# Thread: [SOLVED] New trigonometry problems

1. ## [SOLVED] New trigonometry problems

First, what is the value of $cos20.cos40.cos60.cos80$ = ?
Second, determine of the value $tan9-tan27-tan63-tan81=$
Third, if $sin(A+B).cosC=2sinA .cos (B-C)$ prove that $tanA - \frac{sin(B-A)}{2sinAsinB}$

2. I revise for the third problems :
Third, if $sin(A+B).cosC=2sinA .cos (B-C)$ prove that $tanC = tanA - \frac{sin(B-A)}{2sinAsinB}$

3. Originally Posted by wizard654zzz
First, what is the value of $cos20.cos40.cos60.cos80$ = ?
Second, determine of the value $tan9-tan27-tan63-tan81=$
Third, if $sin(A+B).cosC=2sinA .cos (B-C)$ prove that $tanA - \frac{sin(B-A)}{2sinAsinB}$
ill try the first one....
$cos20.cos40.cos60.cos80=$

$=cos20.\frac{sin20}{sin20}.cos40.cos60.cos80$

$=\frac{cos20.sin20}{sin20}.cos40.\frac{1}{2}.cos80$

$=\frac{\frac{1}{2}sin40}{sin20}.cos40.\frac{1}{2}. cos80$

remember:
$2sinXcosX=sin2X$
so, $sinXcosX=\frac{sin2X}{2}$
$=\frac{1}{2}\frac{1}{2} \frac{sin40.cos40}{sin20}.cos80$

$=\frac{1}{4} \frac{\frac{1}{2}sin80}{sin20}.cos80$

$=\frac{1}{4}\frac{1}{2} \frac{sin80.cos80}{sin20}$

$=\frac{1}{8} \frac{\frac{1}{2}sin160}{sin20}$

$=\frac{1}{16} \frac{sin160}{sin20}$

$=\frac{1}{16} \frac{sin(180-20)}{sin20}$

remember:
$Sin(180-x)=Sin(x)$
$\frac{1}{16} \frac{sin20}{sin20}$

$=\frac{1}{16}.1$

$=\frac{1}{16}$

........hope it help.........

4. Originally Posted by wizard654zzz
First, what is the value of $cos20.cos40.cos60.cos80$ = ?
Second, determine of the value $tan9-tan27-tan63-tan81=$
Third, if $sin(A+B).cosC=2sinA .cos (B-C)$ prove that $tanA - \frac{sin(B-A)}{2sinAsinB}$
HI

For (1)

$\cos 20\cdot \cos 40\cdot \cos 60\cdot \cos 80$

$=\frac{1}{2}(\cos 20\cdot \cos 40\cdot \cos 80)$

$=\frac{1}{2}(\cos 20 \cdot \frac{1}{2}(\cos 120+\cos 40))$

$=\frac{1}{2}(\cos 20\cdot \frac{1}{2}(-\frac{1}{2}+\frac{1}{2}\cos 40))$

$=\frac{1}{2}(-\frac{1}{4}\cos 20+\frac{1}{2}\cos 40\cos 20)$

$=\frac{1}{2}[-\frac{1}{4}\cos 20+\frac{1}{2}(\frac{1}{2}(\cos 60+\cos 20))]$

$=\frac{1}{2}(\frac{1}{8}+\frac{1}{4}\cos 20-\frac{1}{4}\cos 20)$

$
=\frac{1}{16}
$

5. Originally Posted by wizard654zzz
I revise for the third problems :
Third, if $sin(A+B).cosC=2sinA .cos (B-C)$ prove that $tanC = tanA - \frac{sin(B-A)}{2sinAsinB}$

THis is the furthest i can get ..

$\cos C\sin (A+B)=2\sin A\cos (B-C)$

$\cos C[\sin A\cos B+\cos A\sin B]=2\sin A[\cos B\cos C+\sin B\sin C]$

$\sin A\cos B\cos C+\cos A\cos C\sin B=2\sin A\cos B\cos C+2\sin A\sin B\sin C$

Divide both sides by $\cos C$

$\sin A\cos B+\cos A\sin B=2\sin A\cos B+2\sin A\sin B\tan C$

$\sin B\cos A-\cos B\sin A=2\sin A\sin B\tan C$

$\sin (B-A)=2\sin A\sin B\tan C$

$
\tan C=\frac{\sin (B-A)}{2\sin A\sin B}
$

$\tan A$ ??

6. ## Thank you

I would like to say thanks to pencil09 and mathaddict.
mathaddict you are right $tanC = \frac{sin(B-A)}{2sinAsinB}$ i mistyped the third problem..