Ok, You guys never decided on this and I did not want JUST a help forum. I want to make sure that people flex those math muscles. Strictly speaking, this is going to be a thread where I want people to post random math problems they are having trouble with or just want to warm up the day. I searched and it appears this problem was never solved:
2sin²x - 6sinxcosx + 6cos²x=1
x is in [0°, 360°]
I have no idea how to solve this one. Poor Bluet. Let's help him out.
2sin²x - 6sinxcosx + 6cos²x=1
x is in [0°, 360°]
I have no idea how to solve this one. Poor Bluet. Let's help him out.
2 sin²(x) - 6 sin(x)cos(x) + 6 cos²(x)
= 2 sin²(x) - 3 sin(2 x) + 6 cos²(x) [double-angle formula]
= 2 sin²(x) - 3 sin(2 x) + 2 cos²(x) + 4 cos²(x)
= 4 cos²(x) - 3 sin(2 x) + 2 [pythagorean identity]
= 4 [1/2 + cos(2 x)/2] - 3 sin(2 x) + 2 [power-reduction formula]
= 2 cos(2 x) - 3 sin(2 x) + 4
= sqrt(13) sin(2x + π - arcsin(2/sqrt(13))) + 4
=> sqrt(13) sin(2x + π - arcsin(2/sqrt(13))) + 4 = 1.
=> sin(2x + π - arcsin(2/sqrt(13))) + 4 = -3/sqrt(13).
=> 2x + π - arcsin(2/sqrt(13)) = arcsin(-3/sqrt(13)).
=> 2x = arcsin(-3/sqrt(13)) - π + arcsin(2/sqrt(13)).
=> x = arcsin(-3/sqrt(13))/2 - π/2 + arcsin(2/sqrt(13))/2.
simplify? *slits wrists*
= 2 sin²(x) - 3 sin(2 x) + 6 cos²(x) [double-angle formula]
= 2 sin²(x) - 3 sin(2 x) + 2 cos²(x) + 4 cos²(x)
= 4 cos²(x) - 3 sin(2 x) + 2 [pythagorean identity]
= 4 [1/2 + cos(2 x)/2] - 3 sin(2 x) + 2 [power-reduction formula]
= 2 cos(2 x) - 3 sin(2 x) + 4
= sqrt(13) sin(2x + π - arcsin(2/sqrt(13))) + 4
=> sqrt(13) sin(2x + π - arcsin(2/sqrt(13))) + 4 = 1.
=> sin(2x + π - arcsin(2/sqrt(13))) + 4 = -3/sqrt(13).
=> 2x + π - arcsin(2/sqrt(13)) = arcsin(-3/sqrt(13)).
=> 2x = arcsin(-3/sqrt(13)) - π + arcsin(2/sqrt(13)).
=> x = arcsin(-3/sqrt(13))/2 - π/2 + arcsin(2/sqrt(13))/2.
simplify? *slits wrists*
Here is an easy one that was on my practice test:
What does this evaluate to?
sin²x + cos²x(cot²x)
Sorry for the trig. Of course, I am taking precal now so most of these WILL be skewed.
What does this evaluate to?
sin²x + cos²x(cot²x)
Sorry for the trig. Of course, I am taking precal now so most of these WILL be skewed.
Aw no takers?
put up some calc!
I'll have some calc to put up pretty quick. who wants some inverse trig derivatives?
Sure :)
More trig:
D[1/(sinx{5+4cos²x})]
D[1/(sinx{5+4cos²x})]
-(sinx{5+4cos^2x})^-2*cosx{5+4cos^2x}*8cosx*-sinx?
Correct! Can you simplify even more?
-(sinx{5+4cos^2x})^-2*8cos^2x{5+4cos^2x}*-sinx
see, this is what fucks me on the calc tests, simplifying rapes me.
see, this is what fucks me on the calc tests, simplifying rapes me.
