Menu

A Difference of Squares

Part of the Trig Identities section of Coddy's Trigonometry journey. Lesson 8 of 48.

In algebra, (a−b)(a+b)=a2−b2\left(a-b\right)\left(a+b\right) = {a}^{2}-{b}^{2}. This is called a difference of squares. It works the same with a trig function in place of b: (1−sin⁡(x))(1+sin⁡(x))=1−sin⁡(x)2\left(1-\sin\left(x\right)\right)\left(1+\sin\left(x\right)\right) = 1-{\sin\left(x\right)}^{2}. Then the identity turns this into cos⁡(x)2{\cos\left(x\right)}^{2}.

Look for a square minus 1, or 1 minus a square. For example, (csc⁡(x)−1)(csc⁡(x)+1)=csc⁡(x)2−1\left(\csc\left(x\right)-1\right)\left(\csc\left(x\right)+1\right) = {\csc\left(x\right)}^{2}-1, which is cot⁡(x)2{\cot\left(x\right)}^{2}.

challenge icon

Challenge

Show that (sec⁡(x)−1)(sec⁡(x)+1)\left(\sec\left(x\right)-1\right)\left(\sec\left(x\right)+1\right) equals tan⁡(x)2{\tan\left(x\right)}^{2}. Rewrite it until it reads tan²x.

Then press Check.

Try it yourself

(sec⁡x−1)(sec⁡x+1)\left(\sec x-1\right)\left(\sec x+1\right)

Rewrite as tan⁡2x\tan^{2} x

All lessons in Trig Identities