1.4: Trigonometric Identities
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In Section ???, we saw the utility of the Pythagorean Identities in Theorem ??? along with the Quotient and Reciprocal Identities in Theorem ???. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our first set of identities is the `Even / Odd' identities.\footnote{As mentioned at the end of Section ???, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section ???.)}
Note: Even / Odd Identities
For all applicable angles θ:
- cos(−θ)=cos(θ)
- sec(−θ)=sec(θ)
- sin(−θ)=−sin(θ)
- csc(−θ)=−csc(θ)
- tan(−θ)=−tan(θ)
- cot(−θ)=−cot(θ)
In light of the Quotient and Reciprocal Identities, Theorem ???, it suffices to show cos(−θ)=cos(θ) and sin(−θ)=−sin(θ). The remaining four circular functions can be expressed in terms of cos(θ) and sin(θ) so the proofs of their Even / Odd Identities are left as exercises. Consider an angle θ plotted in standard position. Let θo be the angle coterminal with θ with 0≤θo<2π. (We can construct the angle θo by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θo are coterminal, cos(θ)=cos(θo) and sin(θ)=sin(θo).
We now consider the angles −θ and −θo. Since θ is coterminal with θo, there is some integer k so that θ=θo+2π⋅k. Therefore, −θ=−θo−2π⋅k=−θo+2π⋅(−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θo. Hence, cos(−θ)=cos(−θo) and sin(−θ)=sin(−θo). Let P and Q denote the points on the terminal sides of θo and −θo, respectively, which lie on the Unit Circle. By definition, the coordinates of P are (cos(θo),sin(θo)) and the coordinates of Q are (cos(−θo),sin(−θo)). Since θo and −θo sweep out congruent central sectors of the Unit Circle, it follows that the points P and Q are symmetric about the x-axis. Thus, cos(−θo)=cos(θo) and sin(−θo)=−sin(θo). Since the cosines and sines of θo and −θo are the same as those for θ and −θ, respectively, we get cos(−θ)=cos(θ) and sin(−θ)=−sin(θ), as required. The Even / Odd Identities are readily demonstrated using any of the `common angles' noted in Section ???. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities.
Note: Sum and Difference Identities for Cosine
For all angles α and β:
- cos(α+β)=cos(α)cos(β)−sin(α)sin(β)
- cos(α−β)=cos(α)cos(β)+sin(α)sin(β)
We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles α and β to angles αo and βo, coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and αo are coterminal, as are β and βo, it follows that α−β is coterminal with αo−βo. Consider the case below where αo≥βo.
Since the angles POQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B.\footnote{In the picture we've drawn, the \underline{tri}angles POQ and AOB are congruent, which is even better. However, αo−βo could be 0 or it could be π, neither of which makes a triangle. It could also be larger than π, which makes a triangle, just not the one we've drawn. You should think about those three cases.} The distance formula, Equation ???, yields
√(cos(αo)−cos(βo))2+(sin(αo)−sin(βo))2=√(cos(αo−βo)−1)2+(sin(αo−βo)−0)2
Squaring both sides, we expand the left hand side of this equation as
(cos(αo)−cos(βo))2+(sin(αo)−sin(βo))2=cos2(αo)−2cos(αo)cos(βo)+cos2(βo)+sin2(αo)−2sin(αo)sin(βo)+sin2(βo)=cos2(αo)+sin2(αo)+cos2(βo)+sin2(βo)−2cos(αo)cos(βo)−2sin(αo)sin(βo)
From the Pythagorean Identities, cos2(αo)+sin2(αo)=1 and cos2(βo)+sin2(βo)=1, so
(cos(αo)−cos(βo))2+(sin(αo)−sin(βo))2=2−2cos(αo)cos(βo)−2sin(αo)sin(βo)
Turning our attention to the right hand side of our equation, we find
(cos(αo−βo)−1)2+(sin(αo−βo)−0)2=cos2(αo−βo)−2cos(αo−βo)+1+sin2(αo−βo)=1+cos2(αo−βo)+sin2(αo−βo)−2cos(αo−βo)
Once again, we simplify cos2(αo−βo)+sin2(αo−βo)=1, so that
(cos(αo−βo)−1)2+(sin(αo−βo)−0)2=2−2cos(αo−βo)
Putting it all together, we get 2−2cos(αo)cos(βo)−2sin(αo)sin(βo)=2−2cos(αo−βo), which simplifies to: cos(αo−βo)=cos(αo)cos(βo)+sin(αo)sin(βo). Since α and αo, β and βo and α−β and αo−βo are all coterminal pairs of angles, we have cos(α−β)=cos(α)cos(β)+sin(α)sin(β). For the case where αo≤βo, we can apply the above argument to the angle βo−αo to obtain the identity cos(βo−αo)=cos(βo)cos(αo)+sin(βo)sin(αo). Applying the Even Identity of cosine, we get cos(βo−αo)=cos(−(αo−βo))=cos(αo−βo), and we get the identity in this case, too.
