If 4,3 is the last point of the angle a, find the value of cos a 3 tan a 4 sin 3 a cos a 5 2

-4,3) is a point on the terminal edge of the angle a.

Derived. a is the second quadrant angle.

sina=3/5

cos(a-3π)=cos(a-3π+4π)=cos(a+π)=-cosa

tan(a-4π)=tana

sin(3π-a)=-sin(a-3π)=-sin(a-3π+4π)=-sin(a+π)=sina

cos(a+5π/2)=cos(a+5π/2-2π)=cos(a+π/2)=-sina

cos(a-3π)*tan(a-4π)/sin(3π-a)*cos(a+5π/2)

-cosa)tana]/[sina*(-sina)]

sina/-sin²a

1/sina

So angle a is the second quadrant angle.

So the original formula = cos[-(3 -a)]*tana) sina*(-sina).

cosa*tana/sin^2a=1/sina

According to the title, p is in quadrant 2, so a is the obtuse angle of Paibi, then sin a=-4 feast number 5, cosa=3 5, tan a=-4 dust auspicious qin3

The angle a is the second quadrant angle of the destroyer Kai, according to the inscription of the Zhaosouqiao, sina = 4 5, cosa = -3 5, tana = -4 3

Original = (4 5-3 5) * (4 3).

The terminal edge of the angle a passes through the stupid dust point (-3 5, 4 5), sina = 4 5 cosa = -3 shen chain 5 tana = -4 with filial piety 3

sin(煀 2+a) + cos(煀-a) + tan (2 v-a).

cosa-cosa+-tana-tana

The first. Isn't sin( +a)+cos(3 2-a) a simple trigonometric transformation? The result is obviously equal to 0, which has nothing to do with the previous conditions, and is always true.

The second one is to be used.

You give. conditions, but I really don't understand.

Noe Horn? Satisfied.

-4,3) is the angle a

In the end, he regretted and buried.

On a little bit of blue worms.

Draw up the defeat of the reed. A Yes.

Second quadrant angle.

sina=3/5

cos(a-3π)=cos(a-3π+4π)=cos(a+π)cosa

tan(a-4π)=tana

sin(3π-a)=-sin(a-3π)=sin(a-3π+4π)=sin(a+π)sina

cos(a+5π/2)=cos(a+5π/2-2π)=cos(a+π/2)=-sina

cos(a-3π)*tan(a-4π)/sin(3π-a)*cos(a+5π/2)

-cosa)tana]/[sina*(-sina)]

sina/-sin²a

1/sina