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Pythagorean Identities

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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}}cos2θ+sin2θ1+tan2θ1+cot2θ​===​1sec2θcsc2θ​​

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Proof Using the Unit Circle
Let P(x,y)P(x,y)P(x,y) be a point on a circle centered at (0,0)(0,0)(0,0) with radius rrr (shown below).

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


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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}a.b.c.​x2+y2r2x2​+r2y2​(rx​)2+(ry​)2(cosθ)2+(sinθ)2cos2θ+sin2θcos2θ+sin2θ(cos2θcos2θ​)+(cos2θsin2θ​)1+tan2θcos2θ+sin2θ(sin2θcos2θ​)+(sin2θsin2θ​)cot2θ+1​===========​r211111(cos2θ1​)sec2θ1(sin2θ1​)csc2θ​​sinθ=ry​, cosθ=rx​Pythagorean Identitytanθ=cosθsinθ​, sec=cosθ1​Pythagorean Identitycotθ=sinθcosθ​, csc=sinθ1​Pythagorean Identity​​

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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\theta1−sin2θcosθ​=secθ

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}LHSPythagorean IdentityReciprocal Identity​1−sin2θcosθ​cos2θcosθ​cosθ1​secθ​RHSsecθsecθsecθsecθ​​

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)(1−cos2θ)(cscθ) into one trigonometric function.

A) sin⁡2θcos⁡θ\dfrac{\sin^2\theta}{\cos\theta}cosθsin2θ​ 

B) csc⁡θ\csc\thetacscθ 

C) csc⁡θ\csc\thetacscθ 

D) sin⁡θ\sin\thetasinθ 

I don't know

Practice: Pythagorean Identities
cos⁡2θ+cos⁡2θtan⁡2θ\cos^2\theta+\cos^2\theta\tan^2\thetacos2θ+cos2θtan2θ 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}1−cosθsinθ​+1+cosθsinθ​?

Extra Practice

Pythagorean Identities

Practice: Pythagorean Identities
Which of the following is equivalent to sin⁡θsec⁡θ−1+sin⁡θ1+sec⁡θ\dfrac{\sin\theta}{\sec\theta-1}+\dfrac{\sin\theta}{1+\sec\theta}secθ−1sinθ​+1+secθsinθ​?

Pythagorean Identities

Practice: Pythagorean Identities
Which of the following is equivalent to cos⁡θ1−sin⁡θ+cos⁡θ1+sin⁡θ\dfrac{\cos\theta}{1-\sin\theta}+\dfrac{\cos\theta}{1+\sin\theta}1−sinθcosθ​+1+sinθcosθ​?

Pythagorean Identities

Practice: Pythagorean Identities
tan⁡2θ−tan⁡2θsec⁡2θ\tan^2\theta-\tan^2\theta\sec^2\thetatan2θ−tan2θsec2θ is equivalent to which of the following?

Pythagorean Identities

Practice: Pythagorean Identities
sin⁡2θ−sin⁡2θcsc⁡2θ\sin^2\theta-\sin^2\theta\csc^2\thetasin2θ−sin2θcsc2θ is equivalent to which of the following?

Pythagorean Identities

Practice: Pythagorean Identities
Simplify (sec⁡2θ−tan⁡2θ)(cot⁡2θ−csc⁡2θ)(\sec^2\theta-\tan^2\theta)(\cot^2\theta-\csc^2\theta)(sec2θ−tan2θ)(cot2θ−csc2θ) into one trigonometric function.

Pythagorean Identities

Practice: Pythagorean Identities
Simplify (1−sin⁡2θ)(sec⁡θ)(1-\sin^2\theta)(\sec\theta)(1−sin2θ)(secθ) into one trigonometric function.