Series in Trigonometry

**mrmohamed** · December 1st 2009, 11:10 AM

Hi all mathlinkers
here i will post trigonometry problems and i wish that it will be good for al

(1)

with my best wishes

**Soroban** · December 1st 2009, 12:21 PM

Hello, mrmohamed!

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

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

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

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

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

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

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

**mrmohamed** · December 1st 2009, 12:27 PM

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

still there another methode

i will wait
mrmohamed

**Jameson** · December 1st 2009, 12:30 PM

Originally Posted by mrmohamed

thank u mr soroban
very good solution

still there another methode

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.

**mrmohamed** · December 1st 2009, 12:35 PM

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

**mrmohamed** · December 2nd 2009, 03:11 AM

Hi all
my methode

**mrmohamed** · December 2nd 2009, 03:18 AM

Hi all

( 2 )

with my best wishes
mrmohamed

**simplependulum** · December 2nd 2009, 05:17 AM

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

**mrmohamed** · December 2nd 2009, 07:18 AM

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

Hi all

I wish the idea will be new
unusual methode

with my best wishes
mrmohamed

**mrmohamed** · December 10th 2009, 01:50 AM

Hi all

(3)

with my best wishes
mrmohamed

**simplependulum** · December 10th 2009, 02:04 AM

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

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

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

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

similar to the others

finally we have

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

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

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