1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Verify the Trigonometric Identity

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data

    Problem 1
    : [itex]csc(tan^{-1}\dfrac{x}{2})=\sqrt{\dfrac{x^{2}+4}{x}}[/itex]

    Problem 2: [itex]\sqrt{\dfrac{1-sinx}{1+sinx}}=\dfrac{|cosx|}{1+sinx}[/itex]

    2. Relevant equations

    Quotient Identities
    [itex]tan\theta=\dfrac{sin\theta}{cos\theta}[/itex]

    [itex]cos\theta=\dfrac{cos\theta}{sin\theta}[/itex]

    Pythagorean Identites
    [itex]sin^{2}\theta+cos^{2}=1[/itex]

    [itex]cot^{2}\theta+1=csc^{2}\theta[/itex]

    [itex]tan^{2}+1=sec^{2}\theta[/itex]

    Recirpocal Identites
    [itex]sin\theta=\dfrac{1}{csc\theta}[/itex]

    [itex]cos\theta=\dfrac{1}{sec\theta}[/itex]

    [itex]tan\theta=\dfrac{1}{cot\theta}[/itex]

    [itex]csc\theta=\dfrac{1}{sin\theta}[/itex]

    [itex]sec\theta=\dfrac{1}{cos\theta}[/itex]

    [itex]cot\theta=\dfrac{1}{tan\theta}[/itex]

    Cofunction Identities
    (maybe useful?)
    [itex]sin(\dfrac{\pi}{2}-\theta)=cos\theta[/itex]

    [itex]cos(\dfrac{\pi}{2}-\theta)=sin\theta[/itex]

    [itex]tan(\dfrac{\pi}{2}-\theta)=cot\theta[/itex]

    [itex]csc(\dfrac{\pi}{2}-\theta)=sec\theta[/itex]

    [itex]sec(\dfrac{\pi}{2}-\theta)=csc\theta[/itex]

    [itex]cot(\dfrac{\pi}{2}-\theta)=tan\theta[/itex]

    Note: [itex]\sqrt{x^{2}}=|x|[/itex]

    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.

    [itex]csc(tan^{-1}\dfrac{x}{2})=\sqrt{\dfrac{x^{2}+4}{x}}[/itex]

    Maybe manipulate the left side?

    [itex]csc(cot\theta)[/itex]

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

    Problem 2:

    [itex]\sqrt{\dfrac{1-sinx}{1+sinx}}=\dfrac{|cosx|}{1+sinx}[/itex]

    I manipulated the left side

    [itex]\dfrac{\sqrt{1-sinx}}{\sqrt{1+sinx}}\cdot\dfrac{\sqrt{1-sinx}}{\sqrt{1-sinx}}[/itex]

    [itex]\dfrac{1-sinx}{\sqrt{(1+sinx)(1-sinx)}}[/itex]

    [itex]\dfrac{1-sinx}{\sqrt{(1+sin^{2}x)}} <--\sqrt{x^{2}}=|x|[/itex]

    [itex]\dfrac{1-sinx}{|cosx|}[/itex]

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

    [itex]\dfrac{|cosx|}{1+sinx}[/itex] = [itex]\dfrac{|cosx|}{1+sinx}[/itex]
     
  2. jcsd
  3. Nov 21, 2016 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Nov 21, 2016 #3

    lurflurf

    User Avatar
    Homework Helper

    for 1 use
    $$\csc x=\dfrac{\sec x}{\tan x}$$
    and
    $$\sec ^2 x=1+\tan^2 x$$
    for 2 start with
    $$\dfrac{|\cos x|}{1+\sin x}=\sqrt{\left(\dfrac{\cos x}{1+\sin x}\right)^2}$$
     
  5. Nov 21, 2016 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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\,.##
     
  6. Nov 21, 2016 #5
    @Orodruin, if I draw a triangle, csc is [itex]csc\theta=\dfrac{r}{y}[/itex], how do I find r? Is it just [itex]r=(\sqrt{x^{2}+4})[/itex]? Which makes it less confusing actually. But still confused because now I have:
    [itex](\sqrt{x^{2}+4})=\dfrac{\sqrt{x^{2}+4}}{x}[/itex]

    @lurflurf, How does the numerator become [itex]cosx[/itex]? 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?
     
  7. Nov 21, 2016 #6

    lurflurf

    User Avatar
    Homework Helper

    ^It is squared I have used
    $$|x|=\sqrt{x^2}$$
    $$\dfrac{|\cos x|}{1+\sin x}=
    \left|\dfrac{\cos x}{1+\sin x}\right|=
    \sqrt{\left(\dfrac{\cos x}{1+\sin x}\right)^2}=
    \sqrt{\dfrac{\cos^2 x}{(1+\sin x)^2}}$$
     
  8. Nov 21, 2016 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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

    Did you look for that ?
     
  9. Nov 21, 2016 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
  10. Nov 22, 2016 #9
    @SammyS, Oh. I see, where. I had in the denominator with a (+) sign. Here is what it should look like: [itex]\dfrac{1-sinx}{\sqrt{(1-sin^{2}x)}}[/itex]

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

    [itex]\sqrt{\left(\dfrac{cosx}{1+sinx} \right)^{2}}[/itex]

    [itex]\sqrt{\dfrac{cos^{2}x}{(1+sinx)^{2}}}[/itex]

    [itex]\sqrt{\dfrac{1-sin^{2}x}{1+sin^{2}x}}[/itex]

    [itex]\sqrt{\dfrac{1-sinx}{1+sinx}}=\sqrt{\dfrac{1-sinx}{1+sinx}}[/itex]

    Is this correct?

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

    @ehild, Oh, that was an editing mistake, my bad. It's [itex]csc(tan^{-1}\dfrac{x}{2})=\dfrac{\sqrt{x^2+4}}{x}[/itex].
     
  11. Nov 22, 2016 #10

    lurflurf

    User Avatar
    Homework Helper

    ^
    $$\csc x=\frac{1}{\sin x} \\
    \csc x=\frac{1}{\sin x}\cdot\frac{\sec x}{1/\cos x} \\
    \csc x=\frac{\sec x}{\sin x/\cos x} \\
    \csc x=\frac{\sec x}{\tan x}$$
     
  12. Nov 22, 2016 #11
    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.....
     
  13. Nov 22, 2016 #12

    lurflurf

    User Avatar
    Homework Helper

    $$|\sec x|=\sqrt{\sec ^2 x}=\sqrt{1+\tan^2 x}$$
     
  14. Nov 22, 2016 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Use the relation ##\sin(\alpha)=\frac{tan(\alpha)}{\sqrt{1+tan^2(\alpha)}}## with ##\alpha=\tan^{-1}(x/2)##
     
  15. Nov 22, 2016 #14
    I figured it out. It was actually easier than I thought...

    If you plot the tangent of [itex]\dfrac{x}{2}[/itex] in Quadrant one of a triangle, you just find the hypotenuse using the Pythagorean theorem which is [itex]\sqrt{x^{2}+4}[/itex]. Then find the csc which is [itex]csc\theta=\dfrac{hyp}{opp}=\sqrt{\dfrac{x^{2}+4}{x}}[/itex]. Thanks guys!
     
  16. Nov 22, 2016 #15

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is wrong. x should be outside the square root.
     
  17. Nov 22, 2016 #16
    @ehild, Ahhhhh. Thanks for pointing that out, I keep making that editing mistake. It should be this, [itex]csc\theta=\dfrac{hyp}{opp}=\dfrac{\sqrt{x^{2}+4}}{x}[/itex]

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Verify the Trigonometric Identity
Loading...