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Fundamental Trigonometric Identities

Prove equations are true using Reciprocal, Tangent, and other identities.

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Practice Fundamental Trigonometric Identities

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Introduction to Trig Identities

You are given a list of Trig Identities. One of those identities is cos(θ)=cosθ\begin{align*}\cos (- \theta)=\cos \theta\end{align*}. Prove this identity without graphing.

Trigonometric Identities

Trigonometric identities are true for any value of x\begin{align*}x\end{align*} (as long as the value is in the domain). You have learned about secant, cosecant, and cotangent, which are all reciprocal functions of sine, cosine and tangent. These functions can be rewritten as the Reciprocal Identities because they are always true.

Reciprocal Identities: cscθ=1sinθsecθ=1cosθcotθ=1tanθ\begin{align*}\csc \theta=\frac{1}{\sin \theta} \quad \sec \theta=\frac{1}{\cos \theta} \quad \cot \theta=\frac{1}{\tan \theta}\end{align*}

Other identities involve the tangent, variations on the Pythagorean Theorem, phase shifts, and negative angles. We will discover them in this concept.

We know that tanθ=oppositeadjacent\begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. Let's show that tanθ=sinθcosθ\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*}. This is the Tangent Identity.

Whenever we are trying to verify, or prove, an identity, we start with the statement we are trying to prove and work towards the desired answer. In this case, we will start with tanθ=sinθcosθ\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*} and show that it is equivalent to tanθ=oppositeadjacent\begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. First, rewrite sine and cosine in terms of the ratios of the sides.

tanθ=sinθcosθ=oppositehypotenuseadjacenthypotenuse\begin{align*}\tan \theta&=\frac{\sin \theta}{\cos \theta} \\ &=\frac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}}\end{align*}

Then, rewrite the complex fraction as a division problem and simplify.

=oppositehypotenuse÷adjacenthypotenuse=oppositehypotenusehypotenuseadjacent=oppositeadjacent\begin{align*}&=\frac{opposite}{hypotenuse} \div \frac{adjacent}{hypotenuse} \\ &=\frac{opposite}{\cancel{hypotenuse}} \cdot \frac{\cancel{hypotenuse}}{adjacent} \\ &=\frac{opposite}{adjacent}\end{align*}

We now have what we wanted to prove and we are done. Once you verify an identity, you may use it to verify other identities.

Now, let's show that sin2θ+cos2θ=1\begin{align*}\sin^2 \theta+ \cos^2 \theta=1\end{align*} is a true identity.

Change the sine and cosine in the equation into the ratios. In this problem, we will use y\begin{align*}y\end{align*} as the opposite side, x\begin{align*}x\end{align*} is the adjacent side, and r\begin{align*}r\end{align*} is the hypotenuse (or radius), as in the unit circle.

sin2θ+cos2θ(yr)2+(xr)2y2r2+x2r2y2+x2r2=1=1=1=1\begin{align*}\sin^2 \theta+ \cos^2 \theta&=1 \\ \left(\frac{y}{r}\right)^2+\left(\frac{x}{r}\right)^2&=1 \\ \frac{y^2}{r^2}+\frac{x^2}{r^2}&=1 \\ \frac{y^2+x^2}{r^2}&=1\end{align*}

Now, x2+y2=r2\begin{align*}x^2+y^2=r^2\end{align*} from the Pythagorean Theorem. Substitute this in for the numerator of the fraction.

r2r2=1\begin{align*}\frac{r^2}{r^2}=1\end{align*}

This is one of the Pythagorean Identities and very useful.

Finally, let's verify that sin(π2θ)=cosθ\begin{align*}\sin \left(\frac{\pi}{2}- \theta \right)=\cos \theta\end{align*} by using the graphs of the functions.

The function y=sin(π2x)\begin{align*}y=\sin \left(\frac{\pi}{2} - x \right)\end{align*} is a phase shift of π2\begin{align*}\frac{\pi}{2}\end{align*} of the sine curve.

The red function above is y=sinx\begin{align*}y=\sin x\end{align*} and the blue is y=cosx\begin{align*}y=\cos x\end{align*}. If we were to shift the sine curve π2\begin{align*}\frac{\pi}{2}\end{align*}, it would overlap perfectly with the cosine curve, thus proving this Cofunction Identity.

Examples

Example 1

Earlier, you were asked to prove the identity of cos(θ)=cosθ\begin{align*}\cos (- \theta)=\cos \theta\end{align*} without graphing.

First, recall that cosθ=x\begin{align*}\cos\theta=x\end{align*}, where (x,y)\begin{align*}(x, y)\end{align*} is the endpoint of the terminal side of θ\begin{align*}\theta\end{align*} on the unit circle.

Now, if we have cos(θ)\begin{align*}\cos (- \theta)\end{align*}, what is its endpoint? Well, the negative sign tells us that the angle is rotated in a clockwise direction, rather than the usual counter-clockwise. If we make this rotation, we see that cos(θ)=x\begin{align*}\cos (- \theta)=x\end{align*} as well, as illustrated in the following diagram.