To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities
cos(α+β)=cos(α−(−β))=cos(α)cos(−β)+sin(α)sin(−β)=cos(α)cos(β)−sin(α)sin(β)
We put these newfound identities to good use in the following example.
Example 1.4.1: Cosine Sum and Difference
- Find the exact value of cos(15∘).
- Verify the identity: cos(π2−θ)=sin(θ).
Solution
- In order to use Theorem ??? to find cos(15∘), we need to write 15∘ as a sum or difference of angles whose cosines and sines we know. One way to do so is to write 15∘=45∘−30∘.
cos(15∘)=cos(45∘−30∘)=cos(45∘)cos(30∘)+sin(45∘)sin(30∘)=(√22)(√32)+(√22)(12)=√6+√24
- In a straightforward application of Theorem ???, we find
cos(π2−θ)=cos(π2)cos(θ)+sin(π2)sin(θ)=(0)(cos(θ))+(1)(sin(θ))=sin(θ)
The identity verified in Example 1.4.1, namely, cos(π2−θ)=sin(θ), is the first of the celebrated `cofunction' identities. These identities were first hinted at in Exercise ??? in Section ???. From sin(θ)=cos(π2−θ), we get:
sin(π2−θ)=cos(π2−[π2−θ])=cos(θ),
which says, in words, that the `co'sine of an angle is the sine of its `co'mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises.
Note: Cofunction Identities
For all applicable angles θ:
- cos(π2−θ)=sin(θ)
- sin(π2−θ)=cos(θ)
- sec(π2−θ)=csc(θ)
- csc(π2−θ)=sec(θ)
- tan(π2−θ)=cot(θ)
- cot(π2−θ)=tan(θ)
With the Cofunction Identities in place, we are now in the position to derive the sum and difference formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand using the difference formula for cosine
sin(α+β)=cos(π2−(α+β))=cos([π2−α]−β)=cos(π2−α)cos(β)+sin(π2−α)sin(β)=sin(α)cos(β)+cos(α)sin(β)
We can derive the difference formula for sine by rewriting sin(α−β) as sin(α+(−β)) and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader.
Sum and Difference Identities for Sine
For all angles α and β, \index{Difference Identity ! for sine} \index{Sum Identity ! for sine}
- sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
- sin(α−β)=sin(α)cos(β)−cos(α)sin(β)
Example 1.4.1:
- Find the exact value of sin(19π12)
- If α is a Quadrant II angle with sin(α)=513, and β is a Quadrant III angle with tan(β)=2, find sin(α−β).
- Derive a formula for tan(α+β) in terms of tan(α) and tan(β).
