α. β is an acute angle, cos α = 1 / root 10, cos β = 1 / root 5, calculate the value of α + β?

α. β is an acute angle, cos α = 1 / root 10, cos β = 1 / root 5, calculate the value of α + β?

Cos α = 1 / root 10
Sin α = root 1-cos ^ 2 α = root 1 - (1 / root 10) ^ 2 = 3 / root 10 (sine of acute angle must be positive)
Similarly, sin β = 2 / Radix 5
So cos (α + β) = cos α cos β - sin α sin β = 1 / 5 root number 2-6 / 5 root number 2
=-1 / root 2
Obviously α + β = arccos (- 1 / Radix 2) = 135 degrees

Given that a / 2 is the fourth quadrant angle and cosa / 2 = 1 + X / x under the root sign, then the value of sina is

∵ A / 2 is the fourth quadrant angle
∴sin0.5a0
∴sin0.5a=√(1-cos²0.5a)
sina=2sin0.5acos0.5a

A is the second quadrant angle, P (x, Radix 5) is a point on its final edge, cosa = root 2 times 1 / 4 times x, find Sina

cosa=-sin(a-90°)=x/√(x^2+5)
So x / √ (x ^ 2 + 5) = (√ 2 / 4) x, x = - √ 3
Easy to get Sina = √ 10 / 4

If (Sina) ^ 2 + cos root (COSA) ^ 2 = 0 under 1 + Sina root sign, then the quadrant of angle a is?

The original formula = 1 + sina|sina| + cosa|cosa|
∵ original formula = 0
∴sina|sina|+cosa|cosa|=-1
∴sina

If the final edge of the angle alpha passes through the point P (X.2), and cos α is equal to the - fifth root sign five, then x is equal to

x=+-1
Let's first find sin from cos, and then we'll find Tan, which is the slope, which is equal to 2 / X

If Tan alpha = 2, then (2Sin alpha cos alpha + cos square alpha) is 1 =?

tana=2
1/(2sinacosa+cos^2a)
=(sin ^ 2A + cos ^ 2a) / (2sinacosa + cos ^ 2a) divide cos ^ 2a by cos ^ 2A
=(2) Tan (2) a = 2
=5/5
=1

Let vector alpha be equal to (negative root sign 2Sin (2x + / 4), 2cosx), B vector equal to (1,3sinx minus COS) ）X belongs to R. the function f (x) is equal to the vector a * vector B. find the minimum period of the function f (x)

f(x)=-√2sin(2x+∏/4)+2cosx(3sinx-cosx)
=-(sin2x+cos2x)+6sinxcosx-2(cosx)^2
=-sin2x-cos2x+3sin2x-(1+cos2x)
=2sin2x-2cos2x-1
=2√2sin(2x-∏/4)-1,
Its minimum positive period = Π

Given sin alpha cos alpha = radical 2, find the values of (1) Tan alpha + cot alpha (2) sin ^ 3 alpha cos ^ 3 alpha before 12:00~

sinα-cosα=√2
(sinα-cosα)²=2
sin²α+cos²α-2sinαcosα=2
sinαcosα=-1/2
(1) tanα+cotα
=sinα/cosα+cosα/sinα
=(sin²α+cos²α)/(sinαcosα)
=1/(-1/2)=-2
(2)
sin³α-cos³α
=(sinα-cosα)(sin²α+cos²α+sinαcosα)
=√2(1-1/2)
=√2/2

If sin alpha + cos alpha = root 6 / 2, what is alpha equal to

If the sum of the squares of the two is 1 and the square of the sum is 3 / 2, the difference between them is 1 / 2, that is to say, if the sine value of double angle of alpha is 1 / 2, then alfa is 15 degrees

If f (x) = √ 3cos (3x - θ) - sin (3x - θ) is an odd function, then Tan θ is equal to

f(x)=2[sin(π/3)cos(3x-θ)-cos(π/3)sin(3x-θ)]=2 sin(π/3-3x+θ)
It is an odd function and must satisfy the following conditions:
π / 3 + θ = k π, K is any integer
tanθ=tan(kπ-π/3)=-tanπ/3=-√3