f(sinx)=cos2x f'(x)=?

f(sinx)=cos2x f'(x)=?

f(sinx)=cos2x=1-2(sinx)^2
let u=sinx
f(u)=1-2u^2
u=x
so that f'(x)=-4x

Given the vector a = (SiNx, 2), B = (cosx, - 1), when a is parallel to B, find the value of sin ^ 2x-sin2x; find f (x) = (a + b)*

When a is parallel to B
sinx:cosx=2 :(-1)
tanx=-2
sin^2x-sin2x
=(sin^2x-2sinxcosx)/(sin^2x+cos^2x)
=(tan^2x-2tanx)/(tan^2x+1)
=8/5

If f (SiNx + cosx) = sinxcosx, then f [sin (Π / 2 + Π / 6)] =?
Same as above. Thank you

Let a = SiNx + cosxa & sup2; = Sin & sup2; X + 2sinxcosx + cos & sup2; X = 1 + 2sinxcosx, so sinxcosx = (A & sup2; - 1) / 2, so f (a) = (A & sup2; - 1) / 2Sin (π / 2 + π / 6) = √ 3 / 2, so the original formula = f (√ 3 / 2) = [(√ 3 / 2) & sup2; - 1] / 2 = - 1 / 8

How does (√ 2 * SiNx) / (SiNx + cosx) = √ 2 (sin ^ 2x + sinxcosx) change?

After modification, please check

Find the general solution of sin2x-2sinx-cosx + 1 = 0

∵sin2x-2sinx-cosx+1=0 ==>2sinxcosx-2sinx-cosx+1=0
==>2sinx(cosx-1)-(cosx-1)=0
==>(2sinx-1)(cosx-1)=0
==>2sinx-1 = 0, or cosx-1 = 0
==>SiNx = 1 / 2, or cosx = 1
Ψ x = π / 6, or x = 5 π / 6, or x = 0
So the general solution of the original equation is x = 2K π + π / 6, or x = 2K π + 5 π / 6, or x = 2K π (k is an integer)

It is known that cos x = - 1 / 3 and 180 ° < x < 270 ° (1) find the values of sin2x, cos2x and tan2x. (2) find the values of SiNx / 2, cosx / 2 and TaNx / 2

cosx=-1/3 180°＜x＜270°sinx=-2√2/3 180°＜x＜270°360°＜2x＜540°0°＜2x＜180°sin2x=2sinxcosx=2*(-2√2/3 )*(-1/3)=4√2/9cos2x=1-2cos²x=1-2*(-1/3)²=1-2/9=7/9tan2x=sin2x/cos2x=(4√2/9)/(7...

The monotone increasing interval of the function y = - X3 + 3x2 + 3 is______ ．

Y ′ = f ′ (x) = - 3x2 + 6x, Let f ′ (x) = - 3x2 + 6x > 0, the solution is: X ∈ (0,2), so the answer is (0,2)

Decreasing interval of function y = 3 ^ - x2 + 4x-3

Is y = 3 ^ (- X & # 178; + 4x-3)
Exponential function with base 3 increases monotonically when - X & # 178; + 4x-3 decreases monotonically when - X & # 178; + 4x-3 decreases monotonically
So find the decreasing interval of y = - X & # 178; + 4x-3
y=-x²+4x-3=0
y′=-2x+4
When y ′ = 0 - 2x + 4 = 0, x = 2
When x > = 2, y ′

Given the function y = x & # 178; + 6x + 8, then the value range of X which makes the function y > 0 hold is
Ax ＜ - 4 and X ＞ - 2
Bx＜-4
C-4＜x＜-2
DX ＜ - 4 or X ＞ - 2

y=x²+6x+8
=(x+3)²+8-9
=(x+3)²-1
y>0
(x+3)²-1>0
(x+3)²>1
x+3-2
Choose D

Let the minimum value of function f (x) = x & sup2; - 2x-3 in the interval [T, t + 1] be g (T), and write the analytic expression of G (T)

The axis of symmetry x = 1
When t + 1