Solution
- As in Example ???, we need to write the angle 19π12 as a sum or difference of common angles. The denominator of 12 suggests a combination of angles with denominators 3 and 4. One such combination is 19π12=4π3+π4. Applying Theorem ???, we get
sin(19π12)=sin(4π3+π4)=sin(4π3)cos(π4)+cos(4π3)sin(π4)=(−√32)(√22)+(−12)(√22)=−√6−√24
- In order to find sin(α−β) using Theorem ???, we need to find cos(α) and both cos(β) and sin(β). To find cos(α), we use the Pythagorean Identity cos2(α)+sin2(α)=1. Since sin(α)=513, we have cos2(α)+(513)2=1, or cos(α)=±1213. Since α is a Quadrant II angle, cos(α)=−1213. We now set about finding cos(β) and sin(β). We have several ways to proceed, but the Pythagorean Identity 1+tan2(β)=sec2(β) is a quick way to get sec(β), and hence, cos(β). With tan(β)=2, we get 1+22=sec2(β) so that sec(β)=±√5. Since β is a Quadrant III angle, we choose sec(β)=−√5 so cos(β)=1sec(β)=1−√5=−√55. We now need to determine sin(β). We could use The Pythagorean Identity cos2(β)+sin2(β)=1, but we opt instead to use a quotient identity. From tan(β)=sin(β)cos(β), we have sin(β)=tan(β)cos(β) so we get sin(β)=(2)(−√55)=−2√55. We now have all the pieces needed to find sin(α−β):
sin(α−β)=sin(α)cos(β)−cos(α)sin(β)=(513)(−√55)−(−1213)(−2√55)=−29√565
We can start expanding tan(α+β) using a quotient identity and our sum formulas
tan(α+β)=sin(α+β)cos(α+β)=sin(α)cos(β)+cos(α)sin(β)cos(α)cos(β)−sin(α)sin(β)
Since tan(α)=sin(α)cos(α) and tan(β)=sin(β)cos(β), it looks as though if we divide both numerator and denominator by cos(α)cos(β) we will have what we want
tan(α+β)=sin(α)cos(β)+cos(α)sin(β)cos(α)cos(β)−sin(α)sin(β)⋅1cos(α)cos(β)1cos(α)cos(β)=sin(α)cos(β)cos(α)cos(β)+cos(α)sin(β)cos(α)cos(β)cos(α)cos(β)cos(α)cos(β)−sin(α)sin(β)cos(α)cos(β)=sin(α)cos(β)cos(α)cos(β)+cos(α)sin(β)cos(α)cos(β)cos(α)cos(β)cos(α)cos(β)−sin(α)sin(β)cos(α)cos(β)=tan(α)+tan(β)1−tan(α)tan(β)
Naturally, this formula is limited to those cases where all of the tangents are defined.\qed
The formula developed in Exercise ??? for tan(α+β) can be used to find a formula for tan(α−β) by rewriting the difference as a sum, tan(α+(−β)), and the reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for cosine, sine and tangent.
Note
Sum and Difference Identities:} For all applicable angles α and β, \index{Difference Identity ! for tangent} \index{Sum Identity ! for tangent} \index{Difference Identity ! for cosine} \index{Sum Identity ! for cosine} \index{Difference Identity ! for sine} \index{Sum Identity ! for sine}
- cos(α±β)=cos(α)cos(β)∓sin(α)sin(β)
- sin(α±β)=sin(α)cos(β)±cos(α)sin(β)
- tan(α±β)=tan(α)±tan(β)1∓tan(α)tan(β)
In the statement of Theorem ???, we have combined the cases for the sum `$+$' and difference `\)-$' of angles into one formula. The convention here is that if you want the formula for the sum `$+$' of two angles, you use the top sign in the formula; for the difference, `\)-$', use the bottom sign. For example, tan(α−β)=tan(α)−tan(β)1+tan(α)tan(β)
If we specialize the sum formulas in Theorem ??? to the case when α=β, we obtain the following `Double Angle' Identities.
Note Double Angle Identities
For all applicable angles θ, \index{Double Angle Identities}
- cos(2θ)={cos2(θ)−sin2(θ)[5pt]2cos2(θ)−1[5pt]1−2sin2(θ)
- sin(2θ)=2sin(θ)cos(θ)
- tan(2θ)=2tan(θ)1−tan2(θ)
The three different forms for cos(2θ) can be explained by our ability to `exchange' squares of cosine and sine via the Pythagorean Identity cos2(θ)+sin2(θ)=1 and we leave the details to the reader. It is interesting to note that to determine the value of cos(2θ), only \textit{one} piece of information is required: either cos(θ) or sin(θ). To determine sin(2θ), however, it appears that we must know both sin(θ) and cos(θ). In the next example, we show how we can find sin(2θ) knowing just one piece of information, namely tan(θ).
Example 1.4.1:
- Suppose P(−3,4) lies on the terminal side of θ when θ is plotted in standard position. Find cos(2θ) and sin(2θ) and determine the quadrant in which the terminal side of the angle 2θ lies when it is plotted in standard position.
- If sin(θ)=x for −π2≤θ≤π2, find an expression for sin(2θ) in terms of x.
