Verify the Trigonometric Identity

[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}

 

[Orodruin]

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.

 

[lurflurf]

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

 

[SammyS]

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?

 

[lurflurf]

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

 

[SammyS]

...
@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 ?

 

[ehild]

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.

 

[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}.

 

[lurflurf]

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

 

[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...

 

[lurflurf]

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

 

[ehild]

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!

 

[ehild]

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.