The Pythagorean identity is \(\sin^2\theta+\cos^2\theta=1\). We will use this identity along with some other Trig identities to simplify trig expressions.
In order to understand Trig identities, we need to understand why the Pythagorean identity work so we are going to prove the Pythagorean identity.
\(\begin{align}&\sin^2\theta+\cos^2\theta=1\ & \ \ \ \ &\text{1) Given.}\\
&\sin\theta=\large\frac{y}{r},\ \normalsize \cos\theta=\frac{x}{r}\ & \ \ \ \ &\text{2) Definition of sin and cos ratios.}\\
&\left(\frac{y}{r}\right)^2+\left(\frac{x}{r}\right)^2=1\ & \ \ \ \ &\text{3) Substitution.}\\
&\frac{x^2+y^2}{r^2}=\normalsize1\ & \ \ \ \ &\text{4) Simplify.}\\
&x^2+y^2=r^2\ & \ \ \ \ &\text{5) Multiplication Property of Equality}\end{align}\)
Therefore, the Pythagorean identity is derived from the equation of a circle with a center at \(\left(0,0\right)\).
Using this identity, we can derive a few more identities involving \(\sec \theta\) and \(\csc \theta\).
Prove: \(\tan^2\theta+1=\sec^2\theta\)
\(\begin{align}&\sin^2\theta+\cos^2\theta=1\ & \ \ \ \ &\text{1) Pythagorean Identity}\\
&\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}\ & \ \ \ \ &\text{2) Multiply through by}\ \frac{1}{\cos^2\theta}\ \text{which is equivalent to} \sec^2\theta.\\
&\tan^2\theta+1=\sec^2\theta\ & \ \ \ \ &\text{3) Simplify.}\end{align}\)
Prove: \(\cot^2\theta+1=\csc^2\theta\)
\(\begin{align}&\sin^2\theta+\cos^2\theta=1\ & \ \ \ \ &\text{1) Pythagorean Identity}\\
&\frac{\cos^2\theta}{\sin^2\theta}+\frac{\sin^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}\ & \ \ \ \ &\text{2) Multiply through by}\ \frac{1}{\sin^2\theta}\ \text{which is equivalent to} \csc^2\theta.\\
&\tan^2\theta+1=\sec^2\theta\ & \ \ \ \ &\text{3) Simplify.}\end{align}\)
In order to understand Trig identities, we need to understand why the Pythagorean identity work so we are going to prove the Pythagorean identity.
\(\begin{align}&\sin^2\theta+\cos^2\theta=1\ & \ \ \ \ &\text{1) Given.}\\
&\sin\theta=\large\frac{y}{r},\ \normalsize \cos\theta=\frac{x}{r}\ & \ \ \ \ &\text{2) Definition of sin and cos ratios.}\\
&\left(\frac{y}{r}\right)^2+\left(\frac{x}{r}\right)^2=1\ & \ \ \ \ &\text{3) Substitution.}\\
&\frac{x^2+y^2}{r^2}=\normalsize1\ & \ \ \ \ &\text{4) Simplify.}\\
&x^2+y^2=r^2\ & \ \ \ \ &\text{5) Multiplication Property of Equality}\end{align}\)
Therefore, the Pythagorean identity is derived from the equation of a circle with a center at \(\left(0,0\right)\).
Using this identity, we can derive a few more identities involving \(\sec \theta\) and \(\csc \theta\).
Prove: \(\tan^2\theta+1=\sec^2\theta\)
\(\begin{align}&\sin^2\theta+\cos^2\theta=1\ & \ \ \ \ &\text{1) Pythagorean Identity}\\
&\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}\ & \ \ \ \ &\text{2) Multiply through by}\ \frac{1}{\cos^2\theta}\ \text{which is equivalent to} \sec^2\theta.\\
&\tan^2\theta+1=\sec^2\theta\ & \ \ \ \ &\text{3) Simplify.}\end{align}\)
Prove: \(\cot^2\theta+1=\csc^2\theta\)
\(\begin{align}&\sin^2\theta+\cos^2\theta=1\ & \ \ \ \ &\text{1) Pythagorean Identity}\\
&\frac{\cos^2\theta}{\sin^2\theta}+\frac{\sin^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}\ & \ \ \ \ &\text{2) Multiply through by}\ \frac{1}{\sin^2\theta}\ \text{which is equivalent to} \csc^2\theta.\\
&\tan^2\theta+1=\sec^2\theta\ & \ \ \ \ &\text{3) Simplify.}\end{align}\)
The following are trig identities that we will use in this course.
