# TRIG!

posted by hayden

Posted by hayden on Monday, February 23, 2009 at 4:05pm.

sin^6 x + cos^6 x=1 - (3/4)sin^2 2x

work on one side only!

Responses

LS looks like the sum of cubes
sin^6 x + cos^6 x
= (sin^2x)^3 + (cos^2x)^3
= (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x)
= (1)(sin^4x - (sin^2x)(cos^2x) + sin^4x)

Now let's do some "aside"
(sin^2x + cos^2)^2 would be
sin^4x + 2(sin^2x)(cos^2x) + cos^4x

we 'almost' have that above, differing only by the coefficient of the middle term. We can fix that by saying
(sin^4x - (sin^2x)(cos^2x) + sin^4x)
= sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 4(sin^2x)(cos^2x)
= (sin^2x + cos^2x) - 3(sin^2x)(cos^2x)
= 1 - 3(sin^2x)(cos^2x)
almost there!
recall sin 2A = 2(sinA)(cosA)
so 3(sin^2x)(cos^2x)
= 3(sinxcosx)^2
= 3((1/2)sin 2x)^2
= (3/4)sin^2 2x

so
1 - 3(sin^2x)(cos^2x)
= 1 - (3/4)sin^2 2x
= RS !!!!!!

Q.E.D.

(sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x) I DON'T AGREE WITH THIS

SHOULDNT IT BE
(sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + COS^4 X ??

1. Reiny

of course, you are right,

also check up on this part:

we 'almost' have that above, differing only by the coefficient of the middle term. We can fix that by saying
(sin^4x - (sin^2x)(cos^2x) + sin^4x)
= sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 4(sin^2x)(cos^2x)

should say:
we 'almost' have that above, differing only by the coefficient of the middle term. We can fix that by saying
(sin^4x - (sin^2x)(cos^2x) + cos^4x)
= sin^4x + 2(sin^2x)(cos^2x) + cos^4x - 3(sin^2x)(cos^2x)
= (sin^2x + cos^2x)^2 - 3(sin^2x)(cos^2x)

sorry about the typos, one has to be so careful with this crazy trig stuff

2. Anonymous

cos^2x+3sinxcosx=2

## Similar Questions

1. ### trig

Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The quantity …
2. ### trig

it says to verify the following identity, working only on one side: cotx+tanx=cscx*secx Work the left side. cot x + tan x = cos x/sin x + sin x/cos x = (cos^2 x +sin^2x)/(sin x cos x) = 1/(sin x cos x) = 1/sin x * 1/cos x You're almost …
3. ### Math repost - girly93

Posted by girly93 on Monday, February 9, 2009 at 2:55am. i don't understand the sine and cosine rules. 1/2 ab sinC and cosA= b2 + c2 - a2 what?
4. ### Chemistry

At what initial concentration would a solution of acetic acid (ka=1.8x10^-5) be 2% ionized?

sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only!
6. ### trig

The expression 4 sin x cos x is equivalent to which of the following?
7. ### math

Posted by connie on Monday, July 20, 2009 at 11:56pm. Find the indicated outputs for f(x) = 2x²-4x f(0)= ____?
8. ### Math(Geometry)

To Lauren - Ms. Sue, Monday, September 28, 2009 at 10:24pm What post?
9. ### trig 26

simplify to a constant or trig func. 1. sec ²u-tan ²u/cos ²v+sin ²v change expression to only sines and cosines. then to a basic trig function. 2. sin(theta) - tan(theta)*cos(theta)+ cos(pi/2 - theta) 3. (sec y - tan y)(sec y + …
10. ### pre calc trig check my work please

sin x + cos x -------------- = ? sin x sin x cos x ----- + ----- = sin x sin x cos x/sin x = cot x this is what i got, the problem is we have a match the expression to the equation work sheet and this is not one of the answers. need

More Similar Questions