- \label{doubleanglesinewtan} Verify the identity: sin(2θ)=2tan(θ)1+tan2(θ).
- Express cos(3θ) as a polynomial in terms of cos(θ).
Solution
- Using Theorem ??? from Section ??? with x=−3 and y=4, we find r=√x2+y2=5. Hence, cos(θ)=−35 and sin(θ)=45. Applying Theorem ???, we get cos(2θ)=cos2(θ)−sin2(θ)=(−35)2−(45)2=−725, and sin(2θ)=2sin(θ)cos(θ)=2(45)(−35)=−2425. Since both cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III.
- If your first reaction to `$\sin(\theta) = x$' is `No it's not, cos(θ)=x$!′thenyouhaveindeedlearnedsomething,andwetakecomfortinthat.However,contextiseverything.Here,‘$x$′isjustavariable−itdoesnotnecessarilyrepresentthe\(x-coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x represents the quantity sin(θ), and what we wish to know is how to express sin(2θ) in terms of x. We will see more of this kind of thing in Section ???, and, as usual, this is something we need for Calculus. Since sin(2θ)=2sin(θ)cos(θ), we need to write cos(θ) in terms of x to finish the problem. We substitute x=sin(θ) into the Pythagorean Identity, cos2(θ)+sin2(θ)=1, to get cos2(θ)+x2=1, or cos(θ)=±√1−x2. Since −π2≤θ≤π2, cos(θ)≥0, and thus cos(θ)=√1−x2. Our final answer is sin(2θ)=2sin(θ)cos(θ)=2x√1−x2.
We start with the right hand side of the identity and note that 1+tan2(θ)=sec2(θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms of cos(θ) and sin(θ):
2tan(θ)1+tan2(θ)=2tan(θ)sec2(θ)=2(sin(θ)cos(θ))1cos2(θ)=2(sin(θ)cos(θ))cos2(θ)=2(sin(θ)cos(θ))cos(θ)cos(θ)=2sin(θ)cos(θ)=sin(2θ)
- In Theorem ???, one of the formulas for cos(2θ), namely cos(2θ)=2cos2(θ)−1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to find such an identity for cos(3θ). Using the sum formula for cosine, we begin with
cos(3θ)=cos(2θ+θ)=cos(2θ)cos(θ)−sin(2θ)sin(θ)
Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute cos(2θ)=2cos2(θ)−1 and sin(2θ)=2sin(θ)cos(θ) which yields
cos(3θ)=cos(2θ)cos(θ)−sin(2θ)sin(θ)=(2cos2(θ)−1)cos(θ)−(2sin(θ)cos(θ))sin(θ)=2cos3(θ)−cos(θ)−2sin2(θ)cos(θ)
Finally, we exchange sin2(θ) for 1−cos2(θ) courtesy of the Pythagorean Identity, and get
cos(3θ)=2cos3(θ)−cos(θ)−2sin2(θ)cos(θ)=2cos3(θ)−cos(θ)−2(1−cos2(θ))cos(θ)=2cos3(θ)−cos(θ)−2cos(θ)+2cos3(θ)=4cos3(θ)−3cos(θ)
and we are done.
In the last problem in Example ???, we saw how we could rewrite cos(3θ) as sums of powers of cos(θ). In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity cos(2θ)=2cos2(θ)−1 for cos2(θ) and the identity cos(2θ)=1−2sin2(θ) for sin2(θ) results in the aptly-named `Power Reduction' formulas below.
Power Reduction Formulas
For all angles θ, \index{Power Reduction Formulas}
- cos2(θ)=1+cos(2θ)2
- sin2(θ)=1−cos(2θ)2
Example 1.4.1:
Rewrite sin2(θ)cos2(θ) as a sum and difference of cosines to the first power.
Solution
We begin with a straightforward application of Theorem ???
sin2(θ)cos2(θ)=(1−cos(2θ)2)(1+cos(2θ)2)=14(1−cos2(2θ))=14−14cos2(2θ)
Next, we apply the power reduction formula to cos2(2θ) to finish the reduction
sin2(θ)cos2(θ)=14−14cos2(2θ)=14−14(1+cos(2(2θ))2)=14−18−18cos(4θ)=18−18cos(4θ)
Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to \(\cos^{2}\left(\frac{\theta}{2}\right)$
cos2(θ2)=1+cos(2(θ2))2=1+cos(θ)2.