\(\tan\theta=\Large\frac{\sin\theta}{\cos\theta}\) \(\cot\theta=\Large\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}\) \(\csc\theta=\Large\frac{1}{\sin\theta}\) \(\sec\theta=\Large\frac{1}{\cos\theta}\)
\(\sin^2\theta+\cos^2\theta=1\)
\(\tan^2\theta+1=\sec^2\theta\)
\(\cot^2\theta+1=\csc^2\theta\)
\(\tan\theta=\Large\frac{\sin\theta}{\cos\theta}\) \(\cot\theta=\Large\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}\) \(\csc\theta=\Large\frac{1}{\sin\theta}\) \(\sec\theta=\Large\frac{1}{\cos\theta}\)
\(\sin^2\theta+\cos^2\theta=1\)
\(\tan^2\theta+1=\sec^2\theta\)
\(\cot^2\theta+1=\csc^2\theta\)
Example 1: Simplify the expression: \(\Large\frac{1}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}\).
Using the identities, rewrite trig functions in the expression then simplify.
\(\sec^2\theta-\tan^2\theta\)
\(1\)
Even Odd Properties
The definition of an even function is \(f\left(-x\right)=f\left(x\right)\). The definition of an odd function is \(f\left(-x\right)=-f\left(x\right)\). Based on these properties, the function \(f\left(x\right)=\sin\left(x\right)\) is an odd function and the function \(f\left(x\right)=\cos\left(x\right)\) is an even function. Therefore \(\sin\left(-\theta\right)=-\sin\theta\) and \(\cos\left(-\theta\right)=\cos\theta\).
We can look at the graph of the unit circle to understand these properties.
\(\sin\left(\theta\right)=\Large\frac{y}{r}\) and \(\sin\left(-\theta\right)=\Large\frac{-y}{r}=-\frac{y}{r}\)
Therefore \(\sin\left(-\theta\right)=-\sin\theta\)
\(\cos\left(\theta\right)=\Large\frac{x}{r}\) and \(\cos\left(-\theta\right)=\Large\frac{x}{r}\)
Therefore \(\cos\left(-\theta\right)=\cos\theta\)
Example 2: Determine the following:
a) If \(\sin\left(25^{\circ}\right)=.423\) then \(\sin\left(-25^{\circ}\right)\) = ?
Since the sine function is odd, then \(\sin\left(-25^{\circ}\right)=-.423\).
b) If \(\cos\left(36^{\circ}\right)=.809\) then \(\cos\left(-36^{\circ}\right)=\) ?
Since the cosine function is even, then \(\cos\left(-36^{\circ}\right)=.809\).
Using the identities, rewrite trig functions in the expression then simplify.
\(\sec^2\theta-\tan^2\theta\)
\(1\)
Even Odd Properties
The definition of an even function is \(f\left(-x\right)=f\left(x\right)\). The definition of an odd function is \(f\left(-x\right)=-f\left(x\right)\). Based on these properties, the function \(f\left(x\right)=\sin\left(x\right)\) is an odd function and the function \(f\left(x\right)=\cos\left(x\right)\) is an even function. Therefore \(\sin\left(-\theta\right)=-\sin\theta\) and \(\cos\left(-\theta\right)=\cos\theta\).
We can look at the graph of the unit circle to understand these properties.
\(\sin\left(\theta\right)=\Large\frac{y}{r}\) and \(\sin\left(-\theta\right)=\Large\frac{-y}{r}=-\frac{y}{r}\)
Therefore \(\sin\left(-\theta\right)=-\sin\theta\)
\(\cos\left(\theta\right)=\Large\frac{x}{r}\) and \(\cos\left(-\theta\right)=\Large\frac{x}{r}\)
Therefore \(\cos\left(-\theta\right)=\cos\theta\)
Example 2: Determine the following:
a) If \(\sin\left(25^{\circ}\right)=.423\) then \(\sin\left(-25^{\circ}\right)\) = ?
Since the sine function is odd, then \(\sin\left(-25^{\circ}\right)=-.423\).
b) If \(\cos\left(36^{\circ}\right)=.809\) then \(\cos\left(-36^{\circ}\right)=\) ?
Since the cosine function is even, then \(\cos\left(-36^{\circ}\right)=.809\).