# Series in Trigonometry

• December 1st 2009, 11:10 AM
mrmohamed
Series in Trigonometry
here i will post trigonometry problems and i wish that it will be good for al
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(1)

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with my best wishes
• December 1st 2009, 12:21 PM
Soroban
Hello, mrmohamed!

Quote:

1) Prove that: . $\cos(15^o-x)\sec15^o - \sin(15^o-x)\csc15^o \:=\:4\sin x$

We have: . $\frac{\cos(15-x)}{\cos15} - \frac{\sin(15-x)}{\sin15}$

. . . . . $=\;\; \frac{\cos15\cos x + \sin15\sin x}{\cos15} - \frac{\sin15\cos x - \cos15\sin x}{\sin 15}$

. . . . . $=\;\; \frac{\sin15(\cos15\cos x + \sin15\sin x) - \cos15(\sin15\cos x - \cos15\sin x)}{\sin15\cos15}$

. . . . . $=\;\; \frac{{\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \sin^2\!15\sin x - {\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \cos^2\!15\sin x}{\sin15\cos15}$

. . . . . $=\;\; \frac{\overbrace{(\sin^2\!15 + \cos^2\!15)}^{\text{This is 1}}\sin x}{\sin15\cos15} \;\;=\;\;\frac{\sin x}{\sin15\cos15}$

. . . . . $=\;\;\frac{\sin x}{\frac{1}{2}\sin30} \;\;=\;\;\frac{\sin x}{\frac{1}{2}\cdot\frac{1}{2}} \;\;=\;\;4\sin x$

• December 1st 2009, 12:27 PM
mrmohamed
Quote:

Originally Posted by Soroban
Hello, mrmohamed!

We have: . $\frac{\cos(15-x)}{\cos15} - \frac{\sin(15-x)}{\sin15}$

. . . . . $=\;\; \frac{\cos15\cos x + \sin15\sin x}{\cos15} - \frac{\sin15\cos x - \cos15\sin x}{\sin 15}$

. . . . . $=\;\; \frac{\sin15(\cos15\cos x + \sin15\sin x) - \cos15(\sin15\cos x - \cos15\sin x)}{\sin15\cos15}$

. . . . . $=\;\; \frac{{\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \sin^2\!15\sin x - {\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \cos^2\!15\sin x}{\sin15\cos15}$

. . . . . $=\;\; \frac{\overbrace{(\sin^2\!15 + \cos^2\!15)}^{\text{This is 1}}\sin x}{\sin15\cos15} \;\;=\;\;\frac{\sin x}{\sin15\cos15}$

. . . . . $=\;\;\frac{\sin x}{\frac{1}{2}\sin30} \;\;=\;\;\frac{\sin x}{\frac{1}{2}\cdot\frac{1}{2}} \;\;=\;\;4\sin x$

thank u mr soroban
very good solution (Clapping)
still there another methode(Wink)
i will wait
mrmohamed
• December 1st 2009, 12:30 PM
Jameson
Quote:

Originally Posted by mrmohamed
thank u mr soroban
very good solution (Clapping)
still there another methode(Wink)
i will wait
mrmohamed

If you already know the solution and want to post a problem just to challenge others, we have a particular forum for that. I'm moving this thread there.
• December 1st 2009, 12:35 PM
mrmohamed
Quote:

Originally Posted by Jameson
If you already know the solution and want to post a problem just to challenge others, we have a particular forum for that. I'm moving this thread there.

hi sir
at first im sorry to put the topic in the wrong place
second
i dislike to challenge others but im looking for more ideas to learn more
thanks sir and u cane move the topic as u liks
mrmohamed
• December 2nd 2009, 03:11 AM
mrmohamed
• December 2nd 2009, 03:18 AM
mrmohamed
Hi all

( 2 )
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with my best wishes
mrmohamed

(Clapping)(Clapping)(Clapping)(Clapping)(Clapping) (Clapping)
• December 2nd 2009, 05:17 AM
simplependulum
Note that

$\cos(\frac{5\pi}{8}) = \cos( \pi - \frac{3\pi}{8} ) = -\cos(\frac{3\pi}{8})$

Similarly ,

$\cos(\frac{7\pi}{8} = -\cos(\frac{\pi}{8})$

so we have to consider

$( 1 + u)(1+v)(1-v)(1-u)$

$u = \cos(\frac{\pi}{8}) , v = \cos(\frac{3\pi}{8})$

$= (1 - u^2)(1 - v^2 ) = \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8})$

$= \frac{1}{4} ( 1 - \cos(\frac{\pi}{4}))(1 - \cos(\frac{3\pi}{4})) = \frac{1}{8}$

But i guess you have a better and interesting method to solve it (Happy)
• December 2nd 2009, 07:18 AM
mrmohamed
Quote:

Originally Posted by simplependulum
Note that

$\cos(\frac{5\pi}{8}) = \cos( \pi - \frac{3\pi}{8} ) = -\cos(\frac{3\pi}{8})$

Similarly ,

$\cos(\frac{7\pi}{8} = -\cos(\frac{\pi}{8})$

so we have to consider

$( 1 + u)(1+v)(1-v)(1-u)$

$u = \cos(\frac{\pi}{8}) , v = \cos(\frac{3\pi}{8})$

$= (1 - u^2)(1 - v^2 ) = \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8})$

$= \frac{1}{4} ( 1 - \cos(\frac{\pi}{4}))(1 - \cos(\frac{3\pi}{4})) = \frac{1}{8}$

But i guess you have a better and interesting method to solve it (Happy)

Hi all
I wish the idea will be new
unusual methode

with my best wishes
mrmohamed
• December 10th 2009, 01:50 AM
mrmohamed
Hi all

(3)
(Clapping)(Clapping)(Clapping)(Clapping)(Clapping) (Clapping)(Clapping)(Clapping)(Clapping)(Clapping)
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with my best wishes
mrmohamed
• December 10th 2009, 02:04 AM
simplependulum

$\cos(\frac{A}{2} ) \cos(\frac{B-C}{2} )$

$= \cos(\frac{ \pi - (B+C)}{2} ) \cos(\frac{B-C}{2} )$

$= \sin(\frac{B+C}{2})\cos(\frac{B-C}{2} )$

$= \frac{1}{2} [ \sin(B) + \sin(C) ]$

similar to the others

finally we have

$\frac{1}{2} [ \sin(B) + \sin(C) + \sin(A) + \sin(C) + \sin(B) + \sin(A )$

$= \sin(A) + \sin(B) + \sin(C)$