Trigonometric identities are equations that we can use to help us to solve trigonometric equations and simplify trigonometric expressions. In this lesson, we are going to cover reciprocal identities, Pythagorean identities, quotient identities, and the range of trig functions.
We already went over reciprocal identities in the last lesson, but let’s go over them again.
Reciprocal Identities
sin\theta = \frac{1}{csc\theta}, \ \ cos\theta = \frac{1}{sec\theta}, \ \ tan\theta = \frac{1}{cot\theta}csc\theta = \frac{1}{sin\theta}, \ \ sec\theta = \frac{1}{cos\theta} \ \ cot\theta = \frac{1}{tan\theta}Pythagorean Identities
1. \ sin^2\theta \ + \ cos^2\theta = 1 \\ 2. \ tan^2\theta + 1 = sec^2\theta \\ 3. \ cot^2\theta + 1 = csc^2\theta
So let’s take the first identity for example and divide it by cosine squared theta. When we do that, we get the second identity. Sine divided by cosine is tangent, and 1 over cosine is secant.
\frac{sin^2\theta}{cos^2\theta} + \frac{cos^2}{cos^2\theta} = \frac{1}{cos^2\theta}tan^2\theta + 1 = sec^2\theta
Now, let’s take the first identity and divide it by sine. Sine over Sine is 1, and cosine over sine is cotangent. Finally, 1 over sine is cosecant.
\frac{sin^2\theta}{sin^2\theta} + \frac{cos^2\theta}{sin^2} = \frac{1}{sin^2\theta} \\ 1 + cot^2\theta = csc^2\theta
Quotient Identities
tan\theta = \frac{sin\theta}{cos\theta} cot\theta = \frac{cos\theta}{sin\theta}Signs of the Functions in All 4 Quadrants
Range of the Trig Functions
sin\theta, cos\theta \ [-1,1]
tan\theta, cot\theta \ \ [-\infty,\infty]
csc\theta, sec\theta, \ (-\infty, -1] \cup [1, \infty)
Example Problems
1.
tan\theta = 6
\text{find} \ cot\theta = \frac{1}{tan\theta} = \frac{1}{6}1/6 is the answer to the problem.
2.
\text{identify the quadrants satisfying the conditions:} \ sin\alpha < 0, \ cos\alpha < 0We know that sin is negative in quadrants 3 and 4, and cos is negative in quadrants 2 and 3. The only quadrant that satisfies the conditions where both sine and cosine alpha are negative is quadrant 3.
Quadrant 3 is the answer.
4.
\text{Find} \ csc\theta \ \text{given} \ cot\theta = -\frac{1}{6}\ \text{and} \ \theta \ \text{is in quadrant IV}.If we draw this out on a graph, we can clearly see that y has to be -6 and x must be 1 in order for angle theta to fall in quadrant 4 instead of quadrant 2. Therefore, we can get tan theta with this information.
tan\theta = \frac{-6}{1} = \frac{\text{y}}{\text{x}}Next, let’s try to find the r by using our formula taking the square root of x squared plus y squared.
r = \sqrt{1^2 + (-6)^2}r = \sqrt{1 + 36}r = \sqrt{37}Once we get r, we know that cosecant is simply r over y, so we can plug in our values to get our answer.
csc\theta = \frac{\text{r}}{\text{y}} = \frac{\sqrt{37}}{-6}5.
\text{Find the remaining trig functions of} \ \theta \\ tan\theta = -\frac{8}{15} \ (\frac{y}{x}), \ \theta \ \text{is in Quadrant II}Now, if we are familiar with our Pythagorean triples, we can look at this and know that our r in this situation must be 17, but in case you want to double check or you don’t know the triple, let’s continue forward.
I always draw a picture on a graph so I can visually see my answers and that they make sense. We don’t have to do this though. We know that for theta to fall in quadrant 2, x (15) must be negative and y (8) positive.
Now let’s solve for r.
\text{r} = \sqrt{8^2 + (-15)^2} \\ \text{r} = \sqrt{64 + 225} \\ \text{r} = \sqrt{289} \\ \text{r} = 17Since we have our x, y, and r values, we can now simply plug them into all the functions.
cot\theta = \frac{\text{x}}{\text{y}} = \frac{-15}{8} \\ \\ \\ sin\theta = \frac{\text{y}}{\text{r}} = \frac{8}{17} \\ \\ \\ \\ csc\theta = \frac{\text{r}}{\text{y}} = \frac{17}{8} \\ cos\theta = \frac{\text{x}}{\text{r}} = \frac{-15}{17} \\ sec\theta = \frac{\text{r}}{\text{x}} = -\frac{17}{15}6.
\text{Find the remaining trig functions of} \ \theta: \\ sec\theta = -11 \ \ sin\theta > 0 \\ \text{sec is negative} \ \ \ \text{sin is positive} \\ \text{Q2, Q3} \ \ \ \text{Q1, Q2}From the beginning, we are given enough information to figure this out. We are already given secant, which is just -11, and the quadrant we can find theta in. Based off the given information, theta must be in quadrant II.
The easiest function to find second would be the inverse of secant – cosine.
cos\theta = \frac{\text{x}}{\text{r}} = \frac{-1}{11}Let’s recall the Pythagorean Identities. We can use those to help us determine what the other functions are.
We can start by plugging the cosine that we found into the first Pythagorean identity.
sin^2\theta \ + \ cos^2\theta = 1 \\ sin^2\theta + (\frac{-1}{11})^2 = 1 \\ sin^2\theta + \frac{1}{121} = 1 \\ sin^2\theta = \frac{121}{121} - \frac{1}{121} \\ sin^2\theta = \frac{120}{121} \\ sin\theta = \pm \sqrt{\frac{120}{121}} \\
sin\theta = \frac{2\sqrt{30}}{11}Now that we have secant, cosine, and sine, we just need to find the cosecant, tangent, and cotangent.
We know that the inverse of the sine is going to give us our cosecant, so let’s do that next.
csc\theta = \frac{11}{2\sqrt{30}}This answer is technically in the wrong format. We can not have radicals in the denominator, so we must rationalize the denominator.
csc\theta = \frac{11}{2\sqrt{30}} \times \frac{\sqrt{30}}{\sqrt{30}}We cross multiply to get 11 times the square root of 30, and 2 times the square root of 30 times the square root of 30. The square root of something times itself is just that number.
csc\theta = \frac{11\sqrt{30}}{60}And there we have our cosecant. Next, let’s plug the secant into the second Pythagorean identity to find our tangent and finally our cotangent.
tan^2\theta \ + \ 1 = sec^2\theta \\ tan^2\theta \ + \ 1 = (-11)^2 \\ tan^2\theta = 121 \ - \ 1 \\ tan^2\theta = 120 \\ tan\theta = -\sqrt{120} \\ tan\theta = -2\sqrt{30}We know that tangent is negative in quadrant 2, so we know that our answer will come out negative. Here is how we simplified the radical.
\sqrt{120} \\ =\sqrt{4\times30} \\ = 2\sqrt{30}Finally, let’s find our cotangent.
cot\theta = \frac{1}{-2\sqrt{30}} \ \times \ \frac{\sqrt{30}}{\sqrt{30}} \ = \ \ - \frac{\sqrt{30}}{60}7.
sec(3\theta- 10\degree) = \frac{1}{cos(4\theta-18\degree)}sec(3\theta - 10\degree) = sec(4\theta-18\degree) \\ 3\theta - 10\degree = 4\theta - 18\degree \\ -\theta - 10\degree = -18\degree \\ -\theta = -8\degree \\ \theta = 8\degree