What is the solution to finding the sum of square sin() functions?

AI Thread Summary
The discussion centers on calculating the sum of squared sine functions from sin²(10°) to sin²(90°). Participants suggest using the identity sin²(x) = (1 - cos(2x))/2 to simplify the problem. They also note the relationship sin(x) = cos(90° - x), which helps in pairing terms effectively. By applying these identities, the problem can be reduced to manageable pairs, ultimately leading to a total sum of 5. The conversation highlights the utility of trigonometric identities in solving such summation problems.
Aikon
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Hi all,
I work as monitor for a pre-university course and one student showed me this problem:

Homework Statement


To find how much is y:
y=sin2(10)+sin2(20)+sin2(30)+...+sin2(80)+sin2(90)

Homework Equations


I don't know. I thought about sen2x+cos2x=1


The Attempt at a Solution


To use the equation given above and to write a series of cos()'s, but i don't think it get better than before.
 
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Aikon said:

Homework Equations


I thought about sin2x+cos2x=1

You are on the right track. Use also that sin(x)=cos(90-x). For example, sin(80°)=cos(10°).

ehild
 
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There's another identity that I think will be helpful:
sin^2(x) = \frac{1 - cos(2x)}{2}

Also, cos(x) = -cos(\pi - x)
 
Thank you all the answers.

I liked this:
ehild said:
You are on the right track. Use also that sin(x)=cos(90-x). For example, sin(80°)=cos(10°).

ehild

With this identity it becomes almost trivial, because (sin210 +sin280)=1 and it goes like this for other pairs, it gives 5 in the end.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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