Finding exact values for sin, cos, tan.

Question

Find the exact values of sin 2θ, cos2θ, tan 2θ, and the quadrant in which 2θ lies.

cosθ=-21/29, θ in quadrant III

Answer 1

If θ is in quadrant III, then π < θ < 3π/2 and then 2π<2θ< 3π (after multiplying by 2).

So 2θ is in quadrant I or II (still to be finally determined).

Cos θ in standard orientation on a circle with center at (0,0) and radius r can be represented by the formula

cos θ = x/r where, r is always positive.

Given cos θ = -21/29, we can take r = 29 and then x = -21. In quadrant III, y would also be negative.

Then with x, y, and r forming a right triangle, x^2 + y^2 = r^2 and solving for y gives

y = ± √(r^2-x^2) = - √(29^2-(-21)^2 = -√(841-441) = -20.

That makes sin θ = y/r = -20/29.

We can now make use of double angle formulas:

sin 2θ = 2 sin θ cos θ = 2 x -20/29 x -21/29 = 840/841        //

cos 2θ = cos^2 θ - sin^2 θ = (-21/29)^2 - (-20/29)^2 = 41/841  //

tan 2θ = sin 0 / cos 0 = (840/841)/(41/841) = 840/41.         //

Since all of these are positive, 2θ is indeed in quadrant I.      //

Answer 2

cos θ = -21 / 29 in Q III, sin θ = ?

Now use the formulas.