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In section Section TheOtherCircularFunctions, we first encountered the concept of an identity when discussing Theorem recipquotid. Recall that an identity is an
equation which is true regardless of the choice of variable. Identities are important in mathematics because they
facilitate changing forms. (We’ve seen the utility of changing form throughout the text, most recently when
we completed the square in Chapter TheConicSections to put general quadratic equations into standard form in order to graph
them.)
We take a moment to generalize Theorem recipquotid below.
Reciprocal and Quotient Identities: The following relationships hold for all angles \(\theta \) provided each side of each equation is
defined.
It is important to remember that the equivalences stated in Theorem recipquotidfull are valid only when all quantities described therein are
defined. As an example, \(\tan (0) = 0\), but \(\tan (0) \neq \frac {1}{\cot (0)}\) since \(\cot (0)\) is undefined.
When it comes down to it, the Reciprocal and Quotient Identities amount to giving different ratios on the Unit Circle different
names. The main focus of this section is on a more algebraic relationship between certain pairs of the circular functions: the
Pythagorean Identities.
Recall in Definition sinecosineunitcircledefn, the cosine and sine of an angle is defined as the \(x\) and \(y\)-coordinate, respectively, of a point on the Unit
Circle. Since the coordinates of all points \((x,y)\) on the Unit Circle satisfy the equation \(x^2+y^2 = 1\), we get for all angles \(\theta \), \(\left (\cos (\theta )\right )^2 + \left (\sin (\theta )\right )^2 = 1\). An unfortunate (This is unfortunate from a ‘function notation’ perspective. See Section TheInverseTrigonometricFunctions.) convention, which the authors are compelled to
perpetuate, is to write \(\left (\cos (\theta )\right )^2\) as \(\cos ^{2}(\theta )\) and \(\left (\sin (\theta )\right )^2\) as \(\sin ^{2}(\theta )\). Rewriting the identity using this convention results in the following theorem, which is
without a doubt one of the most important results in Trigonometry.
The Pythagorean Identity: For any angle \(\theta \), \(\cos ^{2}(\theta ) + \sin ^{2}(\theta ) = 1\).
The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation
for a circle are ultimately derived. (See Sections AppCartesianPlane and Circles for details.) The word ‘Identity’ reminds us that, regardless of the
angle \(\theta \), the equation in Theorem cosinesinepythid is always true.
If one of \(\cos (\theta )\) or \(\sin (\theta )\) is known, Theorem cosinesinepythid can be used to determine the other, up to a (\(\pm \)) sign. If, in addition, we know where the
terminal side of \(\theta \) lies when in standard position, then we can remove the ambiguity of the (\(\pm \)) and completely determine the
missing value. (See the illustration following Example circularfunctionsex to refresh yourself which circular functions are positive in which
quadrants.) We illustrate this approach in the following example.
Use Theorem cosinesinepythid and the given information to find the indicated value.
If \(\theta \) is a Quadrant II angle with \(\sin (\theta ) = \frac {3}{5}\), find \(\cos (\theta )\).
If \(\pi < t < \frac {3\pi }{2}\) with \(\cos (t) = -\frac {\sqrt {5}}{5}\), find \(\sin (t)\).
If \(\sin (\theta ) = 1\), find \(\cos (\theta )\).
Solution.
When we substitute \(\sin (\theta ) = \frac {3}{5}\) into The Pythagorean Identity, \(\cos ^{2}(\theta ) + \sin ^{2}(\theta ) = 1\), we obtain \(\cos ^{2}(\theta ) + \frac {9}{25} = 1\). Solving, we find \(\cos (\theta ) = \pm \frac {4}{5}\). Since \(\theta \) is a Quadrant II angle,
we know \(\cos (\theta )<0\). Hence, we select \(\cos (\theta ) = - \frac {4}{5}\).
Here we’re using the variable \(t\) instead \(\theta \) which usually corresponds to a real number variable instead of an
angle. As usual, we associate real numbers \(t\) with angles \(\theta \) measuring \(t\) radians, (See our discussion of
the wrapping function in Section RadianMeasure if you need a review of how we associate real numbers with angles in radian
measure.) so the the Pythagorean Identity works equally well for all real numbers \(t\) as it does for all angles \(\theta \).