I think TTF needs a digital whiteboard app.
heh integrated latex =P
alright, this was on my last test that just completely f'd me up
using the formula (1+x)^k roughly = 1+kx
solve
cube root((1-(1/(2+x)))^2)
if more explanation is needed i can most definitely try to provide some :p
using the formula (1+x)^k roughly = 1+kx
solve
cube root((1-(1/(2+x)))^2)
if more explanation is needed i can most definitely try to provide some :p
solve what?
cube root((1-(1/(2+x)))^2) = 0 ?
we need a complete proposition
cube root((1-(1/(2+x)))^2) = 0 ?
we need a complete proposition
your finding the linearization of that using the approximation (1+x)^k≈1+kx and yes, your assuming x is near 0
sorry i wasn't more exact, i was still quite pissed about that calc test.
sorry i wasn't more exact, i was still quite pissed about that calc test.
well, i guess i had no reason to worry as my answer was right! if anyone's curious i could post the answer
i'd be interested to see your answer, since i still don't quite know what the question was
the answer was 1+(2/3)(-(1/(2+x)))
really, all your looking for is the tangent line. normally you would be using the equation
L(a)=f(a)+f'(a)(x-a)
though in this case since your assuming a=0 you can use the formula (1+x)^k≈1+kx
i guess it would have been easier to see if the equation were written like this:
(1-(1/(2+x))^(2/3)
really, all your looking for is the tangent line. normally you would be using the equation
L(a)=f(a)+f'(a)(x-a)
though in this case since your assuming a=0 you can use the formula (1+x)^k≈1+kx
i guess it would have been easier to see if the equation were written like this:
(1-(1/(2+x))^(2/3)
but the ambiguity remains. can you solve (1-(1/(2+x))^(2/3) without knowing what it is supposed to equal?
'simplification' is boring and meaningless. proofs are everything:
A is a random variable
B is a random variable
a is a scalar.
b is a scalar.
E is the expected value function.
Cov is the covariance function. for a variable A and B: Cov(A,B) = E([A-E(A)][B-E(B)])
Var is the variance function and is defined Var(A)=Cov(A,A).
1. show that E(aA+bB)=aE(A)+bE(B)
2. show that the Var(aA+bB)=a^2 Var(A)+b^2 Var(B)+ ab 2Cov(A,B)
3. if that was easy. when is the var(aA) = var(aA+bB)? there should be two answers to this question.
A is a random variable
B is a random variable
a is a scalar.
b is a scalar.
E is the expected value function.
Cov is the covariance function. for a variable A and B: Cov(A,B) = E([A-E(A)][B-E(B)])
Var is the variance function and is defined Var(A)=Cov(A,A).
1. show that E(aA+bB)=aE(A)+bE(B)
2. show that the Var(aA+bB)=a^2 Var(A)+b^2 Var(B)+ ab 2Cov(A,B)
3. if that was easy. when is the var(aA) = var(aA+bB)? there should be two answers to this question.
no you can't. if it is not specified what x equals you can't linearize it. but on the test he was just implying that x=0 since he stated we had to use (1+x)^k≈1+kx, as that can only be used when x=0.
if x is anything but 0 you must use L(a)=f(a)+f'(a)(x-a).
if x is anything but 0 you must use L(a)=f(a)+f'(a)(x-a).
right, that is a first order Taylor series approximation.
for some reason, this probability statement is confusing the shit out of me.
the probability that a specific user is transmitting is .1 (10 percent). If there are 35 users, the probability that 11 or more are transmitting simultaneously is .0004.
I don't have to do any problems based on it, but I just can't figure out how they arrived at the number.
the probability that a specific user is transmitting is .1 (10 percent). If there are 35 users, the probability that 11 or more are transmitting simultaneously is .0004.
I don't have to do any problems based on it, but I just can't figure out how they arrived at the number.
oooooooooooooooooooooooooooooooo
Anyone want to see me do a Gram-Schmidt problem?
um ok
Yeah, sure, why not?