The maximum and minimum values of the function f (x) = - 2Sin? X + 2cosx Process, thank you!

The maximum and minimum values of the function f (x) = - 2Sin? X + 2cosx Process, thank you!

f(x)=-2sin²x+2cosx
=-2(1-cos²x)+2cosx
=2cos²x+2cosx-2
=2(cosx+1/2)²-5/2
cosx=-1/2 f(x)min=-5/2
cosx=1 f(x)max=2

Let f (x) = 1-2sin ^ 2x + 2cosx, and find the maximum and minimum value of F (x) by derivation

F (x) = 1-2 (1-cos ^ 2x) + 2cosx = 1-2 + 2cos ^ 2x + 2cosx = 2cos ^ 2x + 2cosx-1 using the derivative rule of composite function: [f (g (x))] '= f' (g (x)) * g '(x) f (x) is regarded as the compound of 2x ^ 2 + 2x-1 and cosx f' (x) = (4 * cosx + 2) * - SiNx = - 2 (2cosx + 1) SiNx zero point is cosx = - 1 / 2, SiNx = 0cosx

The maximum and minimum values of the function y = 1-2sin? X + 2cosx are

Very happy to answer for you!
According to the known conditions:
y=1-2sin2x+2cosx=2-2sin2x+2cosx-1=2cos2x+2cosx-1
=2（cosx+1/2）2-3/2
Because - 1

Simplify f (x) = - Sin ^ 2x + SiNx

In fact, it is the simplest, but it can be deformed as needed
f(x)=-sin^2 x+sinx=sin x *(1-sin x)=-(sin x-1/2)^2+1/4.

It was proved that sin (X-Y) / (SiNx siny) = cos [(X-Y) / 2] / cos [(x + y) / 2] （cosy-cosx）/（sinx-siny）=tan[（x+y）/2]

You can convert the denominator SiNx - siny into 2Sin ((X-Y) / 2) cos ((x + y) / 2)
So the answer is obvious

(a) It is proved that cos (X-Y) - cos (x + y) = 2 * SiNx * siny (b). Therefore, it is proved that 2Sin θ (sin θ + sin 3 θ + sin 5 θ + sin 7 θ) = 1-cos 8 θ

(a) It is proved that the left side = cos (X-Y) - cos (x + y) = (cosxcosy + sinxsiny) - (cosxcosy sinxsiny) = 2sinxsiny = right side (b) the conclusion of (1) is that left side = 2Sin θ (sin θ + SIN3 θ + sin5 θ + sin 7 θ) = 2Sin θ + 2Sin θ sin 3 θ + 2Sin θ sin 5 θ + 2Sin θ Si

It is proved that | X-Y | ≥ | SiNx siny|

There are many ways to prove this. You have to indicate whether you are studying in middle school or university
Proof in middle school
When x ≥ 0, SiNx ≤ x [this is often used to prove the unit circle method or function derivation method] when x < 0, SiNx > X
That is, | SiNx | ≤| x | [this conclusion is more general]
|SiNx siny | = 2 | cos [(x + y) / 2] sin [(X-Y) / 2] | [sum difference product formula or direct matching]
=2|cos[（x+y）/2]||sin（x-y）/2]|
≤ 2 | sin (X-Y) / 2] | [using the above conclusion]
≤2|（x-y）/2|
=|x-y|
It is very fast to use the mean value theorem directly in University proof

It is proved that cosx (cosx cosy) + SiNx (SiNx siny) = 2Sin (X-Y) / 2

The title should be "prove that cosx (cosx cosy) + SiNx (SiNx siny) = 2Sin 2 (X-Y) / 2"
Pr：
On the left
cos²x-cosxcosy+sin²x-sinxsiny
=1-(cosxcosy+sinxsiny)
=1-cos(x-y)
=1-cos²(x-y)/2+sin²(x-y)/2
=2sin²(x-y)/2
=Right

Why SiNx * cosy cosx * siny = sin (X-Y)

It's proved in this high school textbook

Proving inequality | siny SiNx|

|siny-sinx|=|2sin((y-x)/2)cos((y+x)/2)|