tan 2θ when cos θ = -8/17, θ in quadrant II Step by Step

Step by step

The question

Question: find tan 2θ when cos θ = −8/17, θ in quadrant II.

We know cos θ, and the quadrant.

tan 2θ =

cos θ = −817, quadrant II

1 Draw the triangle

Take a point (x, y) at distance r from O.

cos θ is x over r.

So x = −8 and r = 17.

r is a distance, always positive, so the minus belongs to x.

cos θ = xr = −817 = −817

We draw it in quadrant II: a right triangle from O to the point.

2 Find the third side

Pythagoras: x² + y² = r².

y² + (−8)² = 17²

−8 squared is 64.

17 squared is 289.

y² + 64 = 289

Take away 64 on both sides: y² = 225.

y² = 289 − 64 = 225

The square root of 225 is 15.

Plus or minus.

y = ±√225 = ±15

In quadrant II, y is positive: above the x-axis.

So y = 15.

y = 15

3 Find tan θ

Tan θ is y/x: 15 over −8.

So tan θ = −15/8.

tan θ = yx = 15−8 = −158

4 Write the rule

The double-angle rule: tan 2θ = 2 tan θ over (1 − tan² θ).

tan 2θ = 2 tan θ1 − tan² θ

Careful!

tan 2θ is not 2 times tan θ.

The 2 is inside: it doubles the angle, not the answer.

Careful: not tan 2θ = 2 tan θ = −154 ✗

5 Work it out

Put in tan θ = −15/8.

tan 2θ = 2 × (−158)1 − (−158)²

On top: 2 times −15/8 is −15/4.

At the bottom: −15/8 squared is 225/64.

tan 2θ = −1541 − 22564

1 is 64/64.

So 1 minus 225/64 is −161/64.

1 − 22564 = 6464 − 22564 = −16164

A fraction over a fraction.

tan 2θ = −154−16164

Dividing by a fraction is times its flip: times −64/161.

tan 2θ = −154 × −64161

Multiply and make it simpler: tan 2θ = 240/161.

tan 2θ = 240161

6 Check

Let's check with the calculator.

The reference angle is tan⁻¹ of 15/8: about 61.93°.

In quadrant II, θ is about 118.07°.

θ ≈ 180° − 61.93° = 118.07°

Twice that is 236.14°.

tan of it is about 1.490.

tan 2θ ≈ tan 236.14° ≈ 1.490

And 240/161 is 1.491.

The same!

So it is right.

240161 = 1.491

Let's look back.

First: draw the triangle.

Then: find the third side.

Then: find tan θ.

Then: write the rule.

Then: work it out.

Last: check.

Answer