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In Section FundamentalTrigonometricIdentities, we saw the utility of identities in finding the values of the circular functions of a given angle as well as 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. We observed the even and odd properties of the circular functions graphically in Sections GraphsofSineandCosine and GraphsofOtherCircularFunctions. Here, we take the time to prove these properties from first principles. We state the theorem below for reference.
We start by proving \(\cos (-\theta ) = \cos (\theta )\) and \(\sin (-\theta ) = -\sin (\theta )\).
Consider an angle \(\theta \) plotted in standard position. Let \(\theta _0\) be the angle coterminal with \(\theta \) with \(0 \leq \theta _0 < 2\pi \). (We can construct the angle \(\theta _0\) by rotating counter-clockwise from the positive \(x\)-axis to the terminal side of \(\theta \) as pictured below.) Since \(\theta \) and \(\theta _0\) are coterminal, \(\cos (\theta ) = \cos (\theta _0)\) and \(\sin (\theta ) = \sin (\theta _0)\).
We now consider the angles \(-\theta \) and \(-\theta _0\). Since \(\theta \) is coterminal with \(\theta _0\), there is some integer \(k\) so that \(\theta = \theta _0 + 2\pi \cdot k\). Hence, \(-\theta = -\theta _0 - 2\pi \cdot k = -\theta _0 + 2\pi \cdot (-k)\). Since \(k\) is an integer, so is \((-k)\), which means \(-\theta \) is coterminal with \(-\theta _0\). Therefore, \(\cos (-\theta ) = \cos (-\theta _0)\) and \(\sin (-\theta ) = \sin (-\theta _0)\).
Let \(P\) and \(Q\) denote the points on the terminal sides of \(\theta _0\) and \(-\theta _0\), respectively, which lie on the Unit Circle. By definition, the coordinates of \(P\) are \((\cos (\theta _0),\sin (\theta _0))\) and the coordinates of \(Q\) are \((\cos (-\theta _0),\sin (-\theta _0))\).
Since \(\theta _0\) and \(-\theta _0\) sweep out congruent central sectors of the Unit Circle, it follows that the points \(P\) and \(Q\) are symmetric about the \(x\)-axis. Thus, \(\cos (-\theta _0) = \cos (\theta _0)\) and \(\sin (-\theta _0) = -\sin (\theta _0)\).
Since the cosines and sines of \(\theta _0\) and \(-\theta _0\) are the same as those for \(\theta \) and \(-\theta \), respectively, we get \(\cos (-\theta ) = \cos (\theta )\) and \(\sin (-\theta ) = -\sin (\theta )\), as required.
As we saw in Section GraphsofOtherCircularFunctions, the remaining four circular functions ‘inherit’ their even/odd nature from sine and cosine courtesy of the Reciprocal and Quotient Identities, Theorem recipquotidfull.
Our next set of identities establish how the cosine function handles sums and differences of angles.
We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles \(\alpha \) and \(\beta \) to angles \(\alpha _0\) and \(\beta _0\), coterminal with \(\alpha \) and \(\beta \), respectively, each of which measure between \(0\) and \(2\pi \) radians. Since \(\alpha \) and \(\alpha _0\) are coterminal, as are \(\beta \) and \(\beta _0\), it follows that \((\alpha - \beta )\) is coterminal with \((\alpha _0 - \beta _0)\). Consider the case below where \(\alpha _0 \geq \beta _0\).
Since the angles \(POQ\) and \(AOB\) are congruent, the distance between \(P\) and \(Q\) is equal to the distance between \(A\) and \(B\). The distance formula, Equation distanceformula, yields
Squaring both sides, we expand the left hand side of this equation as
From the Pythagorean Identities, \(\cos ^2(\alpha _0) + \sin ^2(\alpha _0) = 1\) and \(\cos ^2(\beta _0) + \sin ^2(\beta _0) = 1\), so
Turning our attention to the right hand side of our equation, we find
Once again, we simplify \(\cos ^2(\alpha _0 - \beta _0) + \sin ^2(\alpha _0 - \beta _0)= 1\), so that
Putting it all together, we get \(2 - 2\cos (\alpha _0)\cos (\beta _0) - 2\sin (\alpha _0)\sin (\beta _0) = 2 - 2\cos (\alpha _0 - \beta _0)\), which simplifies to: \(\cos (\alpha _0 - \beta _0) = \cos (\alpha _0)\cos (\beta _0) + \sin (\alpha _0)\sin (\beta _0)\).
