If Tan θ = 2, then sin ^ 2 θ + sin θ cos θ 2cos ^ 2 θ= RT sinθcosθ-2cos^2θ

If Tan θ = 2, then sin ^ 2 θ + sin θ cos θ 2cos ^ 2 θ= RT sinθcosθ-2cos^2θ

The original formula = (sin ^ 2 θ + sin θ cos θ + 2cos ^ 2 θ) / (sin ^ 2 θ + cos ^ 2 θ)
Divide by cos ^ 2 θ
=(tan^2θ+tanθ+2)/(tan^2θ+1)

Given Tan α = 3, find sin α + cos α / sin α - 2cos α

tanα=3
(sinα+cosα)/(sinα-2cosα)
The denominator is divided by cos α
= （tanα+1）/（tanα-2）
= （3+1）（3-2）
= 4

Tan α = 3, find 2cos α / (sin α + cos α)

solution
tana=3
Qi
2cosa / (Sina + COSA) -- denominator divided by cosa
=2/(tana+1)
=2/(3+1)
=2/4
=1/2

If sin (7 π - α) - 3cos (3 π / 2 + α) = 2, then the value of (sin (π - α) + cos (π + α)) / (sin α + cos (- α)) is

3 π / 2 + a) = 2Sin (7 π - a) - 3cos (3 π / 2 + a) = 2Sin (7 π - a) = sin (6 π + π - a) = sin (6 π + π - a) = sin (π - a) = sina-3cos (3 π / 2 + a) = - 3cos (2 π - π / 2 + a) = - 3cos (a-π / 2 / 2 + a) = - 3cos (a-π / 2 / 2-A) = - 3sina, so, sina-3sina = - 2sina = 2sina = 1, cosa = 0sin (π - a) + cos s cos s (COS) cos = 0, cosa = 0 sin (π - a) + cos cos, cos, cos = 0, cosa = 1, cosa(π + a) = Sina cosa =

Given sin α + 3cos α = 0, calculate the values of sin α and cos α

From sin α = - 3cos α, sin α is in the second and fourth quadrants. When α is the second quadrant angle, sin α = 31010, CO is the second quadrant

If α is the angle of the second quadrant, then 2 α cannot be in () The first and second quadrants B. the second and third quadrants C. The third and fourth quadrants D. the first and fourth quadrants 90°+360°k＜a＜180°+360°k Then π + 4K π < 2A < 2 π + 4K π It can only be explained that 2a is the third or fourth quadrant. Why B D is right. Can 2A be in the second, third or fourth quadrant

The answer is a! 2A can only be in the third or fourth quadrant! BD is not the answer because there is a third or fourth quadrant among them, so it is not the answer! Your thinking question is so unique! Ha ha. This is or is not, you know

If a is the second quadrant angle, how many quadrants are a / 2, 2a, a / 3 and 180 + a respectively? Fine fine fine fine fine fine fine fine fine fine fine fine fine fine fine fine fine fine fine fine fine Back to the first floor: please do not limit to 0 ~ 360

A is the second quadrant angle, (2n + 1 / 2) Π (n + 1 / 4) Π (2n + 1) Π < 2A < (2n + 2) Π, 2a is the third and fourth quadrant angle
Others are similar

If angle a is in the first quadrant. What quadrant is the 2 / 2 part a of angle 2A? What if angle a is in quadrant 2.3.4? What about 2a.2 fraction a? Find the situation in the table 2A Two a

A one two three four
2A 1,2 quadrant 3,4 quadrant 1,2 quadrant 3,4 quadrant
A / 2 1,3 quadrant 1,3 quadrant 2,4 quadrant 2,4 quadrant

Given that a is the fourth quadrant angle, find the quadrant where the final edge of a / 2, a / 3, 2a is located

If the limit is 270 a 360
135 < A / 2 < 180, a / 2 in the second quadrant;
90 < A / 3 < 120, a / 3 in the second quadrant;
540 ＜ 2A ＜ 720, 2a in the third or fourth quadrant
If 270 < a < 360 is not defined, each quadrant is uncertain

Given sin = 3 / 4, and the final edge of angle a is in the second quadrant, what quadrant is the final edge of 2A

The final edge of angle a is in the second quadrant
2kπ+π/2π4kπ+π<2A<4kπ+2π
The final edge of 2a is in the third and fourth quadrants