Step by step
The question
Question: find cos 2θ when sin θ = −5/13, θ in quadrant IV.
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 = −5 and r = 13.
r is a distance, always positive, so the minus belongs to y.
We draw it in quadrant IV: a right triangle from O to the point.
2 Find the third side
Pythagoras: x² + y² = r².
−5 squared is 25.
13 squared is 169.
Take away 25 on both sides: x² = 144.
The square root of 144 is 12.
Plus or minus.
In quadrant IV, x is positive: right of the y-axis.
So x = 12.
3 Find cos θ
Cos θ is x/r: 12/13.
So cos θ = 12/13.
4 Write the rule
The double-angle rule: cos 2θ = cos² θ − sin² θ.
Careful!
cos 2θ is not 2 times cos θ.
The 2 is inside: it doubles the angle, not the answer.
24/13 is not between −1 and 1, and a cosine always is.
5 Work it out
Put in cos θ = 12/13 and sin θ = −5/13.
Brackets keep each one together.
Square each one: top times top, bottom times bottom.
12 squared is 144.
−5 squared is 25.
Same bottom, 169: 144 minus 25 is 119.
So cos 2θ = 119/169.
6 Check
Let's check with the calculator.
The reference angle is tan⁻¹ of 5/12: about 22.62°.
In quadrant IV, θ is about 337.38°.
Twice that is 674.76°.
cos of it is about 0.704.
And 119/169 is 0.704.
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.