-9 / 41a ^ 3n-a × B ^ 2n + 3 is a binomial of degree six, and the value of n is obtained

-9 / 41a ^ 3n-a × B ^ 2n + 3 is a binomial of degree six, and the value of n is obtained

3n-a+2n+3=6
5n=6-3+a
5n=3+a
n=(3+a)/5

The reduction results of (1-sin ^ 6 a-cos ^ 6 A) / (1-sin ^ 4 a-cos ^ 4 a)

(1-sin^6 a-cos^6 a)/(1-sin^4 a-cos^4 a)
=[1-(sin^6 a+cos^6 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^2 a+cos^2 a)(sin^4 a-sin^2 acos^2 a+cos^4 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^4 a-sin^2 acos^2 a+cos^4 a)]/(1-sin^4 a-cos^4 a)
=[1-sin^4 a+sin^2 acos^2 a-cos^4 a]/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/[(1-sin^4 a)-cos^4 a]
=1+sin^2 acos^2 a/[(1-sin^2 a)(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a-cos^2a)]
=1+sin^2 acos^2 a/[2sin^2 a*cos^2 a]
=1+1/2
=3/2

（1-sin(a)^6-cos(a)^6）/（1-sin(a)^4-cos(a)^4）
（1-sin(a)^6-cos(a)^6）/（1-sin(a)^4-cos(a)^4）

Then, let (SIN) @ (x) square = 1
Original formula = [1-x Cube - (1-x) cube] / [1-x square - (1-x) square]
=(3x-3x Square) / (2x-2x Square)
=3/2 .

Given sin a cos a = 1 / 3, find the value of (sin a-cos a) ^ 2

（sin a-cos a）^2
=sina^2 - 2sinacosa +cosa^2
= 1- 2sina cosa
= 1-2/3
= 1/3

We know that vector a = (3 / 4, sin θ) and vector b = (COS θ, 1 / 3). If vector a is parallel to vector B, then Tan θ=

I'll make it 1
From a / / b we know sin θ * cos θ = 1 / 4
Then divide both sides by cos ^ 2 (θ) to get Tan θ = 1 / [4cos ^ 2 (θ)] ①
Because sin ^ 2 (θ) + cos ^ 2 (θ) = 1
Similarly, divide both sides by cos ^ 2 (θ) to get Tan ^ 2 (θ) + 1 = 1 / cos ^ 2 (θ) ②
① (2) integrate and eliminate cos ^ 2 (θ) to get Tan ^ 2 (θ) - 2tan θ + 1 = 0, namely (Tan θ - 1) = 0
So tan θ = 1

Why is arcsinx limit equal to 0 when x tends to 0
As the title!

There are many ways to prove this, the more common one is to use the law of lobita
lim(x→0）arctanx/x
=lim(x→0）1/√(1-x^2)
=1
So when x → 0, arctanx and X are equivalent infinitesimals

Finding the limit SiNx ^ n / (SiNx) ^ m x to approach 0

When x → 0, SiNx ^ n → 0, cosx → 1, (SiNx) ^ m → 0, so SiNx ^ n / (SiNx) ^ m is of type 0 / 0. According to the law of lobita, Lim [SiNx ^ n / (SiNx) ^ m] (x → 0) = LIM (SiNx ^ n) '/ [(SiNx) ^ m]' (x → 0) = NX ^ (n-1) cosx / [M (SiNx) ^ M-1] cosx = NX ^ (n-1) / M (SiNx) ^ (m-1) continuous use of lobita

How to find the limit (1-sinx) ^ (1 / x) when x approaches positive infinity?
What if x can only be an integer?

lim[x→+∞] (1-sinx)^(1/x)
=lim[x→+∞] [(1-sinx)^(-1/sinx)]^(-sinx/x)
The second important limit is in the brackets, the limit is e, and the exponential limit outside the brackets is - 1
=1/e
If x takes only integers, it is equivalent to selecting a child column of this limit, and the result is the same
I hope I can help you. If you don't understand, you can ask. If you solve the problem, please click the "select as satisfactory answer" button below. Thank you

What is the right limit of SiNx degree of X at x = 0

Let y = x ^ SiNx
lny=sinx*lnx
=lnx/(1/sinx)
Using the law of lobida
=(1/x)/(-cosx/sin^x)
=-(sinx)^2/xcosx
=2sinxcosx/(cosx-xsinx)
Substitute x = 0 in
=0
So the limit of LNY is 0
So y tends to 1
therefore
The limit of SiNx power of X is 1

The limit of [x + 1 ^ 2 - X-1 ^ 2] / SiNx when x tends to 0

When x tends to zero, Lim [(x + 1) ^ 2 - (x-1) ^ 2] / SiNx
=lim[(x+1+x-1)(x+1-x+1)]/sinx
=lim(4x/sinx)
=4