# Proving Trig Formulas

• May 27th 2010, 01:06 PM
orca432
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

• May 27th 2010, 01:35 PM
harish21
Quote:

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

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

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

multiply and cancel to complete the proof
• May 27th 2010, 01:46 PM
harish21
for (3)

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

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

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

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

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

convert all cosines into sines and your proof will be complete.
• May 27th 2010, 03:13 PM
orca432
Quote:

Originally Posted by harish21
for (3)

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

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

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

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

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

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

When I converted them, I got $\displaystyle 1-(sinx)^2+(sinx)^4$, anyone care to help me?
• May 27th 2010, 03:54 PM
harish21
Quote:

Originally Posted by orca432
When I converted them, I got $\displaystyle 1-(sinx)^2+(sinx)^4$, anyone care to help me?

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

substitute this in what you had earlier on

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

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

finish it...
• May 27th 2010, 03:55 PM
pickslides
Here's a little more.

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

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

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

Now expand this.