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Trigonometry

can you please provide the solution(proof) of the above sum?

Note by Shubham Sharma
4 years, 3 months ago

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I think mathematical induction is a good hint, and you should just work through it.

Alternatively, for a more straightforward solution, show that the initial condition implies that \( \sin^2 \theta = \frac { x} { x+y}, \cos^2 \theta = \frac { y } { x+y} \). Then, the subsequent condition follows directly. Calvin Lin Staff · 4 years, 3 months ago

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@Calvin Lin To prove the second condition is much neater. Aditya Parson · 4 years, 3 months ago

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Wat is trigonometry Akshay K Gowda · 3 years, 3 months ago

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It can be solved by MATHEMATICAL INDUCTION!!!!! Please see this technique uploaded by Sir Calvin in the blog!!! Subhrodipto Basu Choudhury · 4 years, 1 month ago

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At first we take the given expression :- (Sin^{4}ө) /x + (cos^{4} ө)/y = 1/x+y Now write cosө in terms of sinө (Sin^{4} ө) /x + ( 1-sin^{2} ө)^{2} /y =1/x+y Take LCM => [ ysin^{4}ө + x(1-2sin^{2}ө + sin^{4}ө) ] /xy = 1/x+y Now cross multiply => sin^{4}ө (x+y)^{2} + x^{2} + xy – 2x^{2}sinө -2xysin^{2}ө = xy Let sin^ {2}ө = α => α^{2} (x+y)^{2} -2αx^{2} - 2xyα + x^{2} = 0 => α^{2}(x+y)^{2} -2αx ( x+y) +x^{2} = 0 =>[ α(x+y) - x ]^{2} = 0 => α = x / x+y We need to prove that sin^{2(m+1) }ө / x^{m} + cos^{2(m+1) }ө / y^{m} = 1/(x+y)^{m} At first we take the expression given on the left hand side :- => (sin^{2}ө )^{m+1} / x^{m} + (1-sin^{2}ө)^{m+1} / y^{m} But sin^{ 2}ө =α , as we had already assumed in the earlier steps =>( x/x+y)^{m+1} /x^{m} + (1 – x/x+y)^{m+1} /y^{m} => x^{m+1} / (x+y)^{m+1}x^{m} + y^{m+1} / (x+y)^{m+1} y^{m}
=> 1/(x+y)^{m+1} [ x^{m+1} / x^{m} + y^{m+1} / y^{m}] => 1/(x+y)^{m+1} [ x^{m+1} y^{m} + y^{m+1}x^{m} / (xy)^{m}] => 1/(x+y)^{m+1} [ x^{m}y^{m} (x+y) / x^{m}y^{m}] => (x+y) /(x+y)^{m+1} => ( x+y)^{–m} => 1/(x+y)^{m} Mishti Angel · 4 years, 3 months ago

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@Mishti Angel its horrible to read that :P Shubham Sharma · 4 years, 3 months ago

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@Shubham Sharma i have no idea regarding the latex form...so i cudn't convert to latex form bt u can go through this...and ask me fr ny help regarding math

contact me at :- prerana.chakrabarti@gmail.com Mishti Angel · 4 years, 3 months ago

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is it done by mathematical induction.... Riya Gupta · 4 years, 3 months ago

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@Riya Gupta but how? :P Shubham Sharma · 4 years, 3 months ago

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@Shubham Sharma arey....i was asking...from you....if it is done...by mathematical induction....i'll try...to xplain u.... Riya Gupta · 4 years, 3 months ago

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@Riya Gupta but how to post a picture over here?? okay, if u wnt the solutn then mail me ur email addrss on shubhamsharma1172@gmail.com... Shubham Sharma · 4 years, 3 months ago

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@Riya Gupta no induction, simple logic! :D will post the solutn in a few mins.... Shubham Sharma · 4 years, 3 months ago

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@Shubham Sharma Man! I am waiting for your solution... i get stuck on a loop, so I did not manage to get the result! Gabriel Barros · 4 years, 3 months ago

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@Gabriel Barros how to post a picture, eh?? its quite troublesome.... i have the solutn ready bt i just can't expose it due to the lack of attachment option...:P Shubham Sharma · 4 years, 3 months ago

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@Shubham Sharma You can upload on another site then just post the link here... Gabriel Barros · 4 years, 3 months ago

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If m = 1 Then result verified Farough Ahmed Siddiqui · 4 years, 3 months ago

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@Farough Ahmed Siddiqui we don't have to find the value of m...m is any integer for which the equation stands true.. Shubham Sharma · 4 years, 3 months ago

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