10.4: Trigonometric Identities
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In Section 10.3, we saw the utility of the Pythagorean Identities in Theorem 10.8 along with the Quotient and Reciprocal Identities in Theorem 10.6. 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.1
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 10.6, 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 θ0 be the angle coterminal with θ with 0≤θ0<2π. (We can construct the angle θ0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θ0 are coterminal, cos(θ)=cos(θ0) and sin(θ)=sin(θ0).
We now consider the angles −θ and −θ0. Since θ is coterminal with θ0, there is some integer k so that θ=θ0+2π⋅k. Therefore, −θ=−θ0−2π⋅k=−θ0+2π⋅(−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θ0. Hence, cos(−θ)=cos(−θ0) and sin(−θ)=sin(−θ0). Let P and Q denote the points on the terminal sides of θ0 and −θ0, respectively, which lie on the Unit Circle. By definition, the coordinates of P are (cos(θ0),sin(θ0)) and the coordinates of Q are (cos(−θ0),sin(−θ0)). Since θ0 and −θ0 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(−θ0)=cos(θ0) and sin(−θ0)=−sin(θ0). Since the cosines and sines of θ0 and −θ0 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 10.2. 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.
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 α0 and β0, coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and α0 are coterminal, as are β and β0, it follows that α−β is coterminal with α0−β0. Consider the case below where α0≥β0.
Since the angles POQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B.2 The distance formula, Equation 1.1, yields
√(cos(α0)−cos(β0))2+(sin(α0)−sin(β0))2=√(cos(α0−β0)−1)2+(sin(α0−β0)−0)2
Squaring both sides, we expand the left hand side of this equation as
(cos(α0)−cos(β0))2+(sin(α0)−sin(β0))2=cos2(α0)−2cos(α0)cos(β0)+cos2(β0)+sin2(α0)−2sin(α0)sin(β0)+sin2(β0)=cos2(α0)+sin2(α0)+cos2(β0)+sin2(β0)−2cos(α0)cos(β0)−2sin(α0)sin(β0)
From the Pythagorean Identities, cos2(α0)+sin2(α0)=1 and cos2(β0)+sin2(β0)=1, so
(cos(α0)−cos(β0))2+(sin(α0)−sin(β0))2=2−2cos(α0)cos(β0)−2sin(α0)sin(β0)
Turning our attention to the right hand side of our equation, we find
(cos(α0−β0)−1)2+(sin(α0−β0)−0)2=cos2(α0−β0)−2cos(α0−β0)+1+sin2(α0−β0)=1+cos2(α0−β0)+sin2(α0−β0)−2cos(α0−β0)
Once again, we simplify cos2(α0−β0)+sin2(α0−β0)=1, so that
(cos(α0−β0)−1)2+(sin(α0−β0)−0)2=2−2cos(α0−β0)
Putting it all together, we get 2−2cos(α0)cos(β0)−2sin(α0)sin(β0)=2−2cos(α0−β0), which simplifies to: cos(α0−β0)=cos(α0)cos(β0)+sin(α0)sin(β0). Since α and α0, β and β0 and α−β and α0−β0 are all coterminal pairs of angles, we have cos(α−β)=cos(α)cos(β)+sin(α)sin(β). For the case where α0≤β0, we can apply the above argument to the angle β0−α0 to obtain the identity cos(β0−α0)=cos(β0)cos(α0)+sin(β0)sin(α0). Applying the Even Identity of cosine, we get cos(β0−α0)=cos(−(α0−β0))=cos(α0−β0), 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.
- Find the exact value of cos(15∘).
- Verify the identity: cos(π2−θ)=sin(θ).
Solution
- In order to use Theorem 10.13 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 10.13, we find
cos(π2−θ)=cos(π2)cos(θ)+sin(π2)sin(θ)[10pt]=(0)(cos(θ))+(1)(sin(θ))[4pt]=sin(θ)
The identity verified in Example 10.4.1, namely, cos(π2−θ)=sin(θ), is the first of the celebrated ‘cofunction’ identities. These identities were first hinted at in Exercise 74 in Section 10.2. 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.