Since \(\alpha \) and \(\alpha _0\), \(\beta \) and \(\beta _0\), and \((\alpha - \beta )\) and \((\alpha _0- \beta _0)\) are all coterminal pairs of angles, we have established the identity: \(\cos (\alpha - \beta ) = \cos (\alpha ) \cos (\beta ) + \sin (\alpha ) \sin (\beta )\).
For the case where \(\alpha _0 \leq \beta _0\), we can apply the above argument to the angle \(\beta _0 - \alpha _0\) to obtain the identity \(\cos (\beta _0 - \alpha _0) = \cos (\beta _0)\cos (\alpha _0) + \sin (\beta _0)\sin (\alpha _0)\). Using this formula in conjunction with the Even Identity of cosine gives us the result in this case, too:
To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities
We put these newfound identities to good use in the following example.
Solution.
In order to use Theorem cosinesumdifference to find \(\cos \left (15^{\circ }\right )\), we need to write \(15^{\circ }\) as a sum or difference of angles whose cosines and sines we know. One way to do so is to write \(15^{\circ } = 45^{\circ } - 30^{\circ }\). We find:
Using Theorem cosinesumdifference gives:
Per Theorem cosinesumdifference, we know \(\cos (\alpha + \beta ) = \cos (\alpha ) \cos (\beta ) - \sin (\alpha ) \sin (\beta )\). Hence, we need to find the sines and cosines of \(\alpha \) and \(\beta \) to complete the problem.
We are given \(\sin (\alpha ) = \frac {3}{5}\), so our first task is to find \(\cos (\alpha )\). We can quickly get \(\cos (\alpha )\) using the Pythagorean Identity \(\cos ^{2}(\alpha ) = 1 - \sin ^{2}(\alpha ) = 1 - \left (\frac {3}{5}\right )^2 = \frac {16}{25}\). We get \(\cos (\alpha ) = \frac {4}{5}\), choosing the positive root since \(\alpha \) is a Quadrant I angle.
Next, we need the \(\sin (\beta )\) and \(\cos (\beta )\). Since \(\sec (\beta ) = 4\), we immediately get \(\cos (\beta ) = \frac {1}{4}\) courtesy of the Reciprocal and Quotient Identities.
To get \(\sin (\beta )\), we employ the Pythagorean Identity: \(\sin ^{2}(\beta ) = 1 - \cos ^{2}(\beta ) = 1 - \left (\frac {1}{4} \right )^2 = \frac {15}{16}\). Here, since \(\beta \) is a Quadrant IV angle, we get \(\sin (\beta ) = - \frac {\sqrt {15}}{4}\).
Finally, we get: \(\cos (\alpha + \beta ) = \cos (\alpha ) \cos (\beta ) - \sin (\alpha ) \sin (\beta ) = \left ( \frac {4}{5} \right ) \left ( \frac {1}{4} \right ) - \left ( \frac {3}{5} \right ) \left ( - \frac {\sqrt {15}}{4} \right ) = \frac {4+3\sqrt {15}}{20}\). □
The identity verified in Example cosinesumdiffex, namely, \(\cos \left (\frac {\pi }{2} - \theta \right ) = \sin (\theta )\), is the first of the celebrated ‘cofunction’ identities. These identities were first hinted at in Exercise cofunctionforeshadowing in Section AppRightTrig.
From \( \sin (\theta ) = \cos \left (\frac {\pi }{2} - \theta \right ) \), we get: \(\sin \left (\frac {\pi }{2} - \theta \right ) = \cos \left (\frac {\pi }{2} -\left [\frac {\pi }{2} - \theta \right ]\right ) = \cos (\theta )\), 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.
The Cofunction Identities enable us to derive the sum and difference formulas for sine. We first convert to sine to cosine and expand:
We can derive the difference formula for sine by rewriting \(\sin (\alpha - \beta )\) as \(\sin (\alpha + (-\beta ))\) and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader.
We try out these new identities in the next example.
Solution.
As in Example cosinesumdiffex, we need to write the angle \(\frac {19 \pi }{12}\) 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 \(\; \frac {19 \pi }{12} = \frac {4 \pi }{3} + \frac {\pi }{4}\). Applying Theorem sinesumdifference, we get
In order to find \(\sin (\alpha - \beta )\) using Theorem sinesumdifference, we need to find \(\cos (\alpha )\) and both \(\cos (\beta )\) and \(\sin (\beta )\).
