The inclination angle of the straight line 3x + 4y-1 = 0

The inclination angle of the straight line 3x + 4y-1 = 0

Arc Tan (- 3 / 4), this doesn't seem to be a special angle. Do you learn anti trigonometric functions

The equation of straight line passing through the point (2,1) and the inclination angle is 1 / 2 of the inclination angle of the line 3x + 4Y + 1 = 0

The straight line 3x + 4Y + 1 = 0 is reduced to y = (- 3 / 4) X-1 / 4, slope = Tana = - 3 / 4. From the formula tan2 α = (2tan α) / (1-tan ^ 2 α), Tan α = (2tan (α / 2)) / (1-tan ^ 2 (α / 2)) = - 3 / 4. Let Tan (α / 2) = X. The above equation is reduced to 2x / (1-x ^ 2) = - 3 / 4-8x = 3-3x ^ 23x ^ 2-8x-3 = 0 (3

When passing through point a (2,1), its inclination angle is half of that of the straight line L: 3x + 4Y + 5 = 0 How do you want to find two slopes

The slope is 3 and - 1 | 3, the original straight line slope is - 3 | 4, the slope angle is 0 to 180, half of 0 to 90, so the slope of the line is 3, and then the equation is obtained according to the point oblique formula

Given that the final edge of angle α passes through point P (x, - 2) (x ≠ 0), and cos α = (√ 3 / 6) x, find sin α + 1 / Tan α

Cos α = (√ 3 / 6) x = x / bevel = 6 / √ 3
Passing point P (x, - 2) x = - √ ((6 / √ 3) ^ 2 - (√ 2) = - √ 10
cosα=（√3/6）x=√3/6*-√10=--√30/6
tana=-√2/-√10=1/√5
(sina+1)/tana=cosa+1/tana=-√30/6+1/√5= (-5√30+6√5)/30

Given that the final edge of the angle a is in the ray y = x (x is greater than 0), the sine, cosine, and tangent of a are calculated by any trigonometric function definition It's better to give the answer directly. Come on

According to the angle with the same final edge, this angle is equivalent to 45 ° + k * 360 ° and K is an integer, so Sina = root 2 / 2, cosa = root 2 / 2, Tana = 1

Given that the final edge of angle a crosses the point (a, 2a) (a ≠ 0), six trigonometric function values of a are obtained

Well, is this still required?
The opposite side of this angle 2A
Edge a
Hypotenuse √ (a ^ 2 + (2a) ^ 2) = √ 5A
sinA=2/√5=2√5/5
cosA=1/√5=√5/5
tanA=2/1=2
cotA=1/2
secA=√5/1
cscA=√5/2

Given that the final edge of angle α is on the line y = 3x, find cos α, sin α, Tan α

Take any point P (T, 3T) on the final edge, t ≠ 0
Then x = t, y = 3T
∴ r=√10 |t|
（1）t>0
r=√10 t
Then cosa = x / r = 1 / √ 10 = √ 10 / 10
sina=y/r=3/√10=3√10/10
tana=y/x=3
（2）t<0
r=-√10 t
Then cosa = x / r = - 1 / √ 10 = - √ 10 / 10
sina=y/r=-3/√10=-3√10/10
tana=y/x=3

（tan45°-cos60°）/sin60°×tan30° Detailed solution steps!

(√ 3) / √ (3) / √ (3) = (3) / (2)
=1/√3×√3/3
=1/3

The values of the equilateral triangle Tan 30 °, sin 60 °, cos 60 ° and Tan 60 ° are used

TAN 30=0.577 sin60=0.866 cos60 =0.5 tan60=1.732

Sin330 degree is equal to sin330 = sin (360-30) = - sin30 = - 1 / 2 why - sin30 ° can't understand and ask for help Formula Sina (π - a) = Sina can't understand the answer

The period of sin (a) function is t (a) = 360 °, that is, the angle is arbitrarily added or subtracted by 360 ° and the value remains unchanged. To understand sin330 ° = sin (- 30 °), consider sin0 ° = sin 360 °