sin 2θ when sin θ = 8/17, θ in quadrant II Step by Step

Step by step

The question

Question: find sin 2θ when sin θ = 8/17, θ in quadrant II.

We know sin θ, and the quadrant.

sin 2θ =

sin θ = 817, quadrant II

1 Draw the triangle

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

sin θ is y over r.

So y = 8 and r = 17.

sin θ = yr = 817

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

2 Find the third side

Pythagoras: x² + y² = r².

x² + 8² = 17²

8 squared is 64.

17 squared is 289.

x² + 64 = 289

Take away 64 on both sides: x² = 225.

x² = 289 − 64 = 225

The square root of 225 is 15.

Plus or minus.

x = ±√225 = ±15

In quadrant II, x is negative: left of the y-axis.

So x = −15.

x = −15

3 Find cos θ

Cos θ is x/r: −15/17.

So cos θ = −15/17.

cos θ = xr = −1517 = −1517

4 Write the rule

The double-angle rule: sin 2θ = 2 sin θ cos θ.

sin 2θ = 2 sin θ cos θ

Careful!

sin 2θ is not 2 times sin θ.

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

Careful: not sin 2θ = 2 sin θ = 1617 ✗

5 Work it out

Put in sin θ = 8/17 and cos θ = −15/17.

sin 2θ = 2 × 817 × (−1517)

Tops: 2 times 8 times 15 is 240.

Bottoms: 17 times 17 is 289.

One minus sign: the answer is negative.

2 × 8 × 15 = 240, 17 × 17 = 289

So sin 2θ = −240/289.

sin 2θ = −240289

6 Check

Let's check with the calculator.

The reference angle is tan⁻¹ of 8/15: about 28.07°.

In quadrant II, θ is about 151.93°.

θ ≈ 180° − 28.07° = 151.93°

Twice that is 303.86°.

sin of it is about −0.830.

sin 2θ ≈ sin 303.86° ≈ −0.830

And −240/289 is −0.830.

The same!

So it is right.

−240289 = −0.830

Let's look back.

First: draw the triangle.

Then: find the third side.

Then: find cos θ.

Then: write the rule.

Then: work it out.

Last: check.

Answer