To find \(\cos (\alpha )\), we use the Pythagorean Identity \(\cos ^2(\alpha ) = 1 - \sin ^{2}(\alpha ) = 1 - \left (\frac {5}{13}\right )^2 = \frac {144}{169}\). We get \(\cos (\alpha ) = -\frac {12}{13}\), the negative, here, owing to the fact that \(\alpha \) is a Quadrant II angle.
We now set about finding \(\sin (\beta )\) and \(\cos (\beta )\). We have several ways to proceed at this point, but since there isn’t a direct way to get from \(\tan (\beta ) = 2\) to either \(\sin (\beta )\) or \(\cos (\beta )\), we opt for a more geometric approach as presented in Section TheOtherCircularFunctions.
Since \(\beta \) is a Quadrant III angle with \(\tan (\beta ) = 2 = \frac {-2}{-1}\), we know the point \(Q(x,y) = (-1,-2)\) is on the terminal side of \(\beta \) as illustrated below.
We find \(r = \sqrt {x^2 + y^2} = \sqrt {(-1)^2+(-2)^2} = \sqrt {5}\), so per Theorem circularfunctionscircle, \(\sin (\beta ) = \frac {-2}{\sqrt {5}} = - \frac {2\sqrt {5}}{5}\) and \(\cos (\beta ) = \frac {-1}{\sqrt {5}} = - \frac {\sqrt {5}}{5}\) .
At last, we have \(\sin (\alpha - \beta ) = \sin (\alpha )\cos (\beta ) - \cos (\alpha )\sin (\beta ) = \left ( \frac {5}{13} \right )\left ( -\frac {\sqrt {5}}{5} \right ) - \left ( -\frac {12}{13} \right )\left ( - \frac {2 \sqrt {5}}{5} \right ) = -\frac {29\sqrt {5}}{65}\).
We can start expanding \(\tan (\alpha + \beta )\) using a quotient identity and our sum formulas
Since \(\tan (\alpha ) = \frac {\sin (\alpha )}{\cos (\alpha )}\) and \(\tan (\beta ) = \frac {\sin (\beta )}{\cos (\beta )}\), it looks as though if we divide both numerator and denominator by \(\cos (\alpha ) \cos (\beta )\) we will have what we want
Naturally, this formula is limited to those cases where all of the tangents are defined. □
The formula developed in Exercise sinesumanddiffex for \(\tan (\alpha + \beta )\) can be used to find a formula for \(\tan (\alpha - \beta )\) by rewriting the difference as a sum, \(\tan (\alpha + (-\beta ))\) and using the odd property of tangent. (The reader is encouraged to fill in the details.) Below we summarize all of the sum and difference formulas.
In the statement of Theorem circularsumdifference, 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,
If we set \(\alpha = \beta \) in the sum formulas in Theorem circularsumdifference, we obtain the following ‘Double Angle’ Identities:
The three different forms for \(\cos (2\theta )\) can be explained by our ability to ‘exchange’ squares of cosine and sine via the Pythagorean Identity. For instance, if we substitute \(\sin ^{2}(\theta ) = 1 - \cos ^{2}(\theta )\) into the first formula for \(\cos (2\theta )\), we get \(\cos (2\theta ) = \cos ^{2}(\theta ) - \sin ^{2}(\theta ) = \cos ^{2}(\theta ) - (1 - \cos ^{2}(\theta )) = 2 \cos ^{2}(\theta ) - 1\).
It is interesting to note that to determine the value of \(\cos (2\theta )\), only one piece of information is required: either \(\cos (\theta )\) or \(\sin (\theta )\). To determine \(\sin (2\theta )\), however, it appears that we must know both \(\sin (\theta )\) and \(\cos (\theta )\). In the next example, we show how we can find \(\sin (2\theta )\) knowing just one piece of information, namely \(\tan (\theta )\).
Suppose \(P(-3,4)\) lies on the terminal side of \(\theta \) when \(\theta \) is plotted in standard position.
Find \(\cos (2\theta )\) and \(\sin (2\theta )\) and determine the quadrant in which the terminal side of the angle \(2\theta \) lies when it is plotted in standard position.
Solution.
We sketch the terminal side of \(\theta \) below on the left. Using Theorem cosinesinecircle from Section TheCircularFunctionsSineandCosine with \(x = -3\) and \(y=4\), we find \(r = \sqrt {x^2+y^2} = 5\). Hence, \(\cos (\theta ) = -\frac {3}{5}\) and \(\sin (\theta ) = \frac {4}{5}\).
Theorem doubleