Given that SiNx = asin (x + β), Tan (x + β) = sin β / (COS β - a)

Given that SiNx = asin (x + β), Tan (x + β) = sin β / (COS β - a)

It is proved that Tan (x + β) = sin β / (COS β - a) cos β Tan (x + β) - atan (x + β) = sin β cos β sin (x + β) / cos (x + β) / cos (x + β) = sin β cos β sin (x + β) / cos (x + β) - SiNx / cos (x + β) = sin β cos β sin (x + β) - SiN x = sin β cos (x + β) sin (x +...)

Y = 4sin ^ 4x + 4cos ^ 4x-3 is transformed into the form of y = asin (Wx + φ)

Original formula = 4 (sin ^ 2x + cos ^ 2x) ^ 2-8sin ^ 2xcos ^ 2x-3
=1-8sin^2xcos^2x
=1-2(sin2x)^2
=cos4x

It is known that the function y = sin? 2x + 2sinxcosx + 3cos? X. ① find the minimum positive period of the function; ② find the monotone increasing interval of the function; ③ when x takes what value, ③ When x is taken, the function takes the maximum value

y=2sinxcosx+3-2sin²x=sin2x+cos2x+2=√2sin(2x+π/4)+2
1) The minimum positive period π;
2) Monotonically increasing interval: ((- 3 / 8 + k) π, (1 / 8 + k) π);
3) The maximum is obtained when x = (1 / 8 + k) π

How to simplify f (x) = Radix 3sin (Wx + φ) - cos (Wx + φ) = 2Sin (Wx + φ - π / 6)?

F (x) = Radix 3sin (Wx + φ) - cos (Wx + φ) = 2 [Radix 3sin (Wx + φ) / 2-cos (Wx + φ) / 2] = 2 [sin (Wx + φ) cos (π / 6) - cos (Wx + φ) sin (π / 6)] = 2Sin (Wx + φ - π / 6)

How to prove sin 4 α + sin 2 α cos 2 α + cos 2 α = 1

Is 4 the fourth power?
From the identity sin? α + cos? α = 1
Left = sin? α (sin? α + cos? α) + cos? α
=sin²α+cos²α
=1 = right
Proof of proposition

When the angle a is an acute angle, sin? A + cos? A = (), and it is proved Help me, homework!

(Sina) ^ 2 + (COSA) ^ 2 = 1. According to the definition, take P (x, y) Sina = Y / R, cosa = x / R (Sina) ^ 2 = y ^ 2 / R ^ 2, (COSA) ^ 2 = x ^ 2 / R ^ 2 (Sina) ^ 2 + (COSA) ^ 2 = y ^ 2 / R ^ 2 + x ^ 2 / R ^ 2 = (y ^ 2 + x ^ 2) / R ^ 2, and because R ^ 2 = y ^ 2 + x ^ 2, so (Sina) ^ 2 + (COSA) ^ 2 = 1

It is proved that (1-cos 2 θ / sin θ - cos θ) - (sin θ + cos θ / Tan θ - 1) = sin θ + cos θ

prove:
Left = (1 - cos 2 θ) / (sin θ - cos θ) - (sin θ + cos θ) / (Tan θ - 1)
=sin²θ/(sinθ－cosθ)-(sinθ+cosθ)/(sin²θ/cos²θ－1)
=sin²θ/(sinθ－cosθ)-(sinθ+cosθ)cos²θ/(sin²θ-cos²θ)
=sin²θ/(sinθ－cosθ)-(sinθ+cosθ)cos²θ/[(sinθ+cosθ)(sinθ-cosθ)]
=sin²θ/(sinθ－cosθ)-cos²θ/(sinθ-cosθ)
=(sin²θ-cos²θ)/(sinθ-cosθ)
=(sinθ+cosθ)(sinθ-cosθ)/(sinθ-cosθ)
=Sin θ + cos θ = right
The equation proved is valid

Prove the triangle identity: cot? A-cos? A = cot? A · cos? A How do you prove that?

cot²a-cos²a=cos²a/sin²a-cos²a=cos²a(1/sin²a-1)=cos²a(1-sin²a)/sin²a=cos²a*cos²a/sin²a=cot²a·cos²a

Given cot θ = 1 / 2, find sin θ - 2Sin θ cos θ + 3cos θ =?

Sin 2 θ - 2 sin θ cos θ + 3 cos 2 θ = (sin 2 θ - 2 sin θ cos θ + 3 cos 2 θ) / (sin 2 θ + cos 2 θ), = (1-2 cot θ + 3 cot 2 θ) / (1 + cot 2 θ) = (1-1 + 3 * 1 / 4) / (1 + 1 / 4) = 3 / 5

It is proved that: (2-cos? A) (1 + 2cot? A) = (2 + cot? A) (2-sin? A)

On the left = 2 + 4cot? A-cos? A-2cos? Acot? A = 2 + 4cot? A - (1-sin? A) - 2 (1-sin? A) cot? A = 2 + 4cot? A-1 + sin? A-2cot? A + 2Sin? Acot? A = 1 + 2cot? A + sin? A + 2Sin & su