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))[10pt]=cos([π2−α]−β)[10pt]=cos(π2−α)cos(β)+sin(π2−α)sin(β)[10pt]=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.
For all angles α and β,
- sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
- sin(α−β)=sin(α)cos(β)−cos(α)sin(β)
- 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 10.4.1, 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 10.15, we get
sin(19π12)=sin(4π3+π4)[10pt]=sin(4π3)cos(π4)+cos(4π3)sin(π4)[10pt]=(−√32)(√22)+(−12)(√22)=−√6−√24
- In order to find sin(α−β) using Theorem 10.15, 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)[10pt]=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.
The formula developed in Exercise 10.4.2 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.
For all applicable angles α and β,
- 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 10.16, 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 10.16 to the case when α=β, we obtain the following ‘Double Angle’ Identities.
For all applicable angles θ,
- cos(2θ)={cos2(θ)−sin2(θ)2cos2(θ)−11−2sin2(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 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).
- 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.
- [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 10.3 from Section 10.2 with x=−3 and y=4, we find r=√x2+y2=5. Hence, cos(θ)=−35 and sin(θ)=45. Applying Theorem 10.17, 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(θ)=x’ is ‘No it’s not, cos(θ)=x!’ then you have indeed learned something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a variable - it does not necessarily represent the 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 10.6, 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 10.17, 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 10.4.3, 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.
For all angles θ,
- cos2(θ)=1+cos(2θ)2
- sin2(θ)=1−cos(2θ)2
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 10.18
sin2(θ)cos2(θ)=(1−cos(2θ)2)(1+cos(2θ)2)[10pt]=14(1−cos2(2θ))[10pt]=14−14cos2(2θ)
Next, we apply the power reduction formula to cos2(2θ) to finish the reduction
sin2(θ)cos2(θ)=14−14cos2(2θ)[10pt]=14−14(1+cos(2(2θ))2)[10pt]=14−18−18cos(4θ)[10pt]=18−18cos(4θ)
Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to cos2(θ2)
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.
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.
- 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 3 of Example 10.4.3 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[10pt]=√1+√322⋅22=√2+√34=√2+√32
Back in Example 10.4.1, 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 10.19 gives
sin(θ2)=−√1−cos(θ)2=−√1−(−35)2[10pt]=−√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 3 of Example 10.4.3 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 θ2
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,3 are easily verified by expanding each of the right hand sides in accordance with Theorem 10.16 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.
For all angles α and β,
- 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 10.7. These are easily verified using the Product to Sum Formulas, and as such, their proofs are left as exercises.
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)
- 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)][4pt]=12cos(−4θ)+12cos(8θ)[4pt]=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 38 - 43 in Section 10.5, 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.
10.4.1. Exercises
In Exercises 1 - 6, use the Even / Odd Identities to verify the identity. Assume all quantities are defined.
- sin(3π−2θ)=−sin(2θ−3π)
- cos(−π4−5t)=cos(5t+π4)
- tan(−t2+1)=−tan(t2−1)
- csc(−θ−5)=−csc(θ+5)
- sec(−6t)=sec(6t)
- cot(9−7θ)=−cot(7θ−9)
In Exercises 7 - 21, use the Sum and Difference Identities to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well.
- cos(75∘)
- sec(165∘)
- sin(105∘)
- csc(195∘)
- cot(255∘)
- tan(375∘)
- cos(13π12)
- sin(11π12)
- tan(13π12)
- [cos7pi12] cos(7π12)
- tan(17π12)
- sin(π12)
- cot(11π12)
- csc(5π12)
- sec(−π12)
- If α is a Quadrant IV angle with cos(α)=√55, and sin(β)=√1010, where π2<β<π, find
- cos(α+β)
- sin(α+β)
- tan(α+β)
- cos(α−β)
- sin(α−β)
- tan(α−β)
- If csc(α)=3, where 0<α<π2, and β is a Quadrant II angle with tan(β)=−7, find
- cos(α+β)
- sin(α+β)
- tan(α+β)
- cos(α−β)
- sin(α−β)
- tan(α−β)
- If sin(α)=35, where 0<α<π2, and cos(β)=1213 where 3π2<β<2π, find
- sin(α+β)
- cos(α−β)
- tan(α−β)
- If sec(α)=−53, where π2<α<π, and tan(β)=247, where π<β<3π2, find
- csc(α−β)
- sec(α+β)
- cot(α+β)
In Exercises 26 - 38, verify the identity.
- cos(θ−π)=−cos(θ)
- sin(π−θ)=sin(θ)
- tan(θ+π2)=−cot(θ)
- sin(α+β)+sin(α−β)=2sin(α)cos(β)
- sin(α+β)−sin(α−β)=2cos(α)sin(β)
- cos(α+β)+cos(α−β)=2cos(α)cos(β)
- cos(α+β)−cos(α−β)=−2sin(α)sin(β)
- sin(α+β)sin(α−β)=1+cot(α)tan(β)1−cot(α)tan(β)
- cos(α+β)cos(α−β)=1−tan(α)tan(β)1+tan(α)tan(β)
- tan(α+β)tan(α−β)=sin(α)cos(α)+sin(β)cos(β)sin(α)cos(α)−sin(β)cos(β)
- sin(t+h)−sin(t)h=cos(t)(sin(h)h)+sin(t)(cos(h)−1h)
- cos(t+h)−cos(t)h=cos(t)(cos(h)−1h)−sin(t)(sin(h)h)
- tan(t+h)−tan(t)h=(tan(h)h)(sec2(t)1−tan(t)tan(h))
In Exercises 39 - 48, use the Half Angle Formulas to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well.
- cos(75∘) (compare with Exercise 7)
- sin(105∘) (compare with Exercise 9)
- cos(67.5∘)
- sin(157.5∘)
- tan(112.5∘)
- cos(7π12) (compare with Exercise 16)
- sin(π12) (compare with Exercise 18)
- cos(π8)
- sin(5π8)
- tan(7π8)
In Exercises 49 - 58, use the given information about θ to find the exact values of
- sin(2θ)
- sin(θ2)
- cos(2θ)
- cos(θ2)
- tan(2θ)
- tan(θ2)
- sin(θ)=−725 where 3π2<θ<2π
- cos(θ)=2853 where 0<θ<π2
- tan(θ)=125 where π<θ<3π2
- csc(θ)=4 where π2<θ<π
- cos(θ)=35 where 0<θ<π2
- sin(θ)=−45 where π<θ<3π2
- cos(θ)=1213 where 3π2<θ<2π
- sin(θ)=513 where π2<θ<π
- sec(θ)=√5 where 3π2<θ<2π
- tan(θ)=−2 where π2<θ<π
In Exercises 59 - 73, verify the identity. Assume all quantities are defined.
- (cos(θ)+sin(θ))2=1+sin(2θ)
- (cos(θ)−sin(θ))2=1−sin(2θ)
- tan(2θ)=11−tan(θ)−11+tan(θ)
- csc(2θ)=cot(θ)+tan(θ)2
- 8sin4(θ)=cos(4θ)−4cos(2θ)+3
- 8cos4(θ)=cos(4θ)+4cos(2θ)+3
- [sine3theta] sin(3θ)=3sin(θ)−4sin3(θ)
- sin(4θ)=4sin(θ)cos3(θ)−4sin3(θ)cos(θ)
- 32sin2(θ)cos4(θ)=2+cos(2θ)−2cos(4θ)−cos(6θ)
- 32sin4(θ)cos2(θ)=2−cos(2θ)−2cos(4θ)+cos(6θ)
- cos(4θ)=8cos4(θ)−8cos2(θ)+1
- cos(8θ)=128cos8(θ)−256cos6(θ)+160cos4(θ)−32cos2(θ)+1 (HINT: Use the result to 69.)