Substituting \(\cos (t) = -\frac {\sqrt {5}}{5}\) into \(\cos ^{2}(t) + \sin ^{2}(t) = 1\) gives \(\sin (t) = \pm \frac {2}{\sqrt {5}} = \pm \frac {2 \sqrt {5}}{5}\). Since \(\pi < t < \frac {3\pi }{2}\), we know \(t\) corresponds to a Quadrant III angle, so \(\sin (t) <0\). Hence, \(\sin (t) = -\frac {2 \sqrt {5}}{5}\).
When we substitute \(\sin (\theta ) = 1\) into \(\cos ^{2}(\theta ) + \sin ^{2}(\theta ) = 1\), we find \(\cos (\theta ) = 0\). □
As with many tools in mathematics, identities give us a different way to approach and solve problems. (For example,
factoring, completing the square, and the quadratic formula are three different (yet equivalent) ways to solve a quadratic
equation. See Section AppQuadEqus for a refresher.) As always, the key is to determine which approach makes the most sense (is more
efficient, for instance) in the given scenario.
Our next task is to use use the Reciprocal and Quotient Identities found in Theorem recipquotidfull coupled with the Pythagorean Identity
found in Theorem cosinesinepythid to derive new Pythagorean-like identities for the remaining four circular functions.
Assuming \(\cos (\theta ) \neq 0\), we may start with \(\cos ^{2}(\theta ) + \sin ^{2}(\theta ) = 1\) and divide both sides by \(\cos ^{2}(\theta )\) to obtain \(1 + \frac {\sin ^{2}(\theta )}{\cos ^{2}(\theta )} = \frac {1}{\cos ^{2}(\theta )}\). Using properties of exponents along with the Reciprocal
and Quotient Identities, this reduces to \(1 + \tan ^{2}(\theta ) = \sec ^{2}(\theta )\).
If \(\sin (\theta ) \neq 0\), we can divide both sides of the identity \(\cos ^{2}(\theta ) + \sin ^{2}(\theta ) = 1\) by \(\sin ^{2}(\theta )\), apply Theorem recipquotidfull once again, and obtain \(\cot ^{2}(\theta ) + 1 = \csc ^{2}(\theta )\).
These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are
summarized in the following theorem.
If \(\theta \) is a Quadrant IV angle with \(\sec (\theta ) = 3\), find \(\tan (\theta )\).
Find \(\csc (t)\) if \(\pi < t < \frac {3\pi }{2}\) and \(\cot (t) = 2\).
If \(\theta \) is a Quadrant II angle with \(\cos (\theta ) = -\frac {3}{5}\), find the exact values of the remaining circular functions.
Solution.
Per Theorem pythids, \(\tan ^{2}(\theta ) = \sec ^{2}(\theta ) -1\). Since \(\sec (\theta ) = 3\), we have \(\tan ^{2}(\theta ) = (3)^2 - 1 = 8\), or \(\tan (\theta ) = \pm \sqrt {8} = \pm 2 \sqrt {2}\). Since \(\theta \) is a Quadrant IV angle, we know \(\tan (\theta ) < 0\) so \(\tan (\theta ) = -2\sqrt {2}\).
Again, using Theorem pythids, we have \(\csc ^{2}(t) = 1 + \cot ^{2}(t)\), so we have \(\csc ^{2}(t) = 1 + (2)^2 = 5\). This gives \(\csc (t) = \pm \sqrt {5}\). Since \(\pi < t < \frac {3\pi }{2}\), \(t\) corresponds to a Quadrant III angle, so \(\csc (t) = -\sqrt {5}\).
With five function values to find, we have our work cut out for us. From Theorem recipquotidfull, we know \(\sec (\theta ) = \frac {1}{\cos (\theta )}\), so we (quickly)
get \(\sec (\theta ) = \frac {1}{-3/5} = - \frac {5}{3}\).
Next, we go after \(\sin (\theta )\) since between \(\sin (\theta )\) and \(\cos (\theta )\), we can get all of the remaining values courtesy of Theorem recipquotidfull.
From Theorem pythids, we have \(\sin ^{2}(\theta ) = 1 - \cos ^{2}(\theta )\), so \(\sin ^{2}(\theta ) = 1 - \left (\frac {3}{5}\right )^2 = 1 - \frac {9}{25} = \frac {16}{25}\). Hence, \(\sin (\theta ) = \pm \frac {4}{5}\) but since \(\theta \) is a Quadrant II angle, we select \(\sin (\theta ) = \frac {4}{5}\).
Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to
find the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to
simplify otherwise complicated expressions. In the next example, we make good use of the Theorems recipquotidfull and
pythids.
Verify the following identities. Assume that all quantities are defined.
Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to
transform it into the other side of the equation.
Starting with the right hand side of \(\tan (\theta ) = \sin (\theta ) \sec (\theta )\), we use \(\sec (\theta ) = \frac {1}{\cos (\theta )}\) and find:
where the last equality is courtesy of Theorem recipquotidfull.
Expanding the left hand side, we get: \((\tan (t) - \sec (t)) (\tan (t) + \sec (t)) = \tan ^{2}(t) - \sec ^{2}(t)\). From Theorem pythids, we know \(\sec ^{2}(t) - \tan ^{2}(t) = 1\), which isn’t quite what we have. We are off
by a negative sign (\(-\)), so we factor it out:
Starting with the right hand side, (We hope by this point a shift of variable to ‘\(x\)’ instead of ‘\(\theta \)’ or ‘\(t\)’ is a
non-issue.) we notice we have a quantity we can immediately simplify per Theorem pythids: \(1 - \sin ^{2}(x) = \cos ^{2}(x)\) . This increases the
number of factors of cosine, (which is part of our goal in looking at the left hand side), so we proceed:
While both sides of our next identity contain fractions, the left side affords us more opportunities to use our
identities. (Or, to put to another way, earn more partial credit if this were an exam question!) Substituting
\(\sec (t) = \frac {1}{\cos (t)}\) and \(\tan (t) = \frac {\sin (t)}{\cos (t)}\), we get:
At this point, we have at least reduced the number of fractions from two to one, it may not be clear how to
proceed. When this happens, it isn’t a bad idea to start working with the other side of the identity to get some
clues how to proceed.
Using a reciprocal and quotient identity, we find \(6\sec (x) \tan (x) = 6 \left (\frac {1}{\cos (x)}\right ) \left (\frac {\sin (x)}{\cos (x)}\right ) = \frac {6 \sin (x)}{\cos ^{2}(x)}\).
Theorem pythids tells us \(1 - \sin ^{2}(x) = \cos ^{2}(x)\), which means to our surprise and delight, we are much closer to our goal that we may have
originally thought:
It is debatable which side of the identity is more complicated. One thing which stands out is that the
denominator on the left hand side is \(1-\cos (\theta )\), while the numerator of the right hand side is \(1+\cos (\theta )\). This suggests the strategy
of starting with the left hand side and multiplying the numerator and denominator by the quantity \(1+\cos (\theta )\). Theorem pythids
comes to our aid once more when we simplify \(1-\cos ^{2}(\theta ) = \sin ^{2}(\theta )\):
In Example idornotex1 number pythconjex above, we see that multiplying \(1-\cos (\theta )\) by \(1+\cos (\theta )\) produces a difference of squares that can be simplified to one term
using Theorem pythids.
This is exactly the same kind of phenomenon that occurs when we multiply expressions such as \(1 - \sqrt {2}\) by \(1+\sqrt {2}\) or \(3 - 4i\) by \(3+4i\). In algebra, these
sorts of expressions were called ‘conjugates.’ (See Sections AppCmpNums and AppRadEqus.)
For this reason, the quantities \((1-\cos (\theta ))\) and \((1+\cos (\theta ))\) are called ‘Pythagorean Conjugates.’ Below is a list of other common Pythagorean
Conjugates.
Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours
struggling with them just to become proficient in the basics.
Like many things in life, there is no short-cut here – there is no complete algorithm for verifying identities. Nevertheless, a
summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is
provided for you in the Exercises.
Strategies for Verifying Identities
Try working on the more complicated side of the identity.
Use the Reciprocal and Quotient Identities in Theorem recipquotidfull to write functions on one side of the identity in terms
of the functions on the other side of the identity.
Simplify the resulting complex fractions.
Add rational expressions with unlike denominators by obtaining common denominators.
Use the Pythagorean Identities in Theorem pythids to ‘exchange’ sines and cosines, secants and tangents, cosecants
and cotangents, and simplify sums or differences of squares to one term.
Multiply numerator and denominator by Pythagorean Conjugates in order to take advantage of the
Pythagorean Identities in Theorem pythids.
If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and
see if you can find a way to bridge the two parts of your work.
Try something. The more you work with identities, the better you’ll get with identities.
PreCalculus