\theta$

$tan^{2}+1=sec^{2}\theta$

Recirpocal Identites
$sin\theta=\dfrac{1}{csc\theta}$

$cos\theta=\dfrac{1}{sec\theta}$

$tan\theta=\dfrac{1}{cot\theta}$

$csc\theta=\dfrac{1}{sin\theta}$

$sec\theta=\dfrac{1}{cos\theta}$

$cot\theta=\dfrac{1}{tan\theta}$

Cofunction Identities
(maybe useful?)
$sin(\dfrac{\pi}{2}-\theta)=cos\theta$

$cos(\dfrac{\pi}{2}-\theta)=sin\theta$

$tan(\dfrac{\pi}{2}-\theta)=cot\theta$

$csc(\dfrac{\pi}{2}-\theta)=sec\theta$

$sec(\dfrac{\pi}{2}-\theta)=csc\theta$

$cot(\dfrac{\pi}{2}-\theta)=tan\theta$

Note: $\sqrt{x^{2}}=|x|$

3. The attempt at a solution

Problem 1: Any ideas on how to get started? I have no idea what to do with inverse tangent and that stupid x/2 and it makes me mad! I know I need some work for it to be a valid thread so here goes nothing.

$csc(tan^{-1}\dfrac{x}{2})=\sqrt{\dfrac{x^{2}+4}{x}}$

Maybe manipulate the left side?

$csc(cot\theta)$

....yup, I'm totally lost.

Problem 2:

$\sqrt{\dfrac{1-sinx}{1+sinx}}=\dfrac{|cosx|}{1+sinx}$

I manipulated the left side

$\dfrac{\sqrt{1-sinx}}{\sqrt{1+sinx}}\cdot\dfrac{\sqrt{1-sinx}}{\sqrt{1-sinx}}$

$\dfrac{1-sinx}{\sqrt{(1+sinx)(1-sinx)}}$

$\dfrac{1-sinx}{\sqrt{(1+sin^{2}x)}} <--\sqrt{x^{2}}=|x|$

$\dfrac{1-sinx}{|cosx|}$

$\dfrac{|cosx|}{1+sinx}$ Am I allowed to just switch it around like that?

$\dfrac{|cosx|}{1+sinx}$ = $\dfrac{|cosx|}{1+sinx}$

2. Nov 21, 2016

### Orodruin

Staff Emeritus
Problem 1: Try drawing a triangle where one of the angles have tan(theta) = x/2 ...

Problem 2: No, you cannot just switch it, you might want to try multiplying denominator and numerator by something different though.

3. Nov 21, 2016

### lurflurf

for 1 use
$$\csc x=\dfrac{\sec x}{\tan x}$$
and
$$\sec ^2 x=1+\tan^2 x$$
$$\dfrac{|\cos x|}{1+\sin x}=\sqrt{\left(\dfrac{\cos x}{1+\sin x}\right)^2}$$

4. Nov 21, 2016

### SammyS

Staff Emeritus
Regarding problem 2:

$(1-u)(1+u)=1-u^2\$ not $\ 1+u^2\,.$

Therefore, $\ (1+\sin x)(1-\sin x)\ne 1+\sin^2x\,.$

5. Nov 21, 2016

### FritoTaco

@Orodruin, if I draw a triangle, csc is $csc\theta=\dfrac{r}{y}$, how do I find r? Is it just $r=(\sqrt{x^{2}+4})$? Which makes it less confusing actually. But still confused because now I have:
$(\sqrt{x^{2}+4})=\dfrac{\sqrt{x^{2}+4}}{x}$

@lurflurf, How does the numerator become $cosx$? Doesn't it need to be squared to use the Pythagorean identity in order to change its value?

@SammyS, is that when I was finding the common denominator?

6. Nov 21, 2016

### lurflurf

^It is squared I have used
$$|x|=\sqrt{x^2}$$
$$\dfrac{|\cos x|}{1+\sin x}= \left|\dfrac{\cos x}{1+\sin x}\right|= \sqrt{\left(\dfrac{\cos x}{1+\sin x}\right)^2}= \sqrt{\dfrac{\cos^2 x}{(1+\sin x)^2}}$$

7. Nov 21, 2016

### SammyS

Staff Emeritus
I only see one place where you have $\ (1+\sin x)(1-\sin x)\$.

Did you look for that ?

8. Nov 21, 2016

### ehild

You can not verify (1) as it is wrong. $csc(tan^{-1}\dfrac{x}{2})=\dfrac{\sqrt{x^{2}+4}}{x}$
is true.

Last edited: Nov 21, 2016
9. Nov 22, 2016

### FritoTaco

@SammyS, Oh. I see, where. I had in the denominator with a (+) sign. Here is what it should look like: $\dfrac{1-sinx}{\sqrt{(1-sin^{2}x)}}$

@lurflurf, Oh. I see now, is this right?

$\sqrt{\left(\dfrac{cosx}{1+sinx} \right)^{2}}$

$\sqrt{\dfrac{cos^{2}x}{(1+sinx)^{2}}}$

$\sqrt{\dfrac{1-sin^{2}x}{1+sin^{2}x}}$

$\sqrt{\dfrac{1-sinx}{1+sinx}}=\sqrt{\dfrac{1-sinx}{1+sinx}}$

Is this correct?

Also, for problem 2. I'm confused on how you get $cscx=\dfrac{secx}{tanx}$ Did you use an identity?

@ehild, Oh, that was an editing mistake, my bad. It's $csc(tan^{-1}\dfrac{x}{2})=\dfrac{\sqrt{x^2+4}}{x}$.

10. Nov 22, 2016

### lurflurf

^
$$\csc x=\frac{1}{\sin x} \\ \csc x=\frac{1}{\sin x}\cdot\frac{\sec x}{1/\cos x} \\ \csc x=\frac{\sec x}{\sin x/\cos x} \\ \csc x=\frac{\sec x}{\tan x}$$

11. Nov 22, 2016

### FritoTaco

Oh okay, but how does that get me to a value of 4 which is part of what I need for my end result? I did the triangle like someone previously suggested which worked but I was missing an x in the denominator. Hmmm.....

12. Nov 22, 2016

### lurflurf

$$|\sec x|=\sqrt{\sec ^2 x}=\sqrt{1+\tan^2 x}$$

13. Nov 22, 2016

### ehild

Use the relation $\sin(\alpha)=\frac{tan(\alpha)}{\sqrt{1+tan^2(\alpha)}}$ with $\alpha=\tan^{-1}(x/2)$

14. Nov 22, 2016

### FritoTaco

I figured it out. It was actually easier than I thought...

If you plot the tangent of $\dfrac{x}{2}$ in Quadrant one of a triangle, you just find the hypotenuse using the Pythagorean theorem which is $\sqrt{x^{2}+4}$. Then find the csc which is $csc\theta=\dfrac{hyp}{opp}=\sqrt{\dfrac{x^{2}+4}{x}}$. Thanks guys!

15. Nov 22, 2016

### ehild

It is wrong. x should be outside the square root.

16. Nov 22, 2016

### FritoTaco

@ehild, Ahhhhh. Thanks for pointing that out, I keep making that editing mistake. It should be this, $csc\theta=\dfrac{hyp}{opp}=\dfrac{\sqrt{x^{2}+4}}{x}$

Note: The original problem should have the denominator of 'x' outside the square root. I didn't initially put that in the problem, to begin with.