- sec(2θ)=cos(θ)cos(θ)+sin(θ)+sin(θ)cos(θ)−sin(θ)
- 1cos(θ)−sin(θ)+1cos(θ)+sin(θ)=2cos(θ)cos(2θ)
- 1cos(θ)−sin(θ)−1cos(θ)+sin(θ)=2sin(θ)cos(2θ)
In Exercises 74 - 79, write the given product as a sum. You may need to use an Even/Odd Identity.
- cos(3θ)cos(5θ)
- sin(2θ)sin(7θ)
- sin(9θ)cos(θ)
- cos(2θ)cos(6θ)
- sin(3θ)sin(2θ)
- cos(θ)sin(3θ)
In Exercises 80 - 85, write the given sum as a product. You may need to use an Even/Odd or Cofunction Identity.
- cos(3θ)+cos(5θ)
- sin(2θ)−sin(7θ)
- cos(5θ)−cos(6θ)
- sin(9θ)−sin(−θ)
- sin(θ)+cos(θ)
- cos(θ)−sin(θ)
- Suppose θ is a Quadrant I angle with sin(θ)=x. Verify the following formulas
- cos(θ)=√1−x2
- sin(2θ)=2x√1−x2
- cos(2θ)=1−2x2
- Discuss with your classmates how each of the formulas, if any, in Exercise 86 change if we change assume θ is a Quadrant II, III, or IV angle.
- Suppose θ is a Quadrant I angle with tan(θ)=x. Verify the following formulas
- cos(θ)=1√x2+1
- sin(θ)=x√x2+1
- sin(2θ)=2xx2+1
- cos(2θ)=1−x2x2+1
- Discuss with your classmates how each of the formulas, if any, in Exercise 88 change if we change assume θ is a Quadrant II, III, or IV angle.
- If sin(θ)=x2 for −π2<θ<π2, find an expression for cos(2θ) in terms of x.
- If tan(θ)=x7 for −π2<θ<π2, find an expression for sin(2θ) in terms of x.
- If sec(θ)=x4 for 0<θ<π2, find an expression for ln|sec(θ)+tan(θ)| in terms of x.
- Show that cos2(θ)−sin2(θ)=2cos2(θ)−1=1−2sin2(θ) for all θ.
- Let θ be a Quadrant III angle with cos(θ)=−15. Show that this is not enough information to determine the sign of sin(θ2) by first assuming 3π<θ<7π2 and then assuming π<θ<3π2 and computing sin(θ2) in both cases.
- Without using your calculator, show that √2+√32=√6+√24
- In part 4 of Example 10.4.3, we wrote cos(3θ) as a polynomial in terms of cos(θ). In Exercise 69, we had you verify an identity which expresses cos(4θ) as a polynomial in terms of cos(θ). Can you find a polynomial in terms of cos(θ) for cos(5θ)? cos(6θ)? Can you find a pattern so that cos(nθ) could be written as a polynomial in cosine for any natural number n?
- In Exercise 65, we has you verify an identity which expresses sin(3θ) as a polynomial in terms of sin(θ). Can you do the same for sin(5θ)? What about for sin(4θ)? If not, what goes wrong?
- Verify the Even / Odd Identities for tangent, secant, cosecant and cotangent.
- Verify the Cofunction Identities for tangent, secant, cosecant and cotangent.
- Verify the Difference Identities for sine and tangent.
- Verify the Product to Sum Identities.
- Verify the Sum to Product Identities.
