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Pythagorean Identities

The Pythagorean Identities are identities that express the Pythagorean Theorem in terms of trigonometric functions. They can be expressed as:
cos⁡2θ+sin⁡2θ=11+tan⁡2θ=sec⁡2θ1+cot⁡2θ=csc⁡2θ\boxed{\begin{array}{rcl} \cos^2\theta+\sin^2\theta&=&1\\\\ 1+\tan^2\theta&=&\sec^2\theta\\\\ 1+\cot^2\theta&=&\csc^2\theta \end{array}}

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Proof Using the Unit Circle

Let P(x,y)P(x,y) be a point on a circle centered at (0,0)(0,0) with radius rr (shown below).

The equation of the circle is expressed by x2+y2=r2\boxed{x^2+y^2=r^2}.


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We can manipulate the equation of the circle to derive the Pythagorean Identities.

a.x2+y2=r2x2r2+y2r2=1(xr)2+(yr)2=1sin⁡θ=yr, cos⁡θ=xr(cos⁡θ)2+(sin⁡θ)2=1cos⁡2θ+sin⁡2θ=1Pythagorean Identityb.cos⁡2θ+sin⁡2θ=1(cos⁡2θcos⁡2θ)+(sin⁡2θcos⁡2θ)=(1cos⁡2θ)tan⁡θ=sin⁡θcos⁡θ, sec⁡=1cos⁡θ1+tan⁡2θ=sec⁡2θPythagorean Identityc.cos⁡2θ+sin⁡2θ=1(cos⁡2θsin⁡2θ)+(sin⁡2θsin⁡2θ)=(1sin⁡2θ)cot⁡θ=cos⁡θsin⁡θ, csc⁡=1sin⁡θcot⁡2θ+1=csc⁡2θPythagorean Identity\begin{array}{rcl} \textbf{a.}\\\\ &x^2+y^2&=&r^2\\\\ &\dfrac{x^2}{r^2}+\displaystyle\frac{y^2}{r^2}&=&1\\\\ &\Bigg(\displaystyle\frac{x}{r}\Bigg)^2+\Bigg(\displaystyle\frac{y}{r}\Bigg)^2&=&1&&{\color{red}\footnotesize\sin\theta=\frac{y}{r},~\cos\theta=\frac{x}{r}}\\\\ &(\cos\theta)^2+(\sin\theta)^2&=&1\\\\ &\cos^2\theta+\sin^2\theta&=&1&&{\color{Red}\footnotesize\textbf{Pythagorean Identity}}\\\\\hline\\ \textbf{b.}\\\\ &\cos^2\theta+\sin^2\theta&=&1\\\\ &\Bigg(\dfrac{\cos^2\theta}{\cos^2\theta}\Bigg)+\Bigg(\dfrac{\sin^2\theta}{\cos^2\theta}\Bigg)&=&\Bigg(\dfrac{1}{\cos^2\theta}\Bigg)&&{\color{Red}\footnotesize\tan\theta=\frac{\sin\theta}{\cos\theta},~\sec=\frac{1}{\cos\theta}}\\\\ &1+\tan^2\theta&=&\sec^2\theta&&{\color{Red}\footnotesize\textbf{Pythagorean Identity}}\\\\\hline\\ \textbf{c.}\\\\ &\cos^2\theta+\sin^2\theta&=&1\\\\ &\Bigg(\dfrac{\cos^2\theta}{\sin^2\theta}\Bigg)+\Bigg(\dfrac{\sin^2\theta}{\sin^2\theta}\Bigg)&=&\Bigg(\dfrac{1}{\sin^2\theta}\Bigg)&&{\color{Red}\footnotesize\cot\theta=\frac{\cos\theta}{\sin\theta},~\csc=\frac{1}{\sin\theta}}\\\\ &\cot^2\theta+1&=&\csc^2\theta&&{\color{Red}\footnotesize\textbf{Pythagorean Identity}} \end{array}


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Example: Pythagorean Identities


Prove the following identity:

cos⁡θ1−sin⁡2θ=sec⁡θ\dfrac{\cos\theta}{1-\sin^2\theta}=\sec\theta


LHSRHSPythagorean Identitycos⁡θ1−sin⁡2θsec⁡θcos⁡θcos⁡2θsec⁡θReciprocal Identity1cos⁡θsec⁡θsec⁡θsec⁡θ\begin{array}{cc|cc}\\ \textbf{LHS}&&\textbf{RHS}\\\hline\\ \color{red}{\scriptsize\text{Pythagorean Identity}}&\dfrac{\cos\theta}{1-\sin^2\theta}&\sec\theta\\\\\\ &\dfrac{\cos\theta}{\cos^2\theta}&\sec\theta\\\\\\ \color{red}{\scriptsize\text{Reciprocal Identity}}&\dfrac{1}{\cos\theta}&\sec\theta\\\\\\ &\sec\theta&\sec\theta \end{array}

Since LHS = RHS, then the following identity has been proved.

Practice: Pythagorean Identities

Simplify (1−cos⁡2θ)(csc⁡θ)(1-\cos^2\theta)(\csc\theta) into one trigonometric function.

Practice: Pythagorean Identities

cos⁡2θ+cos⁡2θtan⁡2θ\cos^2\theta+\cos^2\theta\tan^2\theta is equivalent to which of the following?

Practice: Pythagorean Identities

Which of the following is equivalent to sin⁡θ1−cos⁡θ+sin⁡θ1+cos⁡θ\dfrac{\sin\theta}{1-\cos\theta}+\dfrac{\sin\theta}{1+\cos\theta}?
Extra Practice