If the value of $\sin^26°+\sin^242°+\sin^266°+\sin^278°$ is expressed in the lowest form as $\frac{p}q$, then find the value of $(p+q-10)$.
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1Welcome to MSE! Show what you have tried so far and use mathjax so that the people on this site can better understand your question. – For the love of maths Jan 05 '18 at 13:34
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1Welcome to stackexchange. You are more likely to get answers rather than downvotes and votes to close if you edit the question to show what you tried and where you are stuck. Please use mathjax for mathematics: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Jan 05 '18 at 13:35
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Have you tried finding the value of $\sin^2 6 + \sin ^2 42 + \sin ^2 66 + \sin ^2 78$? – For the love of maths Jan 05 '18 at 13:35
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Hint: $\frac pq = \frac 94$. Hope that solves your question. – For the love of maths Jan 05 '18 at 13:38
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i need the solution without using the values of the given trigonometric functions. – Aditya Modekurti Jan 05 '18 at 13:40
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What do you mean?????????????????? Why were you given those trigonometric functions then? – For the love of maths Jan 05 '18 at 13:41
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Do you want an answer that shows some proving instead of directly inputting values into a calculator? – For the love of maths Jan 05 '18 at 13:43
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we are supposed to use identities and manipulations to solve. we aren't allowed to use calculators in examinations. – Aditya Modekurti Jan 05 '18 at 13:43
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I would have preferred if you specified that in the question. Nevermind, working on the proof for you! – For the love of maths Jan 05 '18 at 13:45
2 Answers
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Hint:
Use $\cos2x=1-2\sin^2x$
to find the summands to be
$\cos12^\circ,\cos(12+72)^\circ,$
$\cos(12+2\cdot72)^\circ,$
$\cos(12+3\cdot72)^\circ=\cos(360-228)^\circ$
Now use How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
lab bhattacharjee
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When you write up an answer only to find someone else correctly answered before you. It was my good fortune that I checked your answer else my answer would have been flagged as duplicate! Really a neat answer (took less space than mine). +1 – For the love of maths Jan 05 '18 at 13:54
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thank you lab bhattacharjee ! your answer was indeed very helpful. – Aditya Modekurti Jan 05 '18 at 13:58
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$$\sin^26°+\sin^242°+\sin^266°+\sin^278°=$$ $$=\frac{1-\cos12^{\circ}+1-\cos84^{\circ}+1-\cos132^{\circ}+1-\cos156^{\circ}}{2}=$$ $$=2-\frac{1}{2}(\cos12^{\circ}+\cos132^{\circ})-\frac{1}{2}(\cos84^{\circ}+\cos156^{\circ})=$$ $$=2-\frac{1}{2}\cos72^{\circ}+\frac{1}{2}\cos36^{\circ}=2-\frac{2\sin36^{\circ}\cos72^{\circ}-2\sin36^{\circ}\cos36^{\circ}}{4\sin36^{\circ}}=$$ $$=2-\frac{\sin108^{\circ}-\sin36^{\circ}-\sin72^{\circ}}{4\sin36^{\circ}}=2+\frac{1}{4}=\frac{9}{4}.$$ Id est, $$p+q-10=3.$$
Michael Rozenberg
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