Finding period of function y = (SiNx cosx) Square-1

Finding period of function y = (SiNx cosx) Square-1

Y = - 1 of the square of (SiNx cosx)
=1-2sinxcosx-1=-2sinxcosx=-sin2x
T=2π/2=π

Function y = (cosx SiNx) squared. What is the minimum lower period?
If 2Sin (a) = 3sin (A / 2), how small is sinc?

Y = 1-sin (2x), so the minimum positive period is π
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Sin (A / 2) = 0 or cos (A / 2) = 3 / 4
If sin (A / 2) = 0, then sin (a) = 0
If cos (A / 2) = 3 / 4, sin (A / 2) = ± sqrt (7) / 4
sin(a)=±3*sqrt(7)/8
Where sqrt is the square root of arithmetic

It is known that an original function of F (x) is SiNx / X. how to prove that ∫ XF '(x) DX = cosx-2sinx / x + C

f(x)=(sinx/x)'=(xcosx-sinx)/x^2
∫xf'(x)dx
=∫xdf(x)
=xf(x)-∫f(x)dx
=x*(xcosx-sinx)/x^2-sinx/x+C
=cosx-2sinx/x+C

It is known that (SiNx) / X is a primitive function of F (x). To find ∫ XF '(x) DX, the answer is cosx - (2sinx) / x + C

∫ f(x)= (sinx)/x+C
∫xf'(x)dx =∫ xd(f(x)
=xf(x)-∫ f(x)dx
=xf(x)-(sinx)/x+c（*）
And f (x) = [(SiNx) / x + C] '= (cosx * x-sinx) / x ^ 2
Cosx - (2sinx) / x + C is obtained by introducing (*)
There are two wrong ones upstairs

A monotone increasing interval of the function y = | SiNx | is______ Just write one

Draw the graph of the function y = | SiNx | in the coordinate system: according to the graph, an increasing interval of the function is obtained as follows: (π, 3 π 2), so the answer is: (π, 3 π 2)

Given the function f (x) = sinx-1 / 2x, X ∈ (0, pie), find the monotone increasing interval of function f (x)

Using derivative method to solve the problem
f(x)=sinx-1/2 x
∴ f'(x)=cosx-1/2>0
∴ cosx>1/2
∵ 0

Given the function f (x) = SiNx − 12x, & nbsp; X ∈ (0, π). (1) find the monotone increasing interval of function f (x); (2) find the tangent equation of the image of function f (x) at point x = π 3

f′(x)＝cosx−12．… (2 points) (1) from X ∈ (0, π) and f ′ (x) = cosx − 12 ＞ 0, the solution of X ∈ (0, π 3).. the monotone increasing interval of function f (x) is (0, π 3) (6 points) (2) f (π 3) = sin π 3 − 12 × π 3 = 32 − π 6 The slope of tangent k = f '(π 3) = cos π 3 − 12 = 0 The tangent equation is y = 32 − π 6 (13 points)

The function f (x) = the third power of X + SiNx + 1 (x belongs to R), if f (a) = 2, then the value of F (- a)

If f (a) = a ^ 3 + Sina + 1 = 2, then a ^ 3 + Sina = 1
f(-a)=-a^3-sina+1=-(a^3+sina)+1=-1+1=0

The monotone increasing interval of function y = sinx-1 is, and the monotone decreasing interval is

The monotone increasing interval is [- π / 2 + 2K π, π / 2 + 2K π], K ∈ Z
The monotone decreasing interval is [π / 2 + 2K π, 3 π / 2 + 2K π], K ∈ Z

It is proved that f (x) = 2x + SiNx is a strictly increasing function on R
Is it OK to use definition certificate? I'm sorry. I can't understand the derivation right now.

f'(x)=2+cosx
-1