# Trigonometric Identities problem

• August 8th 2010, 07:04 PM
GAVREED2
Trigonometric 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?
• August 10th 2010, 06:55 AM
Ephemerasylum
Failed attempt
okay we get started by listing what may be of help here:
$\sin60^\circ=\frac{\sqrt3}2$;
$\sin30^\circ=\frac12$;
$\sin45^\circ=\frac{\sqrt2}2$;
with the trigonometric identities:
$\sin(a\pm b)=\sin a\cos b\pm\sin b\cos a$
$\sin^2 a+\cos^2 a = 1$

First we assume
$y=\sin20^\circ;$;
therefore we have
$\cos20^\circ=\sqrt{1-y^2}$,
together with the application of the identities introduced above, which results:
$\sin40^\circ=2\sin20^\circ\cos20^\circ=2y\sqrt{1-y^2},$
$\cos40^\circ=\sqrt{1-\sin^240^\circ}=\dots=1-2y^2$
(note that y is less than $\sin45^\circ$)

The identities also yield that
$\sin60^\circ=\sin(20^\circ+40^\circ)$
$=\sin20^\circ\cos40^\circ+\cos20^\circ\sin40^\circ =\dots$
$=-4y^3+3y=\frac{\sqrt3}2$
Therefore we may find the only real value of y (which isn't easy)
$y\approx 0.6428$
Hence
$\sin65^\circ\sin(45^\circ+20^\circ)=\dots=\frac{\s qrt2}2(y+\sqrt{1-y^2})=0.9063$

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.
• August 14th 2010, 03:38 PM
helopticor
I think there is not really a nice answer to this problem. GAVREED2, are you sure there's no typo?
• August 14th 2010, 03:49 PM
skeeter
Quote:

Originally Posted by helopticor
I think there is not really a nice answer to this problem. GAVREED2, are you sure there's no typo?

see this "second" post ...

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