# Verify the Trigonometric Identity

• FritoTaco
In summary: So, the original problem should read as follows: csc(tan^{-1}\dfrac{x}{2})=\sqrt{\dfrac{x^{2}+4}{x}}In summary, using the given identities, we can simplify Problem 1 to csc(tan^{-1}\dfrac{x}{2})=\dfrac{\sqrt{x^2+4}}{x}. By plotting the tangent of \dfrac{x}{2} in Quadrant 1 of a triangle, we can find the hypotenuse using the Pythagorean theorem, which is \sqrt{x^2+4}. This gives us the simplified form of csc\theta=\dfrac{hyp}{opp}=\dfrac{\sqrt{x^
FritoTaco

## Homework Statement

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

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

## Homework 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: $\sqrt{x^{2}}=|x|$

## 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?

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

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.

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

FritoTaco
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\,.##

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?

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

FritoTaco
FritoTaco said:
...
@SammyS, is that when I was finding the common denominator?
I only see one place where you have ##\ (1+\sin x)(1-\sin x)\ ##.

Did you look for that ?

FritoTaco
FritoTaco said:

## Homework Statement

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

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

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...

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

FritoTaco
FritoTaco said:
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...
Use the relation ##\sin(\alpha)=\frac{tan(\alpha)}{\sqrt{1+tan^2(\alpha)}}## with ##\alpha=\tan^{-1}(x/2)##

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!

FritoTaco said:
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!
It is wrong. x should be outside the square root.

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.

## 1. What is a trigonometric identity?

A trigonometric identity is a mathematical equation that is true for all values of the variables involved. It is often used to simplify or solve complex trigonometric expressions.

## 2. Why is it important to verify trigonometric identities?

Verifying trigonometric identities is important because it ensures the accuracy and validity of mathematical calculations involving trigonometric functions. It also helps in solving complex equations and proving other mathematical theorems.

## 3. What are some common trigonometric identities?

Some common trigonometric identities include the Pythagorean identities, sum and difference identities, double angle identities, half angle identities, and product-to-sum identities.

## 4. How do you verify a trigonometric identity?

To verify a trigonometric identity, you must manipulate one side of the equation using algebraic and trigonometric properties until it is equivalent to the other side. This can involve using trigonometric identities, expanding expressions, and simplifying fractions.

## 5. Are there any tips for verifying trigonometric identities?

Yes, some tips for verifying trigonometric identities include: always start with the more complex side of the equation, look for common patterns and identities to simplify expressions, and use a graphing calculator to check your work. It is also helpful to practice and familiarize yourself with common trigonometric identities.

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