extending trigonometry

Posted July 23rd, 2010 by aloe

Trigonometry questions

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$

$\tan(\alpha+\beta)=\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$\tan(\alpha-\beta)=\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$\sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2}$

$\sin x-\sin y=2\sin \frac{x-y}{2}\cos \frac{x+y}{2}$

$\cos x+\cos y =2\cos \frac{x+y}{2}\cos \frac{x-y}{2}$

$\cos x-\cos y =-2\sin\frac{x+y}{2}\sin \frac{x-y}{2}$

$\sin 2x=2\sin x\cos x$
$\cos 2x=\cos^2 x-\sin^2 x=2\cos^2 x-1=1-2\sin^2 x$

$\sin 3x=3\sin x-4\sin^3 x$
$\cos 3x=4\cos^3 x-3\cos x$

$\tan 2x=\frac{2\tan x}{1-\tan^2x}$

$\tan 3x=\frac{\tan 2x+\tan x}{1-\tan 2x\tan x}=\frac{\frac{2\tan x}{1-\tan^2x}+\tan x}{1-\frac{2\tan x}{1-\tan^2x}\tan x}=\frac{3\tan x-\tan^3 x}{1-3\tan^2x}$

1. given that cot @ = surd 2, find the value of 1/ ( cos square @ - 2sin@cos@)
$\cot x=\sqrt{2}\: than$

$\frac{1}{\cos^2{x}-2\sin x\cos x}=\frac{\cos^2{x}+\sin ^2x}{\cos^2{x}-2\sin x\cos x}=\frac{\frac{\cos^2{x}+\sin ^2x}{\sin^2{x}}}{\frac{\cos^2{x}-2\sin x\cos x}{\sin^2{x}}}=\frac{\cot^2x+1}{\cot^2x-2\cot x}=\frac{2+1}{2-2\sqrt {2}}$

2. given that sin @ cot @ = 60/169, pai/4 <= @ < = pai/2, find the value of tan @
3. given that tan @ + cot @ =9/12, find the value of tan square @ + sec @ cosec @ + cot square @
4. given that tan @ = 2, find values of :
i. (4sin@ -2 cos@) / ( 5sin @ + 3sin @)
ii. 2/3 (sin square @) + 1/4 ( cos square @)
iii. sin @ cos cube @
5. simplify the followings:
i. surd (1-2sin10degreecos10degree) / (cos 10 degree - surd (1-sin square 100))

$\frac{ \sqrt{1-2\sin\frac{\pi}{18}\cos\frac{\pi}{18}}}{\cos\frac{\pi}{18}-\sqrt{1-\sin^2\frac{10\pi}{18}}}=\frac{\cos\frac{\pi}{18}-\sin\frac{\pi}{18}}{\cos\frac{\pi}{18}-\sin\frac{\pi}{18}}=1$

ii. tan (90 degree - @ ) - ((cot squarre @ sin square (90 degree - @)) / (cot @+cos @))
6. find the values of the following
tan (pai/5) + tan (2pai/5) + tan (3pai/5) + tan (4pai/5)

$\tan\frac{\pi}{5}+\tan\frac{2\pi}{5}+\tan\frac{3\pi}{5}+\tan\frac{4\pi}{5}=0$

7. prove the following identities
i. (sec @ -cosec @) / (tan @ - cot @) = (tan @ + cot @)/ (sec @ + cosec @)
ii. ( 1-sin@ + cos @) / (1-sin @) = ( 1 + sin @ + cos @ ) / cos @

p/s: please help me answer these questions...sorry i need to spell some pf math's symbol because i don't know how to write it in symbol... help me as soon as possible...thanks!!!