We can obtain a formula for cos(θ2) by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below.
Half Angle Formulas
For all applicable angles θ:
- cos(θ2)=±√1+cos(θ)2
- sin(θ2)=±√1−cos(θ)2
- tan(θ2)=±√1−cos(θ)1+cos(θ)
where the choice of ± depends on the quadrant in which the terminal side of θ2 lies.
Example 1.4.1:
- Use a half angle formula to find the exact value of cos(15∘).
- Suppose −π≤θ≤0 with cos(θ)=−35. Find sin(θ2).
- Use the identity given in number ??? of Example ??? to derive the identity tan(θ2)=sin(θ)1+cos(θ)
Solution
- To use the half angle formula, we note that 15∘=30∘2 and since 15∘ is a Quadrant I angle, its cosine is positive. Thus we have
cos(15∘)=+√1+cos(30∘)2=√1+√322=√1+√322⋅22=√2+√34=√2+√32
Back in Example ???, we found cos(15∘) by using the difference formula for cosine. In that case, we determined cos(15∘)=√6+√24. The reader is encouraged to prove that these two expressions are equal.
- If −π≤θ≤0, then −π2≤θ2≤0, which means sin(θ2)<0. Theorem ??? gives
sin(θ2)=−√1−cos(θ)2=−√1−(−35)2=−√1+352⋅55=−√810=−2√55
- Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number ??? of Example ??? and manipulate it into the identity we are asked to prove. The identity we are asked to start with is sin(2θ)=2tan(θ)1+tan2(θ). If we are to use this to derive an identity for tan(θ2), it seems reasonable to proceed by replacing each occurrence of θ with \(\frac{\theta}{2}$
sin(2(θ2))=2tan(θ2)1+tan2(θ2)sin(θ)=2tan(θ2)1+tan2(θ2)
We now have the sin(θ) we need, but we somehow need to get a factor of 1+cos(θ) involved. To get cosines involved, recall that 1+tan2(θ2)=sec2(θ2). We continue to manipulate our given identity by converting secants to cosines and using a power reduction formula
sin(θ)=2tan(θ2)1+tan2(θ2)sin(θ)=2tan(θ2)sec2(θ2)sin(θ)=2tan(θ2)cos2(θ2)sin(θ)=2tan(θ2)(1+cos(2(θ2))2)sin(θ)=tan(θ2)(1+cos(θ))tan(θ2)=sin(θ)1+cos(θ)
Our next batch of identities, the Product to Sum Formulas,\footnote{These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows.} are easily verified by expanding each of the right hand sides in accordance with Theorem ??? and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference.
Note: Product to Sum Formulas
For all angles α and β, \index{Product to Sum Formulas}
- cos(α)cos(β)=12[cos(α−β)+cos(α+β)]
- sin(α)sin(β)=12[cos(α−β)−cos(α+β)]
- sin(α)cos(β)=12[sin(α−β)+sin(α+β)]
Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section ???. These are easily verified using the Product to Sum Formulas, and as such, their proofs are left as exercises.
Note: Sum to Product Formulas:
For all angles α and β:
- cos(α)+cos(β)=2cos(α+β2)cos(α−β2)
- cos(α)−cos(β)=−2sin(α+β2)sin(α−β2)
- sin(α)±sin(β)=2sin(α±β2)cos(α∓β2)
Example 1.4.1:
- Write cos(2θ)cos(6θ) as a sum.
- \Write sin(θ)−sin(3θ) as a product.
Solution
- Identifying α=2θ and β=6θ, we find
cos(2θ)cos(6θ)=12[cos(2θ−6θ)+cos(2θ+6θ)]=12cos(−4θ)+12cos(8θ)=12cos(4θ)+12cos(8θ),
where the last equality is courtesy of the even identity for cosine, cos(−4θ)=cos(4θ).
- Identifying α=θ and β=3θ yields
sin(θ)−sin(3θ)=2sin(θ−3θ2)cos(θ+3θ2)=2sin(−θ)cos(2θ)=−2sin(θ)cos(2θ),
where the last equality is courtesy of the odd identity for sine, sin(−θ)=−sin(θ).
The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises ??? - ??? in Section ???, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs.