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+--- Thread: Calc/Trig Help (/showthread.php?tid=31117)
Calc/Trig Help - Corn - 2010-10-02
Simplify y = f(x) = arccos (cos^4x - sin^4x) ; just one out of infinity many answers is good.
Clearer problem.
Calc/Trig Help - 2147483647 - 2010-10-02
Use Wolfram Alpha.
Calc/Trig Help - Kaasoljoyyx - 2010-10-02
Corn Wrote:Simplify y = f(x) = arccos (cos^4x - sin^4x) ; just one out of infinity many answers is good.
Can you distinguish ( ) better on the 4x? It seems that you can use the sin^2 + xos^2 = 1 trick to simplify and see what happens from there
Calc/Trig Help - Corn - 2010-10-02
Oops, sorry, didn't clarify it good enough.
The 4 is the exponent, and the x is a regular old number....so like
I don't know the Wolfram Alpha. I'm also not quite sure you can pull the sin^x + cos^x = 1, as you can't pull out a ^2, can you?
Edit:
Calc/Trig Help - Hazzy - 2010-10-02
how2Wolfram Alpha
I would not call that "simplified".
Calc/Trig Help - Corn - 2010-10-02
Edit: Turns out you can distribute the powers, never mind.
Calc/Trig Help - Russt - 2010-10-02
Wait, what? No you can't. You can use difference of squares, though.
cos^4 - sin^4 = (cos^2)^2 - (sin^2)^2 = (cos^2 + sin^2)(cos^2 - sin^2) = 1(cos^2 - sin^2) = cos 2x.
arccos cos 2x = 2x.
Calc/Trig Help - Corn - 2010-10-02
Ah, I meant difference of squares.
Calc/Trig Help - OB3LISK - 2010-10-02
Dam Rust did it before me. Yeah it turns out y = 2x
Calc/Trig Help - shouri - 2010-10-02
Russt Wrote:Wait, what? No you can't. You can use difference of squares, though.
cos^4 - sin^4 = (cos^2)^2 - (sin^2)^2 = (cos^2 + sin^2)(cos^2 - sin^2) = 1(cos^2 - sin^2) = cos 2x.
arccos cos 2x = 2x.
Just have to be a bit careful with arccos ( cos 2x). Arccos maxes out at pi, where as 2x doesn't. So if you plug in pi to both sides of the equations and work them out the long way, you wont get the same answer.
Notedly you get 0 = 2*PI
Calc/Trig Help - 2147483647 - 2010-10-02
To add to what shouri said, if you graph the function f(x)=arccos(cos x), you end up with a wave, not a line. Subsequently, when you integrate it, you get another wave, not a parabola.
Calc/Trig Help - Corn - 2010-10-02
Another question. Expect more questions as I'm redoing a test and trying to figure out the ways to do the questions.
Say 4* (sin4x) / (sin2x) and I'm trying to find the limit as x->0.
Can I cancel out a sin2x from the numerator and denominator? If not, how do I find the limit?
Calc/Trig Help - OB3LISK - 2010-10-02
If you're allowed to do derivatives, take the derivative of the top and bottom seperately then plug in.
If not... then F3...Sec.
Calc/Trig Help - Corn - 2010-10-02
Can't do derivatives, or L'Hopital's rule for that matter. I basically can only do substitution, factoring, sinc functions, or trig identities.
Calc/Trig Help - Russt - 2010-10-02
Yeah, L'Hopital makes it nice:
4 sin 4x / sin 2x => 16 cos 4x / 2 cos 2x => 16 cos 0 / 2 cos 0 = 16/2 = 8.
But there's also another way, using the sine double angle identity:
4 sin 4x / sin 2x => 4 sin 2(2x) / sin 2x => 4 (2 sin 2x cos 2x) / sin 2x => 8 cos 2x => 8 cos 0 = 8.
Calc/Trig Help - Corn - 2010-10-02
Great, three problems out of ten left that I don't get!
2.
Edit: Solved all of them!
Calc/Trig Help - Corn - 2010-10-21
I really don't feel like making a new thread.
"Known (3,2) is a point on the curve 2x^2-^3-10=0. Find the value of y when x = 3.04; use differentials in computing the approximation."
Not algebra. Differentials.
Calc/Trig Help - Stereo - 2010-10-21
I can't read the formula... what is "-^3"?
Anyway, you probably want to basically add dx * d/dx (first approximation), then dx^2/2 * d^2/dx^2 (second approximation) and so forth until they become zero.
That is, if you know the value at a point (x = 3, y = 2) and you know the slope (d/dx = 3) then at x = 3.04, you have 0.04*3 + 2 = 2.12 as an approximation of (3.04, 2.12)
It's called the Taylor expansion, f(x + t) = f(x) + t/1!*f'(x) + t^2/2!*f''(x) + t^3/3!*f'''(x) + ...