1. Limited time only! Sign up for a free 30min personal 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!

Prove this identity

  1. Jan 10, 2013 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    If [itex]sin^{-1}x+sin^{-1}y+sin^{-1}z = \pi [/itex] then prove that [itex]x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz [/itex]

    2. Relevant equations

    3. The attempt at a solution

    I assume the inverse functions to be θ, α, β respectively. Rearranging and taking tan of both sides

    [itex]tan(\theta + \alpha) = tan(\pi - \beta) \\

    tan(\theta + \alpha) = -tan(\beta)

    After simplifying I get something like this
    [itex]x\sqrt{(1-y^2)(1-z^2)}+y\sqrt{(1-x^2)(1-z^2)}+z\sqrt{(1-x^2)(1-y^2)} = xyz[/itex]

    I know it's close but it is not yet the final result.
  2. jcsd
  3. Jan 10, 2013 #2
    Hello again, utkarshakash! Starting from your second step...

    [itex]\frac{tan\theta + tan\alpha}{1-tan\alpha tan\theta} = -tan\beta \\

    tan\theta + tan\alpha = -tan\beta + tan\alpha \ tan\theta \ tan\beta \\

    tan\theta + tan\alpha + tan\beta = tan\alpha \ tan\theta \ tan\beta \\

    tan(arcsinx) + tan(arcsiny) + tan(arcsinz) = tan(arcsinx) \ tan(arcsiny) \ tan(arcsinz) \\

    \frac{x}{\sqrt{1-x^2}} + \frac{y}{\sqrt{1-y^2}} + \frac{z}{\sqrt{1-z^2}} = \frac{xyz}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}}[/itex]

    Here, I think, is where you went wrong. You want each term on the left hand side to be of the form "n√(1-n^2)". Thus, we multiply each term by 1 :biggrin:.

    [itex]\frac{x\sqrt{1-x^2}}{1-x^2} + \frac{y\sqrt{1-y^2}}{1-y^2} + \frac{z\sqrt{1-z^2}}{1-z^2} = \frac{xyz}{\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}}[/itex]

    Got it from here?
  4. Jan 10, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There's probably a more elegant way, but if you simply eliminate z from each side (= x√(1-y2) + y√(1-x2)) then it should become reasonably evident.
  5. Jan 10, 2013 #4


    User Avatar
    Gold Member

    I am still not getting it. What I have to do after the last step?
  6. Jan 10, 2013 #5
    I'm sorry for somewhat hijacking this thread, but how do you guys (the homework helpers and the like) help people with problems such as this on a whim? I just got done with Calc1 and I wouldn't even know where to begin with this proof really. For example, I completely forgot that tan(a+b) = tan(a)+tan(b) / (1 - tan(a)tan(b)). How do you guys keep these identities fresh in your mind? Are you teachers or mathematics degree students?

    I don't mention the OP because it's different learning something and applying it directly in a problem that you know involves applying what you have recently learned; I'm talking about learning something and being able to retain it long after you have learned it.
    Last edited: Jan 10, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Prove identity Date
Proving Reciprocal Identities Feb 15, 2017
Proving Identities: trig. Aug 16, 2016
Proving identities Aug 14, 2016
Prove the identity Mar 25, 2016
Proving trigonometric identities in a belt and pulley proble Mar 20, 2016