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Ask a question, very urgent Oh/ If the function y = (m-2) x ^ (m ^ 2-m-3) is an inverse proportional function of X 1. Find the value of M 2. Write the analytic trial and the proportional coefficient of the function
Y = (m-2) x ^ (m ^ 2-m-3) is an inverse proportional function of X
be
m^2-m-3=-1
M-2 is not equal to 0
The solution
m=-1
Analytic trial and scale coefficient of function: - 3
y=-3/x
X SiNx = 1 / 6 x belongs to [0, π] What is the method
Firstly, the value range is calculated according to the definition domain,
Secondly, change the position of X and y
Finally, the value domain becomes the definition domain, and the definition domain becomes the value domain
The inverse trigonometric function SiNx = - 2 / 3 x of sin belongs to (- π, 0)
sinx=-2/3
Then x is in the second or third quadrant
X belongs to (- π, 0)
Then x is in the third quadrant, and X belongs to (- π, - 0.5 π)
sinx=-2/3
Then sin-x = 2 / 3
-X = arcsin2 / 3 ± 2n π or - x = π - arcsin2 / 3 ± 2n π
X = - arcsin2 / 3 ± 2n π or x = arcsin2 / 3 ± 2n π - π
Arcsin (2 / 3) belongs to (0,0.5 π)
Then - arcsin2 / 3 belongs to (- 0.5 π, 0)
Then n = 0
X = - arcsin2 / 3 or x = arcsin2 / 3 - π
Limx tends to infinite N2 (arctan (A / N) - arctan [A / (n + 1)] to find the limit! A South Korean table mate asked me - don't want to lose face, t t t
n→∞,
limn²[arctan(a/n)－arctan(a/(n+1))]
＝limn²[arctan(a/n－a/(n+1))/
(1+a²/n(n+1))]
＝limn²arctan[a/(n(n+1)+a²)]
= Liman 2 / [a 2 + n (n + 1)] (equivalent infinitesimal)
＝a
The value of arctan (1 / 3) + arctan (1 / 2) is?
Tan (arctan (1 / 3) + arctan (1 / 2)) = (tanarctan (1 / 3) + tanarctan (1 / 2)) / (1-tanarctan (1 / 3) * tanarctan (1 / 2)) = (1 / 3 + 1 / 2) / (1-1 / 3 * 1 / 2) = (5 / 6) / (5 / 6) = 1, so arctan (1 / 3) + arctan (1 / 2) = Pai / 4
Calculate: arctan (1-x) + arctan (1 + x)
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan(arctan(1-x)+arctan(1+x))
= (1-x+1+x)/(1-(1-x)(1+x))
= 2/x^2
arctan(1-x)+arctan(1+x)
= arctan(2/x^2)
Wu is the PI and arctan is the sign of anti trigonometric function Why is LN 2 / wuarctanx equal to 2 / wuarctanx-1? I can prove that it is equal through special cases, but I can't figure out how to prove it. It's still not possible to find the derivation on both sides. Please have a look at the students who are better at mathematics. They should have learned in high school mathematics Did I get it wrong? Two days ago, I saw a problem to find the limit. The problem is this: the limit of LIM (2 / π arctanx) ^ x (x approaches), which gives the step equal to e ^ LIM (2 / π arctanx - 1) X. This problem is to find 0 ^ OO (infinite type of 0) in the limit. It should be equal to e ^ Lim ln (2 / π arctanx) ^ X. after simplification, it will be e ^ Lim ln XLN (2 / π arctanx). Let's take a look at the steps it gives us
Constructor f (x) = LN 2 / π arctanx - (2 / π arctanx-1)
If the two are equal, then there should be f '(x) = 0
f'(x)=[ln 2/πarctanx-(2/πarctanx-1)]'
=1/[2/πarctanx]*(2/πarctanx)'-(2/πarctanx)'
Obviously, f '(x) is not equal to 0, that is, f (x) is not a constant
Therefore, the two are not identical
I want to know the algorithm of anti trigonometric function. For example, cosx = 3, then x = arccos3, what is the specific angle? I mean, if cosx = 1 / 2, then the angle X means x = arccos1 / 2. What I want to know is how to calculate the angle X,
It can be calculated in Casio calculator. Press shlft, press COS-1, input 1 / 2, equal to the value, and then press. It is equal to the angle. First of all, the value must be between 1-1
Solution of inverse trigonometric function X belongs to (π, 3 π / 2) cosx=-3/5 Remember to change the X range to 0, π first, and forget how,
x∈(π,3π/2)
Then x - π∈ (0, π / 2)
cos(x-π）=-cosx=3/5
∴x-π=arccos3/5
x=π+arccos3/5
Solution of inverse trigonometric function A / sin θ + B / cos θ = L
Using universal formula
Sin θ = 2tan (θ / 2) / [1 + the square of Tan (θ / 2)]
Cos θ = [1-tan (θ / 2) squared] / [1 + Tan (θ / 2) squared]
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