sin^2(5) + sin^2(10) +............+sin^2(90)/(1+cos^2(5) + cos^2(10)......cos^2(90)) = ?
\[\frac{\sin^2(5) + \sin^2(10) +............+\sin^2(90)}{1+\cos^2(5) + \cos^2(10)......\cos^2(90) }= ?\]
yes
@UnkleRhaukus help
i can't see a good first step yet. maybe use a trig formula
are they ment to be more terms in the numerator than the denominator?
there is 1 in deno
i mean more in the denominator than the numerator, actually i guess the is an invisible \(\sin(0)\) in the numerator so they have the same number of terms
the 1 is a really a \(\cos(0)\)
no
sorry but there is sin^2(85) in numerator and cos^2(85) in denominator
@UnkleRhaukus
\[=\frac{\sum\limits_{n=0}^{18}\sin(5n)}{\sum\limits_{n=0}^{18}\cos(5n)}\]
Add a sin^2 (0) to numerator. And in denominator, 1=cos^2 (0). Now, Sin^2 theta = cos^2 (90-theta) Proceed yourself further.
sin(5) = cos(85) they are symmetric about 45 ... use these to get 1's
so wat will we get
(1 + 1 + 1 + ... +1 + 1/sqrt(2))/(1 + 1 + 1 + ... +1 + 1/sqrt(2)) = 1 my guess ... haven't worked out ... must be 9 1's
you are in good hands, can't help you at the moment
the asnwer should be 1
@experimentX
yeah i guessed that ... i also gave you hint how to solve ... it's too long to type..
The answer is one. you have already got the process.. Its amazing to see how similar you are to shameer and openstudy1.
Try it yourself.
s i will try
i got answer as 1/2..........@experimentX
@experimentX
you shouldn't get that sin^2(5) + sin^2(10) +............+sin^2(90 = sin^2(0) + sin^2(90) + sin^2(5) + sin^2(85) +............+sin^2(40)+sin^2(50) + sin^2(45) first arrange like this and get 9 + 1/sqrt(2) do same for denominator
@experimentX 9 + 1/2 for the numerator and the denominator. You forgot to square 1/sqrt(2).
Oh ... thanks for correcting prof!!
lol ... silly me!!
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