Solve sin 2θ when sin θ = 3/5, θ in quadrant I Easy

The question

Question: find sin 2θ when sin θ = 3/5, θ in quadrant I.

We know sin θ, and the quadrant.

1 Draw the triangle

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

sin θ is y over r.

So y = 3 and r = 5.

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

2 Find the third side

Pythagoras: x² + y² = r².

3 squared is 9.

5 squared is 25.

Take away 9 on both sides: x² = 16.

The square root of 16 is 4.

Plus or minus.

In quadrant I, x is positive: right of the y-axis.

So x = 4.

3 Find cos θ

Cos θ is x/r: 4/5.

So cos θ = 4/5.

4 Write the rule

The double-angle rule: 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.

6/5 is not between −1 and 1, and a sine always is.

5 Work it out

Put in sin θ = 3/5 and cos θ = 4/5.

Tops: 2 times 3 times 4 is 24.

Bottoms: 5 times 5 is 25.

So sin 2θ = 24/25.

6 Check

Let's check with the calculator.

The reference angle is tan⁻¹ of 3/4: about 36.87°.

In quadrant I, θ is about 36.87°.

Twice that is 73.74°.

sin of it is about 0.960.

And 24/25 is 0.960.

The same!

So it is right.

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.

▶ Watch it solved

sin 2θ when cos θ = -4/5, θ in quadrant II Medium

The question

Question: find sin 2θ when cos θ = −4/5, θ in quadrant II.

We know cos θ, and the quadrant.

1 Draw the triangle

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

cos θ is x over r.

So x = −4 and r = 5.

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

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

2 Find the third side

Pythagoras: x² + y² = r².

−4 squared is 16.

5 squared is 25.

Take away 16 on both sides: y² = 9.

The square root of 9 is 3.

Plus or minus.

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

So y = 3.

3 Find sin θ

Sin θ is y/r: 3/5.

So sin θ = 3/5.

4 Write the rule

The double-angle rule: 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.

6/5 is not between −1 and 1, and a sine always is.

5 Work it out

Put in sin θ = 3/5 and cos θ = −4/5.

Tops: 2 times 3 times 4 is 24.

Bottoms: 5 times 5 is 25.

One minus sign: the answer is negative.

So sin 2θ = −24/25.

6 Check

Let's check with the calculator.

The reference angle is tan⁻¹ of 3/4: about 36.87°.

In quadrant II, θ is about 143.13°.

Twice that is 286.26°.

sin of it is about −0.960.

And −24/25 is −0.960.

The same!

So it is right.

Let's look back.

First: draw the triangle.

Then: find the third side.

Then: find sin θ.

Then: write the rule.

Then: work it out.

Last: check.

▶ Watch it solved

tan 2θ when tan θ = 5/12, θ in quadrant III Harder

The question

Question: find tan 2θ when tan θ = 5/12, θ in quadrant III.

We know tan θ, and the quadrant.

1 Write the rule

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

Careful!

tan 2θ is not 2 times tan θ.

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

2 Work it out

Put in tan θ = 5/12.

On top: 2 times 5/12 is 5/6.

At the bottom: 5/12 squared is 25/144.

1 is 144/144.

So 1 minus 25/144 is 119/144.

A fraction over a fraction.

Dividing by a fraction is times its flip: times 144/119.

Multiply and make it simpler: tan 2θ = 120/119.

3 Check

Let's check with the calculator.

The reference angle is tan⁻¹ of 5/12: about 22.62°.

In quadrant III, θ is about 202.62°.

Twice that is 405.24°.

tan of it is about 1.008.

And 120/119 is 1.008.

The same!

So it is right.

Let's look back.

First: write the rule.

Then: work it out.

Last: check.

▶ Watch it solved