Cos (3pi / 2 + 2a) given cosa = 1 / 3

Cos (3pi / 2 + 2a) given cosa = 1 / 3

cosa=1/3
sina=±√(1-cos^2a)=±2√2/3
cos(3pi/2+2a)
=sin2a
=2sinαcosα
=2*(1/3)(±2√2/3)
=±4√2/9
Given that cosa = - 5 / 13, a belongs to (π, 3 π / 2), what is the value of sin (π / 3-A) and COS (π / 3-A)?
Cosa = - 5 / 13, a belongs to (π, 3 π / 2),
sina=-12/13
sin(π/3-a)=√3/2cosa-1/2sina
=(12-5√3)/26
cos(π/3-a)=1/2cosa+√3/2sina
=(5-12√3)/26
How to get the derivative of LNX square,
lnx^2=2lnx
(lnx^2)'=(2lnx)'=2/x
[(lnx)^2]'=2lnx/x
Let u = LNX be the square of y = U
First, find the derivation y '= 2U of Y with respect to u, and then multiply by the derivation u' = 1 / X of u with respect to X
So the derivation of Y with respect to X is y '= (2lnx) / X
Find the minimum value of the function y = SiNx + 2sinxcosx + 3cosx and the set of X at this time
Y = SiNx + 2sinxcosx + 3cosx
=1+sin2x+2cosx*cosx
=2+sin2x+cos2x
=2 + (2 square root) sin (2x + Pai / 4)
You can calculate the result
How to find the derivative of the square of (LNX) / x,
[(Inx)^2/x]'=[2Inx*(1/x)*x-(Inx)^2]/x^2=[2Inx-(Inx)^2]/x^2
The formula (A / b) '= (a'b-ab') / b ^ 2 should be used
Pay attention to the chain rule!
"'" means derivation and "^ 2" means square
Square of (LNX) / x = [(LNX's Square) 'X - (x)' (LNX)] / x square = [2xlnx (LNX) '(LNX's Square)] / x square = [2lnx - (LNX's Square)] / x square
The derivative of compound function, I suggest you read a book
First, the derivative of the outer circumference is multiplied by the derivative of the inner circumference function to INX / X
What is the minimum value of the known function FX = root of half three sin2x - half (cosx square minus SiNx Square) minus one
-Can you give me the formula and calculation process?
Derivative of LNX square
From the derivative method of compound function
y'=2lnx*(lnx)'=(2lnx)/x
It is necessary to find the maximum and minimum of the square X of the function y = 2-5 / 4sinx cos
y=2-4/5sinx-(1-sin²x)
=2-4/5sinx-1+sin²x
=sin²x-4/5sinx+1
=(sinx-2/5)²+21/25
When SiNx = 2 / 5, the minimum value of Y is 21 / 25
When SiNx = - 1, the maximum value of y = (- 1-2 / 5) &# 178; + 21 / 25 = 49 / 25 + 21 / 25 = 14 / 5
There are too many processes to type. First of all, you can see that y is greater than zero. You square both sides at the same time, find the minimum and maximum after the square, and then square
(2-4sinx / 5-cos ^ 2x) = (2-4sinx / 5-cos ^ 2x) = (2-4sinx / 5-cos ^ 2x) = (2-4sinx / 5) cosx-2cosx (- 4 / 5) cosx-2cosx (- 4 / 5) cosx + 2 SiNx cosx let y (39; = (2-4sinx / 5 / 5) cosx = (2-4sinx / 5) when cosx = 0, SiNx = ± 1, the above equation holds, at this time, y = 2 - (4 / 5) SiNx = 2 - (4 / 5 (4 / 5) SiNx = 6 / 5 or 14 / 5 or 14 / 5, when cosx is not equal to 0, the above equation can get SiNx = -2-2 = -2 / 5-2 / 5, at this time, the cos ^ 2x = 1-2x = 21 / 25, then y = 2 = 2-4 / 25, y = 2-4 / 4 / 4 / based on the above conclusions, ymax = 14 / 5, ymax = 6 / 5 & nbsp;
Finding the derivative of F (x) = x square * SiNx * LNX
A:
f(x)=x²(sinx)lnx
Derivation:
f'(x)=2x(sinx)lnx+x²[(sinx)lnx]'
=2x(sinx)lnx+x²[(cosx)lnx+(sinx)(1/x)]
=2x(sinx)lnx+x²(cosx)lnx+x(sinx)
It is x * SiNx + 2x * LNX * SiNx + LNX * cosx * x ^ 2
Find the maximum and minimum of the function y = - cos ^ 2x-4sinx + 8
Thank you very much!
y= -cos^2 x-4sinx+8=sin^2 x-1-4sinx+8=sin^2 x-4sinx+7
Let SiNx = Z, so - 1
y=(sinx)^2-4sinx+7=(sinx-2)^2+3
Maximum 12, minimum 4
Y = - cos ^ 2 x-4sinx + 8 = sin ^ 2x-4sinx + 7 let t = SiNx (- 1)