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1.3: Trigonometric Identities

Difficulty Level: At Grade Created by: CK-12
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Fundamental Identities

Reciprocal, Quotient, Pythagorean

This section reviews the definitions of the trig functions and the Trigonometric Pythagorean Theorem. The Pythagorean identities, you’ll recall, were first covered in lesson 1.6, but here we see a slightly different way of deriving them. It may be useful to reinforce knowledge of the identities by walking through the derivations, but for many students this will simply be review.

You can point out that another way of expressing the reciprocal trig functions is as follows: \begin{align*}\csc \theta = \frac{c}{b}; \sec \theta = \frac{c}{a}; \cot \theta = \frac{a}{b}\end{align*}cscθ=cb;secθ=ca;cotθ=ab. This way of expressing them is only useful for angles in triangles, though; for angles of rotation it may be more useful to think of them as \begin{align*}\frac{1}{\sin \theta}\end{align*}1sinθ and so on.

Confirm Using Analytic Arguments

The diagram here with the vertical line representing a distance of t units may confuse students a little, but it simply demonstrates in a slightly unusual way the fact that any real number can correspond to a distance traveled around the unit circle, and therefore to an angle on that circle. Again, this should be review for most students.

There is a slight error in the last paragraph : where it reads “for points in the third and fourth quadrants we use angles formed by the radius that meets that point and the y axis,” it should say “x axis.” This is an opportunity to remind students about reference angles: the reference angle is always the angle made with the closest portion of the \begin{align*}x-\end{align*}xaxis.

Confirm Using Technological Tools

Calculators generally do not have sec, csc, and cot keys, so instead one must use the cos, sin, and tan keys and then take the reciprocals of those functions using the reciprocal key (marked \begin{align*}x^{-1}\end{align*}x1 or \begin{align*}\frac{1}{x}\end{align*}1x). Stress once again that the keys marked \begin{align*}\sin^{-1}, \cos^{-1}\end{align*}sin1,cos1, and \begin{align*}\tan^{-1}\end{align*}tan1 will not yield the reciprocal functions \begin{align*}\csc, \sec,\end{align*}csc,sec, and \begin{align*}\tan,\end{align*}tan, but rather the \begin{align*}\mathrm{arcsin}, \;\mathrm{arccos}\end{align*}arcsin,arccos, and \begin{align*}\mathrm{arctan}\end{align*}arctan (that is, inverse sine, cosine, and tangent) functions.

Alternative Forms

We see here that knowing just one trig function of an angle does not uniquely determine the angle, as there are always two quadrants it could be in (which two depends on whether the value of the trig function is positive or negative). Generally, knowing a second trig function of the angle, or at least knowing its sign, will narrow down which quadrant the angle could be in—but note that this won’t be the case if the second function is just the reciprocal of the first function. So, for example, knowing the signs of the tangent and cotangent functions won’t tell us what quadrant the angle is in, because the tangent and cotangent always have the same sign whatever the quadrant—and the same is true for the sine and cosecant, or the cosine and secant. But knowing the signs of, say, the tangent and secant functions will tell us which quadrant the angle is in, and the same is true for any two trig functions that are not each other’s reciprocals.

Verifying Identities

Working with Trigonometric Identities

Students may get a little confused by the fact that the equations we’re working with reduce to seemingly obvious identities. The point here, of course, is that we start with an equation that declares two complex expressions to be equivalent, and then prove the equation is true by showing it can be changed into a form that is much more obviously true.

Generally, as the text states, we only change one side of the equation at once. This is because we are not really changing the equation, merely changing how the equation is expressed. To take a simpler example: if we were asked to prove that \begin{align*}17 - (\frac{99}{3}) + 2(40 + 7) = 78\end{align*}17(993)+2(40+7)=78, we would not try to perform the same operations on both sides at once, because we don’t need to; one side is already as simple as it can get. Instead, we would simplify the left-hand side, step by step, until we had shown that it does indeed equal \begin{align*}78\end{align*}78, and so is the same as the right-hand side.

