Let f (x) = 2Sin (π - x) cosx be a trigonometric function of an angle That is, y = asin (Wx + a) + B

Let f (x) = 2Sin (π - x) cosx be a trigonometric function of an angle That is, y = asin (Wx + a) + B

f(x)=2sin(π-x)cosx
=2sinx*cosx
=sin2x
Induction formula and double angle formula are used

How does the trigonometric function | (cosx) ^ 2 - (√ 3 / 2) * sin2x be transformed into cos (2x + Π / 3) + 1 / 2? How does (cosx) ^ 2 - (√ 3 / 2) * sin2x be transformed into cos (2x + Π / 3) + 1 / 2? Please elaborate

(cosx)^2-(√3/2)*sin2x
=（1/2）[2(cosx)^2-1]+1/2-(√3/2)*sin2x
=（1/2）cos2x-(√3/2)*sin2x+1/2
(1 / 2) cos2x - (√ 3 / 2) * sin2x is obtained by the auxiliary angle formula
The original formula = cos (2x + Π / 3) + 1 / 2
Note: the formula of auxiliary angle is
Asina + bcosa = radical (a ^ 2 + B ^ 2) * sin (a + ±)
(is any real number)

Why is 1-cosx equal to 1-cosx = 2Sin 2 (x / 2) 1-cosx =1-[1-2sin²(x/2)] =2sin²(x/2) In addition, SiNx + cosx = 1 Why is 1-cosx equal to 2Sin 2 (x / 2)

Double angle cosine formula
cos2x=1-2sin^2x
So cosx = 1-2sin ^ 2 (x / 2)
So 1-cosx = 2Sin 2 (x / 2)
SiNx + cosx = 1 is wrong
It's sin ^ 2x + cos ^ 2x = 1

Trigonometric function solution: given cos θ = 3 / 5, θ belongs to (0, π / 2) It is known that cos θ = 3 / 5, and θ belongs to (0, π / 2) (1) Find sin (θ + π / 4) （2）sin(2θ+π/2）.

（1）sinθ=3/5
So sin (θ + π / 4)
=√2/2x3/5+√2/2x4/5
=√2/2x7/5
=7√2/10
(2) sin(2θ+π/2)
=cos2θ
=2cos²θ-1
=2x9/25-1
=-7/25

It is known that cos (a + 5 π / 12) = 1 / 3 and - π Is it odd change even invariable, symbol look quadrant?

Odd change even invariable, symbol look at quadrant this is right
Cos (π / 2 - π / 12 + a) = cos (π / 2 - (π / 12-A) = sin (π / 12-A) = 1 / 3 is also correct

If cos (5 л / 12 + α) = 1 / 3 and - л < α < - л / 2, then cos (л / 12 - α) is equal to 1 / 3

∵cos(л/12-α)
=cos(π/2-（5л/12+α））
=sin（5л/12+α）
∵-л＜α＜-л/2
∴－π/2＜5л/12+α＜－π/12
∵cos(5л/12+α)=1/3
∴sin（5л/12+α）=－2√2/3
∴cos(л/12-α)=－2√2/3

Let x ^ 2Sin α - y ^ cos α = 1 denote ellipse Let x ^ 2Sin α - y ^ 2cos α = 1 denote ellipse (1) If the focus of the ellipse is on the x-axis, find the range of α (2) If the focus of the ellipse is on the y-axis, find the range of α The standard equation of a circle is not x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 or the denominator is reversed

Look at the picture below
The equation of ellipse is transformed into the form of standard equation,
If the focus is on the x-axis, the long axis is on the x-axis, a > B;
If the focus is on the y-axis, the long axis is on the y-axis, b > a

If a is the inclination angle of a straight line and the equation x ^ 2Sin α - y ^ 2cos α = 1 denotes an ellipse with the focus on the Y axis, then the range of a is

x^2sinα-y^2cosα=1
Into standard form
x²/(1/sinα)+y²/(-1/cosα)=1
Ellipse with focus on y-axis
So - 1 / cos α > 1 / sin α
α is the inclination angle, and α is known to be ≠ 0
So - Tan α > 1
So tan α

If θ is an inner angle of △ ABC and sin θ cos θ = - 1 / 8, then the value of sin θ - cos θ is A. - root 3 / 2 b. root 3 / 2 C. - root 5 / 2 d. root 5 / 2

sinθcosθ0,cosθ0
(sinθ-cosθ)^2=sinθ^2+cosθ^2-2sinθcosθ
=1-2*(-1/8)
=5/4
Sin θ - cos θ = radical 5 / 2
Choose D

In △ ABC, the opposite sides of angles a, B and C are a, B, C. given the vector M = (2cos A / 2, sin a / 2), vector n = (COS A / 2, - 2Sin A / 2), Mn = - 1. (1) find the value of cosa; (2) if a = 2 √ 3, B = 2, find the value of C

Vector M = (2cos A / 2, sin a / 2), vector n = (COS A / 2, - 2Sin A / 2),
mn＝2(cosA/2)^2-2(sinA/2)^2=2cosA=-1,
(1)cosA=-1/2.
(2) From the cosine theorem,
12=4+c^2-4c*(-1/2),
∴c^2+2c-8=0,c>0,
∴c=2.