10.4.2. Answers
- cos(75∘)=√6−√24
- sec(165∘)=−4√2+√6=√2−√6
- sin(105∘)=√6+√24
- csc(195∘)=4√2−√6=−(√2+√6)
- cot(255∘)=√3−1√3+1=2−√3
- tan(375∘)=3−√33+√3=2−√3
- cos(13π12)=−√6+√24
- sin(11π12)=√6−√24
- tan(13π12)=3−√33+√3=2−√3
- cos(7π12)=√2−√64
- tan(17π12)=2+√3
- sin(π12)=√6−√24
- cot(11π12)=−(2+√3)
- csc(5π12)=√6−√2
- sec(−π12)=√6−√2
-
- cos(α+β)=−√210
- sin(α+β)=7√210
- sin(α−β)=√22
- tan(α−β)=−1
- tan(α+β)=−7
- cos(α−β)=−√22
-
- cos(α+β)=−4+7√230
- sin(α+β)=28−√230
- sin(α−β)=−28+√230
- tan(α−β)=28+√24−7√2=−63+100√241
- tan(α+β)=−28+√24+7√2=63−100√241
- cos(α−β)=−4+7√230
-
- sin(α+β)=1665
- cos(α−β)=3365
- tan(α−β)=5633
-
- csc(α−β)=−54
- sec(α+β)=125117
- cot(α+β)=11744
- cos(75∘)=√2−√32
- sin(105∘)=√2+√32
- cos(67.5∘)=√2−√22
- sin(157.5∘)=√2−√22
- tan(112.5∘)=−√2+√22−√2=−1−√2
- cos(7π12)=−√2−√32
- sin(π12)=√2−√32
- cos(π8)=√2+√22
- sin(5π8)=√2+√22
- tan(7π8)=−√2−√22+√2=1−√2
-
- sin(2θ)=−336625
- sin(θ2)=√210
- cos(2θ)=527625
- cos(θ2)=−7√210
- tan(2θ)=−336527
- tan(θ2)=−17
-
- sin(2θ)=25202809
- sin(θ2)=5√106106
- cos(2θ)=−12412809
- cos(θ2)=9√106106
- tan(2θ)=−25201241
- tan(θ2)=59
-
- sin(2θ)=120169
- sin(θ2)=3√1313
- \cos(2\theta) = -\dfrac{119}{169}
- \cos\left(\frac{\theta}{2}\right) = -\dfrac{2\sqrt{13}}{13}
- \tan(2\theta) = -\dfrac{120}{119}
- \tan\left(\frac{\theta}{2}\right) = -\dfrac{3}{2}
-
- \sin(2\theta) = -\dfrac{\sqrt{15}}{8}
- \sin\left(\frac{\theta}{2}\right) =\dfrac{\sqrt{8+2\sqrt{15}}}{4} \\ \phantom{\tan\left(\frac{\theta}{2}\right) = 4+\sqrt{15}}
- \cos(2\theta) = \dfrac{7}{8}
- \cos\left(\frac{\theta}{2}\right) = \dfrac{\sqrt{8-2\sqrt{15}}}{4} \\ \phantom{\tan\left(\frac{\theta}{2}\right) = 4+\sqrt{15}}
- \tan(2\theta) = -\dfrac{\sqrt{15}}{7}
- \tan\left(\frac{\theta}{2}\right) = \sqrt{\dfrac{8+2\sqrt{15}}{8-2\sqrt{15}}} \\ \tan\left(\frac{\theta}{2}\right) = 4+\sqrt{15}
-
- \sin(2\theta) = \dfrac{24}{25}
- \sin\left(\frac{\theta}{2}\right) = \dfrac{\sqrt{5}}{5}
- \cos(2\theta) = -\dfrac{7}{25}
- \cos\left(\frac{\theta}{2}\right) = \dfrac{2\sqrt{5}}{5}
- \tan(2\theta)=-\dfrac{24}{7}
- \tan\left(\frac{\theta}{2}\right) = \dfrac{1}{2}
-
- \sin(2\theta) = \dfrac{24}{25}
- \sin\left(\frac{\theta}{2}\right) = \dfrac{2\sqrt{5}}{5}
- \cos(2\theta) = -\dfrac{7}{25}
- \cos\left(\frac{\theta}{2}\right) = -\dfrac{\sqrt{5}}{5}
- \tan(2\theta)=-\dfrac{24}{7}
- \tan\left(\frac{\theta}{2}\right) = -2
-
- \sin(2\theta) = -\dfrac{120}{169}
- \sin\left(\frac{\theta}{2}\right) = \dfrac{\sqrt{26}}{26}
- \cos(2\theta) = \dfrac{119}{169}
- \cos\left(\frac{\theta}{2}\right) = -\dfrac{5\sqrt{26}}{26}
