Given that θ is the second quadrant angle and 25 (sin θ) ^ 2 + sin θ - 24 = 0, what is the value of COS (θ / 2)?

Given that θ is the second quadrant angle and 25 (sin θ) ^ 2 + sin θ - 24 = 0, what is the value of COS (θ / 2)?

25sin²θ+sinθ-24=0
（25sinθ-24）（sinθ+1）=0
The solution is: sin θ = 24 / 25 or - 1
∵ θ is the second quadrant angle
∴sinθ=24/25
∵ sin θ + cos θ = 1, and θ is the second quadrant angle
∴cosθ=-7/25
cos（θ/2）=√[（cosθ+1）/2]=3/5

If sin (COS θ) * cos (sin ⊙) is known to be the symbol of Tan (sin θ)

The first question was said by the landlord above. I said the second question directly. The answer is detailed and written separately

If 3sin (π + α) + cos (- α) / 4sin (- α) - cos (9 π + α) = 2, then Tan α= If 2-tan α = α, then sin (- 5 π - α) cos (3 π + α)=

1,3sin(π+α)+cos(-α)/4sin(-α)-cos(9π+α)=2,【-3sinα+cosα】/【-4sinα+cosα】=2,-3sinα+cosα=-8sinα+2cosα5sinα=cosαtan α=sinα/cosα=1/52,tan α=αsin（-5π-α）cos（3π+α）=sin（π-α）co...

If sin ACOS B = 1, then cos (a + b)= emergency

Sinacos B = 1sina = 1 or CoSb = 1cosa = 0, SINB = 0cos (a + b) = cosacosb sinasinb = 0-0 = 0sina = - 1, CoSb = - 1cosa = 0, SINB = 0, cos (a + b) = cosacosb sinasinb = 0-0 = 0

Cos (a + b) * cos (a-b) = 1 / 5 to find cos ^ 2-sin ^ 2

If cos (a + b) cos (a-b) = 1 / 5, then (COSA) ^ 2 - (SINB) =? Cos (a + b) cos (a-b) = (cosacosb sinasinb) (cosacosb + sinasinb) = (cosacosb) ^ 2 - (sinainb) ^ 2 = cosa ^ 2 [1 - (sinab) ^ 2] = (COSA) ^ 2 - (SINB) ^ 2 = 1 / 5

1. It is known that sin θ = asin β, Tan θ = btan β, where θ is an acute angle. It is proved that cos θ = √ ((a ^ 2-1) / (b ^ 2-1)) 2. Known 1 + cos α - sin β + sin α * sin β = 0 and 1-cos α - cos β + sin α * cos β = 0 Please write the specific process, the solution should be simple!

（sinθ×asinβ－tanθ）＋cosα÷2 +1-1

It is proved that the equality is constant sin ^ A + sin ^ b-sin ^ asin ^ B + cos ^ ACOS ^ B = 1 sin^a+sin^b-sin^asin^b+cos^acos^b=1 We only need to prove sin ^ a (1-sin ^ b) + sin ^ B + cos ^ ACOS ^ B = 1 Just prove sin ^ ACOS ^ B + cos ^ ACOS ^ B + sin ^ B = 1 I don't understand from here to here

To prove sin ^ A + sin ^ b-sin ^ B + cos ^ ACOS ^ B = 1, we only need to prove sin ^ a (1-sin ^ b) + sin ^ B + cos ^ ACOS ^ B = 1 -- (merge similar items) only need to prove cos ^ B (sin ^ A + cos ^ a) + sin ^ B = 1 -

It is known that sin α = asin β, Tan α = btan β and α are acute angles. It is proved that: (COS α) ^ 2 = (a ^ 2-1) / (b ^ 2-1)

(sinβ)^2=(sinα)^2/a^2,(cosβ)^2=1-(sinβ)^2=[a^2-(sinα)^2]/a^2.
(tanβ)^2=(sinα)^2/[a^2-(sinα)^2].
(tanα)^2=b^2*(tanβ)^2=b^2(sinα)^2/[a^2-(sinα)^2].
1/(cosα)^2=b^2/[a^2-(sinα)^2].
b^2(cosα)^2=a^2-1+(cosα)^2.
(cosα)^2=(a^2-1)/(b^2-1).

If sin ACOS B = 1 / 2, then the value range of COS asin B

In order to express conveniently, a and B in the question are expressed by α and β respectively ∵ sin α cos β = [sin (α + β) + sin (α - β)] / 2 = 1 / 2  sin (α + β) + sin (α - β) = 1. From sin (α + β) ≤ 1, sin (α - β) = 1-sin (α + β) ≥ 0, sin (α - β) = 1-sin (α - β) ≤ 1

The vector a = (asin ω x, ACOS ω x) B = (COS θ, sin θ), f (x) = a * B + 1, where a > 0, ω > 0, θ is an acute angle, the distance between two adjacent symmetry centers of an image of F (x) is / 2, and when x = / 12, f (x) takes the maximum value of 3 (1) Find the analytic formula of F (x) (2) If G (x) is an odd function, find the minimum value of φ

f(x)=Asinωxcosθ+Acosωxsinθ+1=Asin(ωx+θ)+1.
(1) From the meaning of the title, a + 1 = 3, then a = 2
If the minimum positive period of F (x) is t = 2 * (π / 2) = π = 2 π / ω, then ω = 2
f(x)=2sin(2x+θ)+1.
f(π/12)=2sin(π/6+θ)+1=3、sin(π/6+θ)=1.
Zero