# Help Prove trig identities

• November 4th 2008, 12:31 PM
getem j
Help Prove trig identities
i have 3 questions that i need help on for my test tommorw, so if you could answer any of them that would be great, thanks

prove each identity so L.S. = R.S.

1) sin x/2 = 1-cosx/2 (right side of equation is all under square root sign)

2)sin[cubed]x + cos[cubed]x = (1-sinxcosx)(sinx+cosx)

3) sin(squared)M [csc(squared)M + sec(squared)M] = sec(squared)M

any help is appreciated
• November 4th 2008, 01:21 PM
Soroban
Hello, getem j!

Quote:

$2)\;\;\sin^3\!x + \cos^3\!x \:=\: (1-\sin x\cos x)(\sin x+\cos x)$

$\underbrace{\sin^3\!x + \cos^3\!x}_{\text{sum of cubes}} \;=\;(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)$

. . . . . . . . . $= \;(\sin x + \cos x)(\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} - \sin x\cos x)$

. . . . . . . . . $=\;(\sin x + \cos x)(1 - \sin x\cos x)$

Quote:

$3)\;\;\sin^2\!x(\csc^2\!x + \sec^2\!x) \:=\: \sec^2\!x$

$\text{Multiply: }\;\sin^2\!x(\csc^2\!x + \sec^2\!x) \;\;=\;\;\underbrace{\sin^2\!x\csc^2\!x}_{\text{Th is is 1}} + \sin^2\!x\sec^2\!x \;\;=\;\;1 \;+\; \sin^2\!x\cdot\frac{1}{\cos^2\!x}$

. . . . . . $= \;\;1 + \frac{\sin^2\!x}{\cos^2\!x} \;\;=\;\;1 + \left(\frac{\sin x}{\cos x}\right)^2 \;\;=\;\;1 + \tan^2\!x \;\;=\;\;\sec^2\!x$

• November 4th 2008, 03:13 PM
getem j
thanks again Soroban

you have ne suggestions for the first one?