Thursday, August 24, 2017

trigonometry - Find the value of lambda in frac3tan2piover71tan2piover7=lambdacospiover7




Find the value of λ in 3tan2π71tan2π7=λcosπ7




The numerator looks similar to expansion of tan3x, so I tried this




3tanπ7tan3π7tanπ7(1tan2π7)=λcosπ7



(3tanπ7tan3π7)(13tan2π7)tanπ7(1tan2π7)(13tan2π7)=λcosπ7



tan3π7(13tan2π7)sinπ7(1tan2π7)=λ



But I'm stuck here. Need help. Thanks in advance.


Answer



Let a=π/7 then we have 1+21tan2a=λcosa




multiplying this with tana we get: tana+tan2a=λsina



so sinacos2a+sin2acosa=λsinacosacos2a



multiplying this with 4 we get 4sin3a=λsin4a



Sinnce sin3a=sin4a we get λ=4.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...