Find cos (2q), sin q = 20/29, q lies in quadrant I.?

Answer 1

cos(2q) = 1 - 2sin²q <-- basic identity

cos(2q) = 1 - 2(20/29)²

cos(2q) = 1 - 2(400/841)

cos(2q) = 1 - 800/841

cos(2q) = 41/841

Answer 2

use the formula, cos 2q = 1 - 2 sin^2 q
= 1 - 2(20/29)^2
= 0.049

Answer 3

cos(2q) = cos(q)cos(q) - sin(q)sin(q)

To find the quadrant draw a circle and assume it has a diameter of 29. Draw a right triangle with 20 along the y axis and then finish drawing the triangle. Using the pyth theorem we get the x axis length as 21.

Then cos(q) = 21/29 = 0.724


(21/29)^2 - (20/29)^2 = 0.048

NOTICE: you can check that this is correct by finding the value of the angle using the inverse sine function of your calculator, invsin(20/29) = 43.6 degrees, and 2 times 43.6 degrees is 87.2 degrees (rounded). The cos(87.2 degrees) = 0.048

Answer 4

cos(2q)=1-2sin²q
=1-2(20/29)² = 1-800/841 = 41/841

Answer 5

cos (2q)
= 1 - 2sin^2(q)
= 1 - 2(20/29)^2
= 41/841

Answer 6

This right-angled triangle must have its third side 21 units.
Hence cos q = 21/29.
cos(2q) = cos(^2)x - sin(^2)x = [21/29](^2) - [20/29](^2)
= 41/[29](^2)