anonymous
  • anonymous
ques...
Trigonometry
schrodinger
  • schrodinger
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anonymous
  • anonymous
Given \[\sin^{-1}\frac{2a}{1+a^2}+\sin^{-1}\frac{2b}{1+b^2}=2\tan^{-1}x\] Prove that \[x=\frac{a+b}{1-ab}\] What I've done: Let \[a=\tan \alpha\]\[b=\tan \beta\] \[\implies \sin^{-1}(\sin(2 \alpha))+\sin^{-1}(\sin(2 \beta))=2 \alpha+2 \beta=2(\alpha+\beta)=2\tan^{-1}x\]\[\alpha+\beta=\tan^{-1}x\]\[x=\tan(\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha .\tan \beta}=\frac{a+b}{1-ab}\] Is this ok?
anonymous
  • anonymous
looks good
Michele_Laino
  • Michele_Laino
here is my reasoning: I have applied the sin() operator to both sides of your equation, namely: \[\Large \begin{gathered} \sin \left( {{{\sin }^{ - 1}}\frac{{2a}}{{1 + {a^2}}} + {{\sin }^{ - 1}}\frac{{2b}}{{1 + {b^2}}}} \right) = \hfill \\ \hfill \\ = \sin \left( {2{{\tan }^{ - 1}}x} \right) \hfill \\ \end{gathered} \] so I got: \[\Large \begin{gathered} \frac{{2a}}{{1 + {a^2}}}\frac{{1 - {b^2}}}{{1 + {b^2}}} + \frac{{1 - {a^2}}}{{1 + {a^2}}}\frac{{2b}}{{1 + {b^2}}} = \frac{{2x}}{{1 + {x^2}}} \hfill \\ \hfill \\ \frac{{2\left( {a - b} \right)\left( {1 - ab} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}} = \frac{{2x}}{{1 + {x^2}}} \hfill \\ \end{gathered} \] since: \[\Large \begin{gathered} \cos \left( {\arcsin \left( {\frac{{2b}}{{1 + {b^2}}}} \right)} \right) = \frac{{1 - {b^2}}}{{1 + {b^2}}} \hfill \\ \hfill \\ \cos \left( {\arcsin \left( {\frac{{2a}}{{1 + {a^2}}}} \right)} \right) = \frac{{1 - {a^2}}}{{1 + {a^2}}} \hfill \\ \hfill \\ \sin \left( {2\arctan x} \right) = \frac{{2x}}{{1 + {x^2}}} \hfill \\ \end{gathered} \] now,using your value for the quantity x, I can write: \[\Large \frac{{2x}}{{1 + {x^2}}} = \frac{{2\left( {a - b} \right)\left( {1 - ab} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}}\]

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anonymous
  • anonymous
But the problem I think in my solution is LET \[a=\tan \alpha\] and \[b=\tan \beta\] a and b are most likely constants and not variables, but can I still let them equal to something?
Michele_Laino
  • Michele_Laino
what is the range of values of constants a and b?
anonymous
  • anonymous
Idk, there is nothing given just If \[\sin^{-1}\frac{2a}{1+a^2}+\sin^{-1}\frac{2b}{1+b^2}=2\tan^{-1}x\] prove that \[x=\frac{a+b}{1-ab}\] That's all there is given
Michele_Laino
  • Michele_Laino
I think that your formulas, namely \[\begin{gathered} a = \tan \alpha \hfill \\ b = \tan \beta \hfill \\ \end{gathered} \] can be accepted
anonymous
  • anonymous
I guess since \[y=\tan^{-1}x\] is defined \[\forall x \in \mathbb{R}\] so whatever the constants a and b(it's not given but they must be real) \[\tan^{-1}a \tan^{-1}b\] would be defined

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