The parameter equation of the line is x = 2 + tsin78, y = - 1-tcos78, then the inclination angle of the line L is

The parameter equation of the line is x = 2 + tsin78, y = - 1-tcos78, then the inclination angle of the line L is

x-2=tsin18
y+1=-tcos18
(y+1)/(x-2)=-tan12
So the tilt angle is 168

What is the inclination angle of the straight line {x = tsin20 degrees - 1, y = 2-tcos20 degrees} (t is the parameter)?

Upstairs, I looked down the title. In x = tsin20 degrees - 1, y = 2-tcos20 degrees, the "- 1" and "2" only translate the straight line without changing the inclination angle, so I just need to look at sin20 and cos20. Translate the straight line to the origin as x = tsin20 degrees, y = - tcos20 degrees, k = Tana = - cos20 / sin20 = - cot20 = - tan70a = 180 +

Sina + cosa = (1 - √ 3) / 2, and X ∈ (π / 2, π), then Sina cosa=

（sina+cosa）²=sin²a+cos²a+2sina×cosa=（2-√3）/2
∵sin²a+cos²a=1
∴2sina×cosa=（2-√3）/2-1=-√3/2
sina-cosa=（1+√3）/2

Simplified radical (1-sinq / 1 + SINQ) + radical (1 + cosq / 1-cosq) (π / 2)

Radical (1-sinq / 1 + SINQ) + radical (1 + cosq / 1-cosq)
=-(1-sinQ)/cosQ+(1+cosQ)/sinQ
=[-sinQ+sin^2(Q)+cosQ+cos^2(Q)]/[sinQcosQ]
=2(1-sinQ+cosQ)/sin(2Q)

(a + SINQ) (B + cosq) can simplify m?

(a+sinQ)(b+cosQ)
=ab+acosQ+bsinQ+sinQcosQ

If SINQ / (1 + cosq) = 1 / 3, then the value of SINQ is___

From the universal formula
sinQ =2tan(Q/2)/(1+tan^2(Q/2))
cosQ=(1-tan^2(Q/2))/(1+tan^2(Q/2))
So the above formula = Tan (Q / 2) = 1 / 3
Substituting SINQ = 2tan (Q / 2) / (1 + Tan ^ 2 (Q / 2))
So SINQ = 3 / 5

If SINQ + cosq = √ 2, then the value of Tan (Q + π / 3) is?

∵sinQ+cosQ=√2
∴(√2)sin(Q+π/4)=√2
∴sin(Q+π/4)=1
∴Q+π/4=π/2+2kπ(k∈Z)
∴Q=π/4+2kπ(k∈Z)
∴tanQ=tan(π/4+2kπ)=tan(π/4)=1
∴tan(Q+π/3)
=[tanQ+tan(π/3)]/[1-tanQtan(π/3)]
=(1+√3)/(1-√3)
=(1+√3)^2/[(1-√3)(1+√3)]
=(1+2√3+3)/(1-3)
=(4+2√3)/(-2)
=-(2+√3).

Using trigonometric function line to determine the definition domain of function f (x) = √ sinx-1 / 2

Using trigonometric function line to determine the definition domain of function f (x) = √ (sinx-1 / 2)
sinx≥1/2
If a straight line y = 1 / 2 is drawn in the unit circle and two intersections A and B of the unit circle are used as rays OA and ob, then the angle X with the end edge falling in ∠ AOB (including the boundary) belongs to the function domain
The same angle as the terminal edge of OA: 2K π + π / 6,
The same angle as ob terminal edge: 2K π + 5 π / 6,
The definition domain of function f (x) = √ (sinx-1 / 2) [2K π + π / 6,2k π + 5 π / 6], K ∈ Z
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Trigonometric function salon chart (5) trigonometric function line:

The function y = LG (1-sinx) - LG (1 = SiNx) is odd, but not even, right?

That's right
Because f (- x) = - f (x)
But it does not satisfy f (- x) = f (x)
And the domain of definition is that x is not equal to K π + π / 2
Origin of Symmetry
So it's an odd function

The parity of the function y = SiNx + TaNx is ()
A. Odd function B. even function C. both odd and even function D. non odd and non even function

The definition domain of function y = f (x) = SiNx + TaNx is {x | x ≠ K π + π 2, K ∈ Z}. It is symmetric about the origin and satisfies f (- x) = sin (- x) + Tan (- x) = - (SiNx + TaNx) = - f (x), so the function is odd, so a