Let function f (x) = cos (2x + π \ 3) + sin ^ 2x. Find the monotonically increasing interval of function f (x)

Let function f (x) = cos (2x + π \ 3) + sin ^ 2x. Find the monotonically increasing interval of function f (x)

f(x)=cos(2x+π\3)+sin^2x
=1/2cos2x-√3/2*sin2x+1-1/2(1+cos2x)
=-√3/2*sin2x+1/2
2x∈[2k∏+∏/2,2k∏+3∏/2]
That is, X ∈ [K Π + Π / 4, K Π + 3 Π / 4]
Monotonic increasing of function f (x)

Partial proof of calculus and derivative It is proved that if f (x) exists at the point of X. (1) if X. is the extreme point, there must be f '(x) = 0 (2) if X. is the inflection point, there must be the derivative part of F' '(x) = 0 calculus Correct the X. It's an X

(1) Fermat's theorem. Some textbooks include it in the proof of Rolle's theorem. If you turn to the book, I'll prove it again;
(2) treat f '' (x) as the derivative of F '' (x), that is, use Fermat's theorem for f '' (x)

Y = (x-1) (X-2) (x-3) (x-4) derivative

y=(x-1)(x-2)(x-3)(x-4)=(x^2-5x)^2+10(x^2-5x)+24
y'=2(x^2-5x)(2x-5)+10(2x-5)
=(2x-5)(2x^2-10x+10)

How to simplify the addition of trigonometric functions, such as SiNx + cosx, and find the period

Generally, it is √ 2
sinx+cosx =√2（√2/2sinx+√2/2cosx)=√2(sin(x+π/4）

Trigonometric function if 0 is less than or equal to 2 π and satisfies cosx at the same time

B must be wrong. For example, when x = π / 3, TaNx = 1.732 > SiNx, it is best to analyze TaNx < SiNx in combination with quadrants. If both are greater than 0 at the same time, TaNx will always be greater than SiNx. (you can draw this conclusion with right triangle. SiNx = A / C, TaNx = A / b. right angle side B will always be less than bevel side C). And when both

How can y = 1-cosx \ SiNx be reduced to a trigonometric function Not clear: should be (1-cosx) \ SiNx

y=1-(1-2(sinX/2)^2)\2(sinx/2)*(cosx/2);
y=sin(x/2)\cos(x/2);
tan(x/2)=y;