Let f (x) = 2sinx * cos square · / 2 + cosx * sin · (0

Let f (x) = 2sinx * cos square · / 2 + cosx * sin · (0

2cos square · / 2 = cos · + 1
F (x) = 2sinx * cos square · / 2 + cosx * sin ·
=sinx*cos¤+cosx*sin¤+sinx
=sin(x+¤)+sinx
=2sin(x+¤/2)cos¤/2
=2sin(π+¤/2)cos¤/2
=-2sin¤/2cos¤/2
=-sin¤ ,¤=π/2

Find the derivative 1 y = (2x + 3) ^ 3.2 y = e ^ x ^ 2-2x.3 y = sin (2x + 4 parts π)

1、y'=2*3（2x+3）^2
=6（2x+3）^2
2、y'=2x*e^x²-2
3、y'=2cos(2x+π/4)

How to find the original function of function sin (x square)? Some people say that this is the method of series for non elementary functions. Can you tell us in detail how to use series to solve integral?

Let the original function be f (x)
According to the meaning of the topic and the definition of the original function, there are:
F'(x)=sin(x^2)
dF(x)=sin(x^2)dx
F(x)=∫sin(x^2)dx
It has been known that this is a transcendental integral and a non integrable function, that is to say, f (x) cannot be expressed in terms of functions that people already know
Introduce: ERF (x) = ∫ [0, x] e ^ (- T ^ 2) DT, [0, x] means that the lower limit of integral is 0 and the upper limit is X
The integral of the function can be expressed as ERF (x) + C
That is: F (x) = ERF (x) + C

The value range of the function y = cos2x SiNx is______ .

The function y = cos2x SiNx = 1-sin2x-sinx = - (SiNx + 1)
2)2+5
4，
So when SiNx = - 1
When 2, the function y has a maximum value of 5
When SiNx = 1, the function y has a minimum value of - 1
So the value range of function y is [− 1,5
4]，
So the answer is: [− 1,5
4]．

F (x) = SiNx ^ 2 - radical 2 * SiNx + 1, find the function range

First of all, draw the image of - 1 to 1. (or else SiNx is equal to - 1 1 1 √ 2 / 2 to find the value of three points) I hope it can help you!

The known function f (x) = Cos2 (x + π) 12)+1 2sin2x． (1) Find the maximum value of F (x); (2) Find the monotone increasing interval of F (x)

(1) F (x) = 12 [1 + cos (2x + π 6)] + 12sin2x (2 points) = 12 [1 + (cos2xcos π 6 − sin2xsin π 6) + sin2x] = 12 (1 + 32cos2x + 12sin2x) (2 points) = 12sin (2x + π 3) + 12. (2) the maximum value of F (x) is 1 and the minimum value is 0; (2) f (x) increases monotonically, so 2x + π

Y = cos ^ 2 angle * sin angle when sin angle = how many y has the maximum value

Y = cos? Xsinxy? = cos? Xcos? Xsin? X = 1 / 2 (COS? Xcos? X * 2Sin? X) ≤ 1 / 2 * [(2cos? X + 2Sin? X) / 3] 3 = 4 / 27 If and only if cos? X = 2Sin? X, 1-sin? X = 2Sin? Xsin

The maximum value of (COS α - cos β) ^ 2 + (sin α - sin β) ^ 2

=cos^2 a + cos^2 b - 2cosacosb + sin^2 a + sin^2 b - 2 sinasinb
=2 - 2(cosacosb+sina+sinb)
=2-2cos(a-b)
Because cos (a-b) belongs to [- 1,1]
Therefore, when cos (a-b) = - 1, the original formula takes the maximum value of 4

It is known that SiNx + cosx = m, the absolute value m ≤ root 2 and the absolute value m is not equal to 1, and X is the second quadrant angle It is known that SiNx + cosx = m, the absolute value m ≤ root 2 and the absolute value m is not equal to 1, and X is the second quadrant angle. Find 1. Sin cubic x + cos cubic x 2. Sin quartic X - cos quartic X

One
SiNx + cosx = m squared
sin^2x+2sinxcosx+cos^2x=m^2
sinxcosx=(m^2-1)/2
sin^3x+cos^3x
=(sinx+cosx)(sin^2-sinxcosx+cos^2x)
=m(1-(m^2-1)/2)
=m(3-m^2)/2
Two
sin^2x+cos^2x=1
sin^2x-2sinxcosx+cos^2x=1-2sinxcosx=1-2*(m^2-1)/2=2-m^2
(sinx-cosx)^2=2-m^2
Because x is the second quadrant angle
So SiNx > 0, cosx0
So SiNx cosx = √ (2-m ^ 2)
sin^4x-cos^4x
=(sin^2x+cos^2x)(sinx+cosx)(sinx-cosx)
=1*m*√(2-m^2)
=m√(2-m^2)

The real number a satisfies | 2005 − a|+ If a − 2006 = a, then the value of the algebraic formula a-20052 is______ .

A kind of
A − 2006 makes sense,
∴a-2006≥0，
The original formula = a-2005+
A − 2006 = a, i.e
a−2006=2005，
∴a-2006=20052，
∴a-20052=2006．
So the answer is: 2006