# Math Help - Proving Trig Formulas

1. ## Proving Trig Formulas

Hey, I have a few questions I just can't seem to get down, looking for some help here...
1) sin2x = 2csc2x-tanx
1-cos2x

2) cos(x+y)cos(x-y)=(cosx)^2 +(cosy)^2 -1

3) (cosx)^6 + (sinx)^6 = 1-3(sinx)^2 + 3(sinx)^4

2. Originally Posted by orca432
Hey, I have a few questions I just can't seem to get down, looking for some help here...
1) sin2x = 2csc2x-tanx
1-cos2x

2) cos(x+y)cos(x-y)=(cosx)^2 +(cosy)^2 -1

3) (cosx)^6 + (sinx)^6 = 1-3(sinx)^2 + 3(sinx)^4

You need to show you work too..

(2) $cos(x+y)cos(x-y) = (cosx.cosy-sinx.siny)(cosx.cosy+sinx.siny)$

$= (cos^2x.cos^2y) - (sin^2x.sin^2y)$

$= (cos^2x.cos^2y) - [(1-cos^2x)(1-cos^2y)]$

multiply and cancel to complete the proof

3. for (3)

$(cosx)^6+(sinx)^6$

$= (cos^{2}x)^3 + (sin^{2}x)^3$

since $(a^3+b^3) = (a+b)(a^2-ab+b^2)$, you can write:

$= (cos^{2}x+sin^{2}x)(cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)$

$= 1 (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)$

convert all cosines into sines and your proof will be complete.

4. Originally Posted by harish21
for (3)

$(cosx)^6+(sinx)^6$

$= (cos^{2}x)^3 + (sin^{2}x)^3$

since $(a^3+b^3) = (a+b)(a^2-ab+b^2)$, you can write:

$= (cos^{2}x+sin^{2}x)(cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)$

$= 1 (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)$

convert all cosines into sines and your proof will be complete.
When I converted them, I got $1-(sinx)^2+(sinx)^4$, anyone care to help me?

5. Originally Posted by orca432
When I converted them, I got $1-(sinx)^2+(sinx)^4$, anyone care to help me?
$cos^{4}x = (cos^{2}x)^2= (1-sin^{2}x)^2=1-2sin^{2}x+sin^{4}x$

substitute this in what you had earlier on

$1 . (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)$

$=1-2sin^{2}x+sin^{4}x-[(1-sin^{2}x). sin^{2}x]+sin^{4}x$

finish it...

6. Here's a little more.

$\cos^{4}x-\cos^{2}x \sin^{2}x+\sin^{4}x$

$\cos^{2}x\cos^{2}x-\cos^{2}x \sin^{2}x+\sin^{4}x$

$(1-\sin^2{x})(1-\sin^2{x})-(1-\sin^2{x}) \sin^{2}x+\sin^{4}x$

Now expand this.