We know that x=x\begin{align*}x = x\end{align*}, so we can set the two expressions equal to one another.

cosθ=cos(θ)\begin{align*}\cos\theta = \cos (- \theta) \end{align*}

We can now flip this identity around to get:

cos(θ)=cosθ\begin{align*}\cos (- \theta)=\cos \theta\end{align*}

Example 2

Prove the Pythagorean Identity: 1+tan2θ=sec2θ\begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*}.

First, let’s use the Tangent Identity and the Reciprocal Identity to change tangent and secant in terms of sine and cosine.

1+tan2θ1+sin2θcos2θ=sec2θ=1cos2θ\begin{align*}1+ \tan^2 \theta&=\sec^2 \theta \\ 1+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta}\end{align*}

Now, change the 1 into a fraction with a base of cos2θ\begin{align*}\cos^2 \theta\end{align*} and simplify.

1+sin2θcos2θcos2θcos2θ+sin2θcos2θcos2θ+sin2θcos2θ1cos2θ=1cos2θ=1cos2θ=1cos2θ=1cos2θ\begin{align*}1+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{\cos^2 \theta}{\cos^2 \theta}+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{\cos^2 \theta+ \sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{1}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta}\end{align*}

In the second to last step, we arrived at the original Pythagorean Identity sin2θ+cos2θ\begin{align*}\sin^2 \theta+\cos^2 \theta\end{align*} in the numerator of the left-hand side. Therefore, we can substitute in 1 for this and the two sides of the equation are the same.

Example 3

Without graphing, show that sin(θ)=sinθ\begin{align*}\sin (- \theta)=- \sin \theta\end{align*}.

First, recall that sinθ=y\begin{align*}\sin \theta=y\end{align*}, where (x,y)\begin{align*}(x, y)\end{align*} is the endpoint of the terminal side of θ\begin{align*}\theta\end{align*} on the unit circle.

Now, if we have sin(θ)\begin{align*}\sin (- \theta)\end{align*}, what is it’s endpoint? Well, the negative sign tells us that the angle is rotated in a clockwise direction, rather than the usual counter-clockwise. By looking at the picture, we see that sin(θ)=y\begin{align*}\sin (- \theta)=-y\end{align*}. Therefore, if sinθ=y\begin{align*}\sin \theta=y\end{align*}, then sinθ=y\begin{align*}- \sin \theta=-y\end{align*} and combining the equations, we have sin(θ)=sinθ\begin{align*}\sin (- \theta)=- \sin \theta\end{align*}.

Review

1. Show that cotθ=cosθsinθ\begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*}.
2. Show that tanθ=secθcscθ\begin{align*}\tan \theta=\frac{\sec \theta}{\csc \theta}\end{align*}.
3. Show that 1+cot2θ=csc2θ\begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*}.
4. Explain why cos(π2θ)=sinθ\begin{align*}\cos \left(\frac{\pi}{2} - \theta \right)=\sin \theta\end{align*} by using the graphs of the two functions.
5. Show that sec(θ)=secθ\begin{align*}\sec (- \theta)=\sec \theta\end{align*}.
6. Explain why \begin{align*}\tan (- \theta)=- \tan \theta\end{align*} is true, using the Tangent Identity and the other Negative Angle Identities.

Verify the following identities.

1. \begin{align*}\cot \theta \sec \theta=\csc \theta\end{align*}
2. \begin{align*}\frac{\cos \theta}{\cot \theta}=\frac{\tan \theta}{\sec \theta}\end{align*}
3. \begin{align*}\sin \theta \csc \theta=1\end{align*}
4. \begin{align*}\cot(- \theta)=- \cot \theta\end{align*}
5. \begin{align*}\tan x \csc x \cos x=1\end{align*}
6. \begin{align*}\frac{\sin^2 \left(-x\right)}{\tan^2 x}= \cos ^2 x\end{align*}

Show that \begin{align*}\sin^2 \theta+ \cos^2 \theta=1\end{align*} is true for the following values of \begin{align*}\theta\end{align*}.

1. \begin{align*}\frac{\pi}{4}\end{align*}
2. \begin{align*}\frac{2 \pi}{3}\end{align*}
3. \begin{align*}- \frac{7 \pi}{6}\end{align*}
4. Recall that a function is odd if \begin{align*}f(-x)=-f(x)\end{align*} and even if \begin{align*}f(-x)=f(x)\end{align*}. Which of the six trigonometric functions are odd? Which are even?

To see the Review answers, open this PDF file and look for section 14.6.

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Color Highlighted Text Notes

Vocabulary Language: English

TermDefinition
cofunction Cofunctions are functions that are identical except for a reflection and horizontal shift. Examples include: sine and cosine, tangent and cotangent, secant and cosecant.
even An even function is a function with a graph that is symmetric with respect to the $y$-axis and has the property that $f(-x) = f(x)$.
identity An identity is a mathematical sentence involving the symbol “=” that is always true for variables within the domains of the expressions on either side.
Odd Function An odd function is a function with the property that $f(-x) = -f(x)$. Odd functions have rotational symmetry about the origin.
proof A proof is a series of true statements leading to the acceptance of truth of a more complex statement.