Here’s one way to explain the difference between proving identities and solving equations: When we start out with a complicated equation like \begin{align*}17x + 5(x - 2) + 3 = 15\end{align*}17x+5(x2)+3=15 and solve it for \begin{align*}x\end{align*}x, what we’re really trying to do is answer the question “For what value(s) of \begin{align*}x\end{align*}x is this equation true?” When we reduce the equation to \begin{align*}x = 1\end{align*}x=1, we’ve just shown that the original equation is true if \begin{align*}x = 1\end{align*}x=1, and false otherwise.

But if we tried to solve an equation like \begin{align*}3(x + 5) = 5x - 2(x + 5) + 25\end{align*}3(x+5)=5x2(x+5)+25, we would find that it reduces to \begin{align*}3x + 15 = 3x + 15\end{align*}3x+15=3x+15, which reduces even further to \begin{align*}0 = 0\end{align*}0=0, which is true no matter what \begin{align*}x\end{align*}x is. So instead of proving the equation is true for certain values of \begin{align*}x\end{align*}x, we’ve discovered it’s true for all values of \begin{align*}x\end{align*}x.

And that’s exactly what we’re doing in proving these trig identities: by showing that both sides of the equation reduce to the same expression, we’re proving that the equation is always true for any value of θ. And once we know it’s true, we can use it to make useful substitutions when solving trig problems in the future.

Technology Note

Calculators can be useful in verifying identities, but it is dangerous to rely on them too much. If the graphs of two expressions look identical, it may mean the expressions are indeed equivalent, but it may also mean that the difference between them is just too small for the graph to show, or that they are only equivalent over this small interval. Since a graphing calculator can only show us part of a graph and can only draw it with limited precision, it cannot tell us for sure if two expressions really are mathematically equivalent.

What a graph can do, though, is tell us for sure if two expressions are not equivalent. If the graphs of the expressions look wildly different—in fact, if they look even a little bit different—then we can safely say that the expressions are not equal.

Sum and Difference Identities for Cosine

Difference and Sum Formulas for Cosine

Students may not immediately see why we are trying to find a formula for \begin{align*}\cos(a-b)\end{align*}cos(ab). The idea is that once we find such a formula, we can use it to find the cosine of an unfamiliar angle if we can express that angle as the difference of two angles we are familiar with. The sum formula will be similarly helpful.

Students who haven’t encountered the distance formula for a while might need to be reminded of it. On the coordinate grid, the distance between two points \begin{align*}(a,b)\end{align*}(a,b) and \begin{align*}(c,d)\end{align*}(c,d) is equal to \begin{align*}\sqrt{(a -c)^2 + (b - d)^2}\end{align*}; in other words, the square of the straight-line distance between the points is equal to the square of the horizontal distance plus the square of the vertical distance. This is derived directly from the Pythagorean Theorem; drawing a right triangle on the coordinate grid whose hypotenuse is the line between the two points will show how it is derived.

Note that two different identities are derived in this section. The diagrams and the table show how we derive the formula for \begin{align*}\cos(a-b)\end{align*}; then the following lines show how we can use this formula to derive, in turn, the formula for \begin{align*}\cos(a+b)\end{align*}. These identities may be easier for students to remember if expressed in words: “The cosine of the difference is the product of the cosines plus the product of the sines” and “The cosine of the sum is the product of the cosines minus the product of the sines.”

Use Cosine of Sum or Difference Identities to Verify Other Identities

Call attention to the labels “Identity A” and “Identity B” here. These labels aren’t official names for these identities, but they will be used to refer back to them later in this lesson.

The identities themselves simply say that the sine of an angle is equal to the cosine of the angle’s complement, and vice versa. We’ve already seen that this is true from working with angles in right triangles, as the sine of one acute angle in a triangle is equal to the cosine of the other, and the two angles are each other’s complements.

Use Cosine of Sum or Difference Identities to Find Exact Values

Here we see that the sum and difference formulas we have just learned can tell us the value of the cosine function for angles we haven’t previously worked with, based on the values we already know for angles that are multiples of \begin{align*}30^\circ\end{align*} or \begin{align*}45^\circ\end{align*}. (Note, though, that this technique will still only tell us the cosines of angles that are multiples of \begin{align*}15^\circ\end{align*}. That’s because when you add or subtract angles that are multiples of \begin{align*}15^\circ\end{align*} (which includes all angles that are multiples of \begin{align*}30^\circ\end{align*} or \begin{align*}45^\circ\end{align*}), the result is always also a multiple of \begin{align*}15^\circ\end{align*}.)

