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Squares Erase the Signs

Part of the Trig Identities section of Coddy's Trigonometry journey. Lesson 2 of 48.

A point on the circle of radius 1 is 35\frac{3}{5} high. How far across is it? Most people say 45\frac{4}{5}. That is only half the answer. A second point at the same height is on the other side.

Two points on the unit circle at the same height 3/5, one right of the y-axis, one left

Subtract sin⁡(x)2{\sin\left(x\right)}^{2} from both sides: cos⁡(x)2=1−sin⁡(x)2=1−925{\cos\left(x\right)}^{2} = 1-{\sin\left(x\right)}^{2} = 1-\frac{9}{25}, which is 1625\frac{16}{25}. Two numbers have the square 16/25: 45\frac{4}{5} and −45\frac{-4}{5}. The identity gives the size. The quadrant gives the sign. In quadrant II, left of the y-axis, cos⁡(x)=−45\cos\left(x\right) = \frac{-4}{5}.

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Challenge

A point on the circle of radius 1 is 2029\frac{20}{29} high. Call how far across it is c. Then c2+20292=1{c}^{2}+{\frac{20}{29}}^{2} = 1. Find every value c can have.

Then press Check.

Try it yourself

c2+20292=1c^{2}+\frac{20}{29}^{2} = 1

Get cc alone

All lessons in Trig Identities