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Find the minimum positive period of function y = (TaNx + Cotx) / SiNx
X = k π / 2 and y = (TaNx + Cotx) / SiNx = 2 / [sin2xsinx] the minimum positive period is 2 π,
Do the function y = TaNx times SiNx?
y=1/tanx*sinx
=cosx 【tanx≠0】
The image is obtained by removing x = k π (K ∈ z) [i.e. all points at the maximum and minimum] from the cosine image
Make the image of the function y = 1 / TaNx * SiNx Can it be directly converted into cosx
You can't directly draw the definition domain first, because TaNx should be meaningful
So x is not equal to K π + π / 4
In addition, TaNx is not equal to 0
So x is not equal to K π
To sum up
x≠1/2 Kπ
The sum of the square of y to the power of M + 5 of 3x and the nth power of y to the third power of X is a monomial, then the m power of n =? eldest brother.
And are monomials, then two are congeners
So the times of X and y are equal
So m + 5 = 3
2=n
m=-2,n=2
The m power of n = the (- 2) power of 2 = 1 / 4
If the sum of the square of y to the M + 5th power of 3x and the n-th power of y to the cube of X is a monomial, what is the m-th power of N
That is, the two are similar items
So m + 5 = 3
2=n
So m = - 2, n = 2
So the original formula = 2 to the power of - 2 = 1 / 4
Calculus for indefinite integral There is the formula cscdx = ln|cscx cotx| + C And I push cscdx = 1 / sinxdx (C and integral symbols are not written) =1/(2sinx/2*cosx/2)dx =1/(2tanx/2*cos^2x/2)dx =1/(2tanx/2)dtanx/2 =1/2ln|tanx/2| So 1 / 2ln|tanx / 2| = ln|cscx cotx|? How? Or did I push it wrong? Ask your teachers
You have one more half of the one above yourself
Using double angle formula sin2x = 2sinxcosx,cos2x = 2cos ² x - 1 OK
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