Technology Note

As before, note that calculators cannot tell us with absolute certainty if two expressions have identical values; they might, for example, be identical up to twenty decimal places, but differ after that point. However, if two expressions seem to have equal values when plugged into a calculator, there is a reasonably good chance that they are really equal—and more importantly, the calculator will clearly tell us if they are not equal at all. Hence, double-checking answers with a calculator is still a good idea.

Sum and Difference Identities for Sine and Tangent

Sum and Difference Identities for Sine

(For reference, Identities A and B are on page 244.)

The angle in example 1, \begin{align*}\frac{5 \pi}{12}\end{align*}, could of course be expressed in many different ways as the sum or difference of two other angles, but the way shown in the text is the most convenient way to express it in terms of angles we are already familiar with. Consider asking your students to explain why this choice is a good one. (You might need to remind them, though, that \begin{align*}\frac{3 \pi}{12}\end{align*} is the same as \begin{align*}\frac{\pi}{4}\end{align*} and \begin{align*}\frac{2 \pi}{12}\end{align*} is the same as \begin{align*}\frac{\pi}{6}\end{align*} in order for them to see why these angles should be easy to work with.)

A shortcut to finding the cosines of the angles in example 2 is to notice that the side lengths are Pythagorean triples. That is, if the sine of angle \begin{align*}\alpha\end{align*} is \begin{align*}\frac{5}{13}\end{align*}, the other leg of the relevant triangle must measure \begin{align*}12\end{align*} (to complete the triple \begin{align*}5-12-13\end{align*}) and so the cosine is \begin{align*}\frac{12}{13}\end{align*}. Similar reasoning holds for angle \begin{align*}\beta\end{align*}.

Sum and Difference Identities for Tangent

Once again, expressing the sum formula for tangent in words may make it easier for students to remember: “The tangent of the sum equals the sum of the tangents over 1 minus the product of the tangents.” There isn’t really a concise way to do this for the sine formulas, though.

Consider whether you want your students to memorize these and other trig identities or not. They definitely need to develop a good sense for when to use which ones, but the formulas themselves are the least important part of that knowledge; in fact, knowing how to derive the formulas may be more useful than simply knowing the formulas themselves. Instead of requiring that all the formulas be memorized for a test, for example, it might be more educational to supply a few of the formulas and require that students re-derive the others from the few given (after making sure, of course, that the formulas supplied are sufficient to derive the others from.)

Of course, some students might find that a little too challenging, but on the other hand, some will find it easier than memorization. Perhaps a good compromise would be to provide a few of the formulas so that students can derive the others if they want to, but not make that derivation a required part of the exam. Since students will presumably need to use any or all of the formulas to solve some of the exam questions, they will still have to either derive the formulas or have them memorized, but can do whichever of those two works best for them.

Double-Angle Identities

Deriving the Double-Angle Identities

There is really only one form of the double-angle identity for sine, but we see here that the double-angle identity for cosine can be expressed in several different ways. Students should be familiar with all of these formulas in case they encounter them “in the wild”; it’s good to be able to recognize when an expression can be converted to \begin{align*}\cos 2a\end{align*}, just as it’s useful to be able to express \begin{align*}\cos 2a\end{align*} in terms of functions of a. Also, as will be seen later, some problems are easier or harder depending on which form of the double-angle identity we choose to work with.

Applying the Double-Angle Identities

Just as in the previous lesson, noticing that \begin{align*}5\end{align*} and \begin{align*}13\end{align*} are part of a Pythagorean triple is a useful shortcut to finding \begin{align*}\cos a\end{align*}.

Finding Angle Values Given Double Angles

(Example 1 should read “\begin{align*}2x\end{align*} is a Quadrant II angle” (rather than “\begin{align*}x\end{align*} is a Quadrant II angle”), both in the introduction and in the second-to-last line of the table.)

