Known vector a = (1 + sin2x, SiNx cosx), vector b = (1, SiNx + cosx), f (x) = vector a * vector B If f（ α）= 8 / 5.4 α Value of

Known vector a = (1 + sin2x, SiNx cosx), vector b = (1, SiNx + cosx), f (x) = vector a * vector B If f（ α）= 8 / 5.4 α Value of

Solution: F (x) = vector a * vector b = 1 + sin (2x) + (SiNx cosx) (SiNx + cosx) = 1 + sin (2x) - ((cosx) ^ 2 - (SiNx) ^ 2) = 1 + sin (2x) - cos (2x) - > F（ α)= 1+sin(2 α)- cos(2 α)= 8/5--> sin(2 α)- cos(2 α)= 3 / 5 -- > square on both sides of the equation: (sin (2) α))^ 2+(cos(2...

Given the vector a = (3-cos2 (x + 4 / π), - 2 √ 2), B = (1, SiNx + cosx), C ∈ [- 3 π / 4, π / 4], and a * b = 8 / 9, find the value of sin2x

A = (3-cos2 (x + π / 4) should be like this
a*b=[3-cos2(x+π/4)]-2√2(sinx+cosx)
=2〔sin(x+π/4)-1〕^2
=8/9
Sin (x + π / 4) = 5 / 3 (rounded off) or sin (x + π / 4) = 1 / 3
sinx+cosx=√2/3 (sinx+cosx)^2=2/9 2sinxcosx=2/9-1=-7/9
Sin2x = - 7 / 9

The known function f (x) = cos (2x-3 π) + the square of SiNx - the square of cosx 1. Find the minimum positive period of function f (x) and the symmetry axis equation of the image 2. Let the square of function g (x) = [f (x)] + F (x), find the value range of G (x)

Known function f (x) = cos (2x - π / 3) + sin ² x-cos ² X = (cos2x) / 2 + (√ 3sin2x) / 2-cos2x = - (cos2x) / 2 + (√ 3sin2x) / 2 = sin (2x - π / 6) axis of symmetry x = π / 3 + K π / 2, K ∈ ZT = π because f (x) ∈ [- 1,1] g (x) = f (x) ²+ f(x)=(f(x)+1/2) ²- 1 / 4 Institute

Derivation of higher numbers: if f (U) is differentiable and y = f (e ^ x), then dy = () If f (U) is differentiable and y = f (e ^ x), then dy = () A.dy=f'(e^x)dx B.dy=f'(e^x)de^x C.dy=[f(e^x)]'de^x D.dy=[f(e^x)]'e^xdx What is the correct answer? A is definitely wrong. B, C and D are right. Why are they right and where are they wrong?

This is the differential of a function, which can be reduced to a derivative
Dy / DX = [f (e ^ x)] '= f' (e ^ x) * e ^ x (derivation of composite function, derivative of outer function multiplied by derivative of inner function)
Double DX
For dy = f '(e ^ x) * (e ^ x) DX, you can put e ^ x into the differential and it becomes dy = f' (e ^ x) de ^ X
So B is right
C: Wrong, it should be dy = [f (e ^ x)]'dx
D: Wrong, it should be dy = f '(e ^ x) e ^ xdx (the single quotation mark inside represents the derivative of the outer function, and the single quotation mark outside the braces represents the derivative of the composite function)
If you don't understand, ask again

High number problem to find the derivative, please write the process: y = ln [x + (a ^ 2 + x ^ 2) ^ 1 / 2] Let u = x + (a ^ 2 + x ^ 2) ^ 1 / 2, so y = 1 / u * u `, and then a ^ 2 + x ^ 2 = w, continue to simplify to 1 / 2 * u ^ (- 1 / 2) * W ` I don't know what's wrong with the idea, and I can't get the answer right after several times

Maybe you finally simplified it wrong [x + (a ²+ x ²)^ (1/2)]'=1+1/[2√(a ²+ x ²)]* (a ²+ x ²)'= 1+x/√(a ²+ x ²)= [√(a ²+ x ²)+ x]/√(a ²+ x ²) So the original formula = 1 / [x + √ (a ²+ x ²)]* [√...

Another high number problem to find the derivative, please write the process: y = arcsin [2T / (1 + T ^ 2)] The final reduction is [2 * (1-T ^ 2)] / [(T ^ 2-1) * (T ^ 2 + 1)]. At this time, it can also be reduced by dividing T ^ 2-1 up and down, but this is wrong. The answer is - 2 / (T ^ 2 + 1) when T ^ 2 > 1; When T ^ 2 < 1, 2 / (T ^ 2 + 1)

When you open the root sign, you don't pay attention to the positive and negative of the number in the root sign: (arcsinx) '= 1 / √ (1-x ^ 2) so: arcsin [2T / (1 + T ^ 2)]' = 1 / √ {1 - [2T / (1 + T ^ 2)] ^ 2} * [2T / (1 + T ^ 2)] 'you will certainly ask. I just say 1 / √ {1 - [2T / (1 + T ^ 2)] ^ 2} look at √ {1 - [2T / (1 + T ^ 2)] ^ 2} = √ {