- \tan(2\theta)=-\dfrac{120}{119}
- \tan\left(\frac{\theta}{2}\right) = -\dfrac{1}{5}
-
- \sin(2\theta) = -\dfrac{120}{169}
- \sin\left(\frac{\theta}{2}\right) = \dfrac{5\sqrt{26}}{26}
- \cos(2\theta) = \dfrac{119}{169}
- \cos\left(\frac{\theta}{2}\right) = \dfrac{\sqrt{26}}{26}
- \tan(2\theta)=-\dfrac{120}{119}
- \tan\left(\frac{\theta}{2}\right) = 5
-
- \sin(2\theta) = -\dfrac{4}{5}
- \sin\left(\frac{\theta}{2}\right) = \dfrac{\sqrt{50-10\sqrt{5}}}{10} \\ \phantom{\tan\left(\frac{\theta}{2}\right) =\dfrac{5-5\sqrt{5}}{10}}
- \cos(2\theta) = -\dfrac{3}{5}
- \cos\left(\frac{\theta}{2}\right)= -\dfrac{\sqrt{50+10\sqrt{5}}}{10} \\ \phantom{\tan\left(\frac{\theta}{2}\right) =\dfrac{5-5\sqrt{5}}{10}}
- \tan(2\theta)=\dfrac{4}{3}
- \tan\left(\frac{\theta}{2}\right) = -\sqrt{\dfrac{5-\sqrt{5}}{5+\sqrt{5}}} \\ \tan\left(\frac{\theta}{2}\right) =\dfrac{5-5\sqrt{5}}{10}
-
- \sin(2\theta) = -\dfrac{4}{5}
- \sin\left(\frac{\theta}{2}\right) = \dfrac{\sqrt{50+10\sqrt{5}}}{10} \\ \phantom{\tan\left(\frac{\theta}{2}\right) =\dfrac{5-5\sqrt{5}}{10}}
- \cos(2\theta) = -\dfrac{3}{5}
- \cos\left(\frac{\theta}{2}\right)= \dfrac{\sqrt{50-10\sqrt{5}}}{10} \\ \phantom{\tan\left(\frac{\theta}{2}\right) =\dfrac{5-5\sqrt{5}}{10}}
- \tan(2\theta)=\dfrac{4}{3}
- \tan\left(\frac{\theta}{2}\right) = \sqrt{\dfrac{5+\sqrt{5}}{5-\sqrt{5}}} \\ \tan\left(\frac{\theta}{2}\right) =\dfrac{5+5\sqrt{5}}{10}
- \dfrac{\cos(2\theta) + \cos(8\theta)}{2}
- \dfrac{\cos(5\theta) - \cos(9\theta)}{2}
- \dfrac{\sin(8\theta) + \sin(10\theta)}{2}
- \dfrac{\cos(4\theta) + \cos(8\theta)}{2}
- \dfrac{\cos(\theta) - \cos(5\theta)}{2}
- \dfrac{\sin(2\theta) + \sin(4\theta)}{2}
- 2\cos(4\theta)\cos(\theta)
- -2\cos \left( \dfrac{9}{2}\theta \right) \sin \left( \dfrac{5}{2}\theta \right)
- 2\sin \left( \dfrac{11}{2}\theta \right) \sin \left( \dfrac{1}{2}\theta \right)
- 2\cos(4\theta)\sin(5\theta)
- \sqrt{2}\cos \left(\theta - \dfrac{\pi}{4} \right)
- -\sqrt{2}\sin \left(\theta - \dfrac{\pi}{4} \right)
- 1 - \dfrac{x^{2}}{2}
- \dfrac{14x}{x^{2} + 49}
- \ln |x + \sqrt{x^{2} + 16}| - \ln(4)
Reference
1 As mentioned at the end of Section 10.2, 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 1.6.)
2 In the picture we’ve drawn, the triangles POQ and AOB are congruent, which is even better. However, \alpha_{0}-\beta_{0} could be 0 or it could be \pi, neither of which makes a triangle. It could also be larger than \pi, which makes a triangle, just not the one we’ve drawn. You should think about those three cases.
3 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.