The seventh line of the table may merit an explanation, as students whose algebra is rusty may wonder where the \begin{align*}``a\text{''}\end{align*} comes from. Explain that it is simply a “dummy variable” that we choose to stand for \begin{align*}\sin^2 x\end{align*} so that the equation we are working with becomes a simple quadratic, which we know how to solve. Later, we will change \begin{align*}a\end{align*} back to \begin{align*}\sin^2 x\end{align*} in order to finish solving for \begin{align*}x\end{align*}.

Simplify Expressions Using Double-Angle Identities

In the example given here, you can show, if you wish, how choosing a different double-angle formula would make the expression we are working with more complicated, and hence why the one used here is the best one to use in this case.

For example, using \begin{align*}\cos 2 \theta = \cos^2 \theta - \sin^2 \theta\end{align*} instead yields the following:

\begin{align*}& \frac{1-\cos 2 \theta}{\sin 2 \theta}\\ & =\frac{1 - (\cos^2 \theta - \sin^2 \theta)} {2 \sin \theta \cos \theta}\\ & = \frac{1 - \cos^2 \theta + \sin^2 \theta}{2 \sin \theta \cos \theta}\\ & = \frac{1}{2 \sin \theta \cos \theta} - \frac{\cos^2 \theta}{2 \sin \theta \cos \theta} + \frac{\sin^2 \theta}{2 \sin \theta \cos \theta}\\ & = \frac{1} {2 \sin \theta \cos \theta} - \frac{\cos \theta}{2 \sin \theta} + \frac{\sin \theta}{2 \cos \theta}\\ & = 2 \csc \theta \sec \theta - \frac{1}{2} \cot \theta + \frac{1}{2} \tan \theta\end{align*}

and at this point it isn’t at all obvious how to show that this is equivalent to \begin{align*}\tan \theta\end{align*}.

Lesson Summary

The first paragraph here describes a useful strategy for getting an idea of what sort of answers are reasonable before attempting to solve a trig problem: figure out, if you can, approximately where the given angles are and approximately what the values of the trig functions should be for those angles, or at least figure out upper or lower bounds on the trig functions based on whether the given angle is greater or less than an angle you are already familiar with, and whether the relevant trig functions are increasing or decreasing in the part of the unit circle where the angle is located.

This particular example contains a slight error , though: \begin{align*}\sin 45^\circ\end{align*} is actually about \begin{align*}0.7\end{align*}, and \begin{align*}\sin 30^\circ\end{align*} is \begin{align*}0.5\end{align*}, so the angle \begin{align*}\theta\end{align*} in the problem is actually between \begin{align*}30^\circ\end{align*} and \begin{align*}45^\circ\end{align*}, and \begin{align*}2 \theta\end{align*} is therefore between \begin{align*}60^\circ\end{align*} and \begin{align*}90^\circ\end{align*}.

Review Questions

Some of these questions involve finding \begin{align*}\tan 2x\end{align*}, which we haven’t derived a formula for. In fact, there isn’t a concise formula for the tangent of a double angle; however, once \begin{align*}\sin 2x\end{align*} and \begin{align*}\cos 2x\end{align*} have been found, \begin{align*}\tan 2x\end{align*} is simply the quotient of the two.

Half-Angle Identities

Deriving the Half-Angle Formulas

In the first and second derivations here, it’s important for students to see that when we substitute a for \begin{align*}2 \theta\end{align*}, we can also substitute for \begin{align*}\theta\end{align*}. This is how we are able to find an expression for the sine or cosine of \begin{align*}\frac{a}{2}\end{align*} based on that of \begin{align*}a\end{align*}.

(In the tangent derivation, though, the text uses \begin{align*}a\end{align*} and \begin{align*}\theta\end{align*} interchangeably. This is because we don’t have to make any actual half-angle substitutions there, but merely plug in the formulas for sine and cosine.)

Expressing the half-angle formula for cosine as \begin{align*}\sqrt{\frac{1 + \cos a}{2}}\end{align*} instead of \begin{align*}\sqrt{\frac{\cos a + 1}{2}}\end{align*} may make it easier to remember, as it then looks more like the formula for sine.

Use Half-Angle Identities to Find Exact Values

You might need to review how we know the sine of \begin{align*}225^\circ\end{align*}. \begin{align*}225^\circ\end{align*} is equal to \begin{align*}180^\circ + 45^\circ\end{align*}, which means it is a third quadrant angle with reference angle \begin{align*}45^\circ\end{align*}, but you may want to draw it to make this clearer.

