What is the monotone increasing interval of the square X of COS and sin

What is the monotone increasing interval of the square X of COS and sin

y=cos²x－sin²x=cos2x
The increasing interval is 2K π - π ≤ 2x ≤ 2K π
The results show that K π - π / 2 ≤ x ≤ K π
The increasing interval is [K π - π / 2, K π], where k ∈ Z

The monotone interval of function y = sin (x + π / 4) is?

Let the whole in brackets be t, then y = sin (T), then the monotone interval is (- π + 2K π, π + 2K π) increasing (π + 2K π, 2 π + 2K π) decreasing, then x + π / 4 is in the above two ranges
So when x is (- 5 π / 4 + 2K π, 3 π / 4 + 2K π), it is increasing and decreasing. You can calculate it yourself

The proof of the derivation of SiNx?

According to the definition of derivative, there are:
(sinX)'=lim(△x→0)[sin(x+△x)-sinx]/(△x)
=lim(△x→0)[sinxcos(△x)+cosxsin(△x)-sinx]/(△x)
=lim(△x→0)[sinx*1+cosxsin(△x)-sinx]/(△x)
=lim(△x→0)[cosxsin(△x)]/(△x)
=[cosx*△x]/(△x)
=Cosx, proved
Here we use LIM (△ x → 0) cos (△ x) = cos0 = 1 and sin △ x →△ x when △ x → 0

How to derive SiNx?

LZ:
Derivative can be obtained according to the definition of derivative. The steps of using the definition to obtain the derivative of function y = f (x) at x0 are as follows: ① find the increment of function Δ y = f (x0 + Δ x) - f (x0), ② find the average rate of change, and ③ take the limit. For SiNx, there is (SiNx) '= Lim [sin (x + △ x) - SiNx] / (△ x), where △ x → 0, expand sin (x + △ x) - SiNx, that is sinxcos △ x + cosxsin △ x-sinx, because △ x → 0, cos △ x → 1, Thus, sinxcos △ x + cosxsin △ x-sinx → cosxsin △ x, then (SiNx) '= LIM (cosxsin △ x) / △ x, here we must use an important limit (pinch criterion). When △ x → 0, LIM (sin △ x) / △ x = 1, then (SiNx)' = cosx.. we hope to adopt it in time

What is the result of N times derivation of SiNx?

What is the sixth derivative of (SiNx / x)?

（sinx/x）'=[(sinx)'x-x'sinx]=(xcosx-sinx)/x^2
(sinx/x)''=[-x^3sinx-2x(xcosx-sinx)]/x^4
(sinx/x)'''=(7x^10sinx-x^11cosx+14x^9cosx-14x^8sinx)/x^16

What is the square of SiNx
All the possibilities

(sinx)^2=1-(cosx)^2

Derivative of sin2x (help!

(sin2x)=cos2x*(2x)'=2cos2x

The n-th derivative of sin2x is

If the order of SiNx is sin (x + n π / 2), then the order of sin2x is 2 ^ n sin (2x + n π / 2)

The derivative of y = x × SiNx + cos2x is

y=xsinx+cos2x
y'=x'sinx+x(sinx)'+(cos2x)'
=sinx+xcosx-2sin2x