Math Is Fun Forum

If you really can't get it I'll explain how to do it, but you really ought to learn to do this for yourself.  I wouldn't really be helping you if I just did the problem for you.

BLASPHEMY!

But I take it you meant, math in itself is useless - untill applied to something.

Also note the identiy for sin(2x) and the three identities for cos(2x).

I'd really really really reccomend printing out a reference sheet of all the indentities. I have a binder of lots of difference indentities and math notes. This is math, not spelling. Understanding why a formula works is more important then remembering it. Just print out a reference sheet.

"We're super excited," said Boone, a chemistry professor. "We've been looking for such a number for a long time."

Whats he excited about? lol. Chemists...

Trigonometric identies:

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

You should print out a reference sheet.

Ok, I'm going to use the four following identies to do this:

1.  sin^2 (x) + cos^2 (x) = 1

2.  sin(2x) = 2 sin x cos x

3.  cos (2x) = 2 cos^2 (x) - 1

4.  cos (A + B) = cos A cos B - sin A sin B

I'll refere to these as identities 1, 2, 3 and 4.

Ok. We begin with cos (3x) and note this can be written as cos (2x + 1) if we use identity 4 and let A = 2x and B = x we have:

cos(2x) cos(x) - sin(2x) sin(x)

Now I'll substitue cos(2x) using identity 3:

(2cos^2 (x) -1) cos (x) - sin(2x) sin(x)

multiplied:

2 cos^3 (x) - cos(x) - sin(2x) sin (x)

Now I'll replace sin(2x) using identity 2.

2 cos^3 (x) -  cos(x) - 2sin x cos x sin x

simplified:

2 cos^3 (x) -  cos(x) - 2 cos (x) sin^2 (x)

in indentiy 1, we can solve for sin^2 (x) to find sin^2(x) = 1 - cos^2(x) we replace sin^2 (x) with (cos^2 (x) -1)

2 cos^3 (x) -  cos(x) - 2 cos (x)(1 -cos^2(x))

multiplied:

2 cos^3 (x) -  cos(x) + - 2 cos (x) + 2 cos^3(x)

Added like terms:

4 cos^3(x) -3 cos(x)    Quod Erat Demonstratum! That which was to be demonstrated:-)

lol. I was just really confused for a split second.

forgive me if I'm wrong but isn't there a x^5/5! missing there?

I'd love to! Saving to photobucket and copying it here is a real pain! But last couple times I tried that it didn't work. Let me try it again:

f''''(x) = (16x^4 - 12) sin (x^2) - 48x^2 cos (x^2)

f'''''(x) = (160x^3) sin (x^2) + (32x^5 -120x) cos (x^2)

f''(x) = -4x^2 sin (x^2)  + 2 cos (x^2)

f'''(x) = -8x^3 cos (x^2) - 12x sin (x^2)

I'll right, I just had some crackers and milk so I'm ready to relive the nightmare again. I'll Find the derivatives one at a time and post them in separate replies, so you guys can check each one individually, (if you care to).

Ok. Here we go:

f(x) = sin(x^2)

f'(x) = 2xcos(x^2)

Oops! Thats it! I wrote Area = (17/4)x^2 -225x + 7225 , the middle coefficient is supposed to be -255, not -225. I checked my answer with the books anwser but I must have missed that since the numbers look so similar. All right I'll try generating that equation again.

Ok, just did the problem again. Ok this is scary, I multiplied 340*(-6)/8 on my calculator and got -255, I copied the number on paper and my hand wrote -225. Oh no..... I thought I had dislexia but now I'm sure of it! x_x