trig identities (2 Viewers)

kazemagic · Aug 21, 2012

Sup guys, how do u do this question? I dont know how to find the value for B >.>
Thanks

SpiralFlex · Aug 21, 2012

dat broken image is cute (srs)

SpiralFlex · Aug 21, 2012

$Draw le triangle and use the Pythagoras$

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

$=\frac{-\sqrt{5}}{3}*\frac{-3}{\sqrt{13}}-\frac{2}{3}*\frac{-2}{\sqrt{13}}$

$\frac{3\sqrt{5}+4}{3\sqrt{13}}$

kazemagic · Aug 21, 2012

SpiralFlex said:
$\cos (A+B)=\cos A \cos B - \sin A \sin B$

$Draw le triangle and use the Pythagoras$

$=\frac{-\sqrt{5}}{3}*\frac{-3}{\sqrt{13}}-\frac{2}{3}*\frac{-2}{\sqrt{13}}$

$\frac{3\sqrt{5}+4}{3\sqrt{13}}$

but for angle B its larger than 180 degrees O_O u still can draw a triangle for that?

edit: kk nvm thanks

SpiralFlex · Aug 21, 2012

Updated with le diagram.

kazemagic · Aug 23, 2012

idk how to do B :/

shravan_872 · Aug 23, 2012

I need to do it on paper but I think you put the pi/8 into the LHS and then simplify as usual...

Sy123 · Aug 23, 2012

kazemagic said:
idk how to do B :/

$\frac{1-\cos 2x}{1+\cos 2x}=\tan^2 x \ \ \ $From a$ \\ \\ x=\frac{\pi}{8} \ \Rightarrow \frac{1-\cos 2 \frac{\pi}{8}}{1+\cos 2 \frac{\pi}{8} } = \tan^2 \frac{\pi}{8} \\ \\ \tan \frac{\pi}{8}=\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}} \\ \\ = \sqrt{\frac{ \sqrt{2}-1}{\sqrt{2}+1 }}$

kazemagic · Aug 23, 2012

shravan_872 said:
I need to do it on paper but I think you put the pi/8 into the LHS and then simplify as usual...

thx, but dunno wat ur on abt lol

kazemagic · Aug 23, 2012

Sy123 said:
$\frac{1-\cos 2x}{1+\cos 2x}=\tan^2 x \ \ \ $From a$ \\ \\ x=\frac{\pi}{8} \ \Rightarrow \frac{1-\cos 2 \frac{\pi}{8}}{1+\cos 2 \frac{\pi}{8} } = \tan^2 \frac{\pi}{8} \\ \\ \tan \frac{\pi}{8}=\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}} \\ \\ = \sqrt{\frac{ \sqrt{2}-1}{\sqrt{2}+1 }}$

O thx, but how do u know cos2pi/8 = 1/root 2?

Carrotsticks · Aug 23, 2012

cos(2pi/8) = cos(pi/4) = 1/root2

Sy123 · Aug 23, 2012

Because $2 \times \frac{\pi}{8} = \frac{\pi}{4}$
And we know that $\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$

We know this due to the exact ratios the right angle triangles make when both their angles are 45 degrees (you surely must know this)

kazemagic · Aug 23, 2012

Carrotsticks said:
cos(2pi/8) = cos(pi/4) = 1/root2

Sy123 said:
Because $2 \times \frac{\pi}{8} = \frac{\pi}{4}$
And we know that $\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$

We know this due to the exact ratios the right angle triangles make when both their angles are 45 degrees (you surely must know this)

kk thx guys

ye i do know exact values lol

kazemagic · Aug 24, 2012

Hey guys, how do u do Q23? I keep on getting that tan alpha = 1/4 :S
EDIT: Nvm

kazemagic · Aug 24, 2012

kazemagic said:
Hey guys, how do u do Q23? I keep on getting that tan alpha = 1/4 :S

NVM, i did a careless mistake omfg

kazemagic · Sep 5, 2012

New question ^.^, how do I prove the below equation without using triple angle formula? Thx!!

SpiralFlex · Sep 5, 2012

$\frac{\sin 3A}{\sin A}+\frac{\cos 3A}{\cos A}$

$=\frac{\sin 3A \cos A + \cos 3A \sin A}{\sin A \cos A}$

$=\frac{\sin 4A}{\frac{1}{2}\sin 2A}$

$=\frac{2\sin 2A \cos 2A}{\frac{1}{2}\sin 2A}$

$=4\cos 2A$

kazemagic · Sep 5, 2012

SpiralFlex said:
$\frac{\sin 3A}{\sin A}+\frac{\cos 3A}{\cos A}$

$=\frac{\sin 3A \cos A + \cos 3A \sin A}{\sin A \cos A}$

$=\frac{\sin 4A}{\frac{1}{2}\sin 2A}$

$=\frac{2\sin 2A \cos 2A}{\frac{1}{2}\sin 2A}$

$=4\cos 2A$

thank you so much!

Sy123 · Sep 5, 2012

$\frac{\sin{A+2A}}{\sin{A}}+\frac{\cos{A+2A}}{\cos {A}} \\ =\frac{\sin A \cos {2A}+\cos A \sin{2A}}{\sin {A}}+\frac{\cos A \cos 2A - \sin A \sin {2A}}{\cos {A}} \\ \\ = \frac{\sin A \cos 2A \cos A + \cos^2 A \sin 2A+ \sin A \cos A \cos 2A -\sin^2 A \sin 2A}{\sin A \cos A} \\ \\ =\frac{\sin 2A \cos 2A+\sin 2A (\cos^2 A - \sin^2 A}{\sin A \ cos A} \\ \\ =\frac{2\sin 2A \cos 2A}{\frac{1}{2}\sin 2A } \\ \\ =4\cos 2A$

EDIT: Beaten, Spiral's method is very elegant

kazemagic · Sep 5, 2012

Sy123 said:
$\frac{\sin{A+2A}}{\sin{A}}+\frac{\cos{A+2A}}{\cos {A}} \\ =\frac{\sin A \cos {2A}+\cos A \sin{2A}}{\sin {A}}+\frac{\cos A \cos 2A - \sin A \sin {2A}}{\cos {A}} \\ \\ = \frac{\sin A \cos 2A \cos A + \cos^2 A \sin 2A+ \sin A \cos A \cos 2A -\sin^2 A \sin 2A}{\sin A \cos A} \\ \\ =\frac{\sin 2A \cos 2A+\sin 2A (\cos^2 A - \sin^2 A}{\sin A \ cos A} \\ \\ =\frac{2\sin 2A \cos 2A}{\frac{1}{2}\sin 2A } \\ \\ =4\cos 2A$

EDIT: Beaten, Spiral's method is very elegant

:O thanks! I'll take a look at it when I finish this question ^^
always good to know more than 1 method

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