1. ## Series in Trigonometry

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

(1)

with my best wishes

2. Hello, mrmohamed!

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

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

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

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

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

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

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

3. Originally Posted by Soroban
Hello, mrmohamed!

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

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

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

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

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

. . . . . $\displaystyle =\;\;\frac{\sin x}{\frac{1}{2}\sin30} \;\;=\;\;\frac{\sin x}{\frac{1}{2}\cdot\frac{1}{2}} \;\;=\;\;4\sin x$
thank u mr soroban
very good solution
still there another methode
i will wait
mrmohamed

4. 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.

5. 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

6. Hi all
my methode

7. Hi all

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with my best wishes
mrmohamed

8. Note that

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

Similarly ,

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

so we have to consider

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

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

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

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

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

9. Originally Posted by simplependulum
Note that

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

Similarly ,

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

so we have to consider

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

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

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

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

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

10. Hi all

(3)

with my best wishes
mrmohamed

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

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

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

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

similar to the others

finally we have

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

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