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Recap: the Circle Identity

Part of the Trig Identities section of Coddy's Trigonometry journey. Lesson 3 of 48.

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Challenge

Show that 8−8cos⁡(x)28-8{\cos\left(x\right)}^{2} equals 8sin⁡(x)28{\sin\left(x\right)}^{2} at every angle. Rewrite it until it reads 8sin⁡(x)28{\sin\left(x\right)}^{2}.

Then press Check.

Try it yourself

8−8cos⁡2x8-8\cos^{2} x

Rewrite as 8sin⁡2x8\sin^{2} x

All lessons in Trig Identities