Given that P (3T, - 4T) (T < 0) is a point on the terminal edge of angle α, then sin α - Tan α =?

Given that P (3T, - 4T) (T < 0) is a point on the terminal edge of angle α, then sin α - Tan α =?

sinα=-4t/√[(3t)²＋(-4t)²]=﹣4/5
tanα=-4t/3t=-4/3

Given that the terminal edge of angle α passes through point P (- 4T, 3T) (T > 0), then sin α + cos α =?

√[﹙-4t)²+(3t)²]=5t
sinα+cosα
=(3t/5t)+(-4t/5t)
=3/5-4/5
=-1/5

tanα=2,sinαXcosα=?

sinαXcosα
=tanα*cos^2 α
=2*1/(1/cos^2α)
=2*1/((sin^2 α+cos^2 α)cos^2 α)
=2/(tan^2 α+1)
=2/(2^2+1)
=2/5

Prove (Tan xsin x) / (Tan x-sin x) - (1 + cos x) / sin X
There's a wrong number on it
(tan xsin x)/(tan x-sin x)=(1+cos x)/sin x

It is proved that because TaNx = SiNx / cosx, the denominator of cosx = SiNx / TaNx (Tan xsin x) / (Tan x-sinx) is divided by TaNx = SiNx / (1-sinx / TaNx) = SiNx / (1-cosx) and multiplied by (1 + cosx) = SiNx (1 + cosx) / [(1-cosx) (1 + cosx)] = SiNx (1 + cosx) / (1-c

What is sin (- 1200 degrees) xcos1290 degrees + cos (- 1020 degrees) xsin (- 1050 degrees) + Tan 945 degrees?
Why is cos (- 1020) = - cos (2x360 + 300) negative?
Isn't the induction formula cos (- angle) = cos positive?

Sin (- 1200 degrees) xcos1290 degrees + cos (- 1020 degrees) xsin (- 1050 degrees) + Tan 945
=sin(-1080-120)xcos(1080+180+30)+cos（-1080+60）xsin（-1080+30）
+tan（720+180+45）
=sin(120)cos(30)+cos60sin30+tan45=2
Why is cos (- 1020) = - cos (2x360 + 300) negative?
Isn't the induction formula cos (- angle) = cos positive?
This is wrong. We should not put forward a minus sign

Simplify cos (A-wu) / sin (wu-a) xsin (A-2 Wu) xcos (2 wu-a) to a detailed process!

Cos (A-wu) / sin (wu-a) xsin (A-2 Wu) xcos (2 wu-a)
=cos(π-a)/sina×sina×cosa
=-cosa×cosa
=-cos²a

Given sin (π / 4 + a) = a, find the value of sin (5 π / 4) / cos (9 π / 4 + a) xcos (7 π / 4-A)

sin(5π/4)/cos(9π/4+a)xcos(7π/4-a)=sin(π+π/4)/cos(2π+π/4+a)xcos(2π-π/4-a)=sin(π+π/4)/cos(2π+π/4+a)xcos(2π-π/4-a)=-sinπ/4/cos(π/4+a)xcos[2π-(π/4+a)]=-sinπ/4/cos(π/4+a)xcos(π/4+a)=-...

It is proved that the new coordinates of the points (x, y) in the plane are (xcos a-ysin a, xsin a-ycos a) by rotating the angle a in the counter clockwise direction
This problem is a point in the plane rectangular coordinate system, which requires the use of geometric methods such as similar shape to prove! (polar coordinate proof has been solved)

The new coordinates should be (xcos a-ysin a, xsin a + ycos a)
prove:
Let the new coordinates be (U, V) and the angle of turning (x, y) counterclockwise from the positive direction of x-axis be B
Then x / (x ^ 2 + y ^ 2) = cos (b)
y/(x^2+y^2)=sin(b)
u/(x^2+y^2)=cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
=cos(a)*x/(x^2+y^2)-sin(a)*y/(x^2+y^2)
v/(x^2+y^2)=sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
=sin(a)*x/(x^2+y^2)+cos(a)*y/(x^2+y^2)
So u = cos (a) * x-sin (a) * y v = sin (a) * x + cos (a) * y
The new coordinates after rotation are (U, V) = (xcos (a) - ysin (a), xsin (a) + ycos (a))

Let r > 0, then the positional relation between the straight line xcos θ + ysin θ = R (θ is a constant) and the circle x = RCOs φ y = rsin φ (φ is a parameter) is ()
A. Intersection B. tangency C. separation D. depending on the size of R

The center of the circle x = RCOs φ y = rsin φ is the coordinate origin, and the radius is R. the distance from the center of the circle to the straight line is rsin2 θ + Cos2 θ = R, so the straight line is tangent to the circle

The maximum distance from point a (1,1) to the line xcos θ + ysin θ - 2 = 0 is ()
A. 1+2B. 2+2C. 1+3D. 2+3

The distance from point a (1,1) to the straight line xcos θ + ysin θ - 2 = 0, d = | cos θ + sin θ − 2 | cos 2 θ + sin 2 θ = 2 − 2Sin (θ + π 4), if and only if sin (θ + π 4) = - 1, D gets the maximum, d = 2 + 2