I would be very grateful if someone could solve this question for me???

Using the identities for sin (A+B) and cos^2(theta)+sin^2(theta)=1

Find sine 65 degrees?

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- Aug 8th 2010, 07:04 PMGAVREED2Trigonometric Identities problem
I would be very grateful if someone could solve this question for me???

Using the identities for sin (A__+__B) and cos^2(theta)+sin^2(theta)=1

Find sine 65 degrees? - Aug 10th 2010, 06:55 AMEphemerasylumFailed attempt
okay we get started by listing what may be of help here:

$\displaystyle \sin60^\circ=\frac{\sqrt3}2$;

$\displaystyle \sin30^\circ=\frac12$;

$\displaystyle \sin45^\circ=\frac{\sqrt2}2$;

with the trigonometric identities:

$\displaystyle \sin(a\pm b)=\sin a\cos b\pm\sin b\cos a$

$\displaystyle \sin^2 a+\cos^2 a = 1$

First we assume

$\displaystyle y=\sin20^\circ;$;

therefore we have

$\displaystyle \cos20^\circ=\sqrt{1-y^2}$,

together with the application of the identities introduced above, which results:

$\displaystyle \sin40^\circ=2\sin20^\circ\cos20^\circ=2y\sqrt{1-y^2},$

$\displaystyle \cos40^\circ=\sqrt{1-\sin^240^\circ}=\dots=1-2y^2$

(note that y is less than $\displaystyle \sin45^\circ$)

The identities also yield that

$\displaystyle \sin60^\circ=\sin(20^\circ+40^\circ)$

$\displaystyle =\sin20^\circ\cos40^\circ+\cos20^\circ\sin40^\circ =\dots$

$\displaystyle =-4y^3+3y=\frac{\sqrt3}2$

Therefore we may find the only real value of y (which isn't easy)

$\displaystyle y\approx 0.6428$

Hence

$\displaystyle \sin65^\circ\sin(45^\circ+20^\circ)=\dots=\frac{\s qrt2}2(y+\sqrt{1-y^2})=0.9063$

The bad news is that, I haven't yet figured out how to solve the equation without a calulator. So good luck on that, and I'll put my findings here as soon as I'm getting something. - Aug 14th 2010, 03:38 PMhelopticor
I think there is not really a nice answer to this problem. GAVREED2, are you sure there's no typo?

- Aug 14th 2010, 03:49 PMskeeter
see this "second" post ...

http://www.mathhelpforum.com/math-he...le-153405.html