How to find tan(2x)

Question

If cos x= 21/29, 0° < x < 90°, what is tan(2x)?

Answer 1

Hello Alyssa,

Given cos(x), we can first find sin(x). Using the well known identity sin²(x) + cos²(x) = 1, we have

sin²(x) + (21/29)² = 1

sin²(x) + 441/841 = 1

sin²(x) = 1 - 441/841

sin²(x) = 400/841

sin(x) = ±20/29.

Note that the above gives two possible values for sin(x), but since 0° < x < 90° (x is in the first quadrant), we choose the positive value. That is,

sin(x) = 20/29

Now we see that

tan(x) = sin(x)/cos(x) = [20/29]/[21/29]

tan(x) = 20/21

Now use the double angle formula for tangent.

tan(2x) = 2tan(x)/[1 -tan²(x)]

tan(2x) = 2(20/21)/[1 - (20/21)²]

tan(2x) = (40/41)/[1 - 400/441] = (40/41)/[41/441] = (40/41)•(441/41)

tan(2x) = 17640/1681

Hope that helps! Let me know if you need any further explanation.

William

Answer 2

cosx = 21/29

sinx = √[1 - cos²x] = 20/29

So, tanx = sinx / cosx = 20/21

tan(2x) = 2tanx / [1 - tan²x] = (40/21) / [1 - 400/441] = (40/21) / (41/441) = 840/41