Proof of COS (3 π - α) = - sin α

Proof of COS (3 π - α) = - sin α

cos(3/2π-α)= cos[(π+(π/2-α)]=
When α is an acute angle, π + (π / 2 - α) is in the third quadrant
cos(3/2π-α)= cos[(π+(π/2-α)]= - cos(π/2-α)= -sinα

It is proved that sin (3 π / 2 - α) = - cos α

sin(3π/2-α)
=sin3π/2cosα-cos3π/2sinα
=-1*cosα-0*sinα
=-cosα

It is proved that: (1) (3 π / 2 - α) = - cos α (2) (3 π / 2 - α) = - sin α

The original title should be: (1) sin (3 π / 2 - α) = - cos α (2) cos (3 π / 2 - α) = - sin α! Prove: (1) left = sin (2 π - π / 2 - α) = sin (- π / 2 - α) = - sin (π / 2 + α) = - cos α = right, so the original equation is proved

The formula of sum and difference of two angles,

As shown above, this is a unit circle with radius of 1
AB^2=OA^2 + OB^2 - 2*OA * OB * cos(α-β) = 1 + 1 - 2*1*1*cos(α-β) = 2 - 2 *cos(α-β)
According to the Pythagorean theorem,
AB^2 = DE^2 + CF^2
         = (OD - OE)^2 + (OF + OC)^2
         = (OA * sinα - OB * sinβ) ^2 + (OB * cosβ - OA * cosα)^2
         = (sinα - sinβ) ^2 + (cosβ - cosα)^2
         = (sinα)^2 + (sinβ)^2 - 2*sinα*sinβ + (cosα)^2 + (cosβ)^2 - 2*cosα*cosβ
         = (sinα)^2 + (cosα)^2 + (sinβ)^2 + (cosβ)^2 - 2*sinα*sinβ - 2*cosα*cosβ
         = 1 + 1 - 2*sinα*sinβ - 2*cosα*cosβ
Combining these two formulas, we can get the following results:
2 - 2 *cos(α-β) = 2 - 2*sinα*sinβ - 2*cosα*cosβ
∴ cos(α-β) = cosα*cosβ + sinα*sinβ 
The cosine formula of the sum of two angles is as follows:
 cos(α+β) = cos(α - (-β))
                = cosα*cos(-β) + sinα*sin(-β)
∵ cos(-β) = cosβ,sin(-β) = - sinβ
∴ cos(α+β) = cosα*cosβ - sinα*sinβ 

Derivation of cosine formula of two angle difference

Vector method: take rectangular coordinate system, make unit circle, take a point a, connect OA, the angle with X axis is a, take a point B, connect ob, the angle with X axis is boa, the angle between B and ob is A-ba (COSA, Sina), B (CoSb, SINB) OA = (COSA, Sina) ob = (CoSb, SINB) OA * ob = (OA | ob | cos (a-b) = cosacosb + sinasinb

What is the formula of two angles and sine?

sin(x+y)
=sinxcosy+cosxsiny

Double angle formula and its deformation formula in high school mathematics

tanA+tanB=sin(A+B)/cosAcosB=tan(A+B)(1-tanAtanB)
tanA-tanB=sin(A-B)/cosAcosB=tan(A-B)(1+tanAtanB)
Sin2A=2SinA•CosA
Cos2A=CosA^2-SinA^2=1-2SinA^2=2CosA^2-1
tan2A=（2tanA）/（1-tanA^2）
(Note: Sina ^ 2 is the square of sina sin2 (a)
sin3a
=sin(2a+a)
=sin2acosa+cos2asina
=2sina(1-sin²a)+(1-2sin²a)sina
=3sina-4sin³a
cos3a
=cos(2a+a)
=cos2acosa-sin2asina
=(2cos²a-1)cosa-2(1-sin²a)cosa
=4cos³a-3cosa
sin3a=3sina-4sin³a
=4sina(3/4-sin²a)
=4sina[(√3/2)²-sin²a]
=4sina(sin²60°-sin²a)
=4sina(sin60°+sina)(sin60°-sina)
=4sina*2sin[(60+a)/2]cos[(60°-a)/2]*2sin[(60°-a)/2]cos[(60°-a)/2]
=4sinasin(60°+a)sin(60°-a)
cos3a=4cos³a-3cosa
=4cosa(cos²a-3/4)
=4cosa[cos²a-(√3/2)²]
=4cosa(cos²a-cos²30°)
=4cosa(cosa+cos30°)(cosa-cos30°)
=4cosa*2cos[(a+30°)/2]cos[(a-30°)/2]*{-2sin[(a+30°)/2]sin[(a-30°)/2]}
=-4cosasin(a+30°)sin(a-30°)
=-4cosasin[90°-(60°-a)]sin[-90°+(60°+a)]
=-4cosacos(60°-a)[-cos(60°+a)]
=4cosacos(60°-a)cos(60°+a)
By comparing the above two formulas
tan3a=tanatan(60°-a)tan(60°+a)

Inflection point and concave convex interval of y = 2x ＾ 2 / (1-x) ＾ 2

Why is this not answered?
If you are a college student, this is done by using the first derivative and the second derivative
If you are a middle school student, do a transformation, change the 2x ^ 2 of the numerator into 2 (x-1) ^ 2-4 (1-x) + 2, and then do it after the numerator and denominator is reduced. In this way, there is no X in the numerator, but only in the denominator. The inflection point of the denominator is the inflection point of the whole y function. The increase and decrease of the denominator is opposite to that of the Y function, and the concave convex interval is obvious, Note that the denominator is meaningless when x = 1

Monotonicity and convexity of y = e ^ - 2x

Monotone reduction, concave function. Xiaoyueyueyou, learn mathematics well

The concave convex interval and inflection point of the function y = 2x / (x + 1) square are determined

y=2x/(x+1)^2
y'=2[(x+1)^2-2(x+1)x]/(x+1)^4=2[x+1-2x]/(x+1)^3=2(1-x)/(x+1)^3
y"=2[-(x+1)^3-3(x+1)^2(1-x)]/(x+1)^6=2[-x-1-3+3x]/(x+1)^4=4(x-2)/(x+1)^4
From Y "= 0, x = 2,
y(2)=4/9
When x > 2, Y "> 0 is a concave interval;
When-1