Find Half-Angle Values Given Angles

For example 2, stress that although we have enough information to figure out the measure of angle \begin{align*}\theta\end{align*}, we don’t actually need to know it to apply the half-angle formula; the formula only requires us to know the cosine of \begin{align*}\theta\end{align*}, which we already have.

You may want to explain, though, how we know that half of a fourth quadrant angle is a second quadrant angle. In general, halving a first or second quadrant angle will yield an angle in the first quadrant, and halving a third or fourth quadrant angle will yield an angle in the second quadrant. We can verify this numerically: half of an angle between \begin{align*}0^\circ\end{align*} and \begin{align*}180^\circ\end{align*} must be between \begin{align*}0^\circ\end{align*} and \begin{align*}90^\circ\end{align*}, and half of an angle between \begin{align*}180^\circ\end{align*} and \begin{align*}360^\circ\end{align*} must be between \begin{align*}90^\circ\end{align*} and \begin{align*}180^\circ.\end{align*}

Using the Half- or Double-Angle Formulas to Verify Identities

Again, you may point out in the second line of this derivation that of the three possible expressions for cos \begin{align*}2\theta\end{align*}, we’ve chosen the one that makes the numerator reduce to the simplest form.

Technology Notes

Another way to demonstrate that \begin{align*}sin \frac{\theta}{2}\end{align*} does not equal \begin{align*}\frac{1}{2} \sin \theta\end{align*} is to pick a familiar pair of angles, such as \begin{align*}90^\circ\end{align*} and \begin{align*}180^\circ,\end{align*} and note that \begin{align*}\sin 90^\circ\end{align*} is definitely not half of \begin{align*}\sin 180^\circ.\end{align*}

Product-and-Sum, Sum-and-Product and Linear Combinations of Identities

Transformations of Sums, Differences of Sines and Cosines, and Products of Sines and Cosines

Remind students not to mix up the formulas here with the sum and difference identities learned earlier: the formula for \begin{align*}\cos (\alpha + \beta)\end{align*} is not at all the same as the formula for \begin{align*}\cos \alpha + \cos \beta\end{align*}!

For an extra challenge, you might ask students to derive the three formulas whose derivations are not shown, by applying similar reasoning to that used in the derivation that is shown.

Transformations of Products of Sines and Cosines into Sums and Differences of Sines and Cosines

The product formulas shown here bear a certain resemblance to the sum formulas they are derived from, and students may be tempted to apply them after applying the sum formulas. For example, they may think, after they have determined that \begin{align*}\cos (a + b) = \cos a \cos b - \sin a \sin b\end{align*}, that it would then be a good idea to plug in the expressions they’ve just learned for \begin{align*}\cos a \cos b\end{align*} and \begin{align*}\sin a \sin b\end{align*}. But this will only give them a complex expression in terms of \begin{align*}\cos (a + b)\end{align*} and \begin{align*}\cos (a - b)\end{align*}, which won’t help them much since the value of \begin{align*}\cos (a + b)\end{align*} is what they were looking for to begin with.

The key thing they need to understand is that each of these identities is a tool to be used in different situations, depending on what knowledge they already have. If they need to find the sine or cosine of an angle, and that angle can be expressed as a sum or difference of two angles of which they already know the sine and cosine, then the sum formula from earlier is useful. If they need to know the product of two sines or cosines, and they don’t know the sines or cosines themselves, but do know the sine and cosine of the sum and difference of those two angles, then the product formula learned here is useful. In general, they should form the habit of writing down exactly what it is they are looking for and then considering which tools in their possession might apply to that particular situation.

Linear Combinations

You may need to define the term “linear combination” for students who haven’t encountered it before. A linear combination of two quantities is simply a multiple of one quantity plus a multiple of the other—so, for example, a linear combination of \begin{align*}\sin x\end{align*} and \begin{align*}\cos x\end{align*} would be any expression of the form a \begin{align*}\sin x + b \cos x\end{align*}, where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are any two numbers (real numbers, unless otherwise specified). Students will use linear combinations in their study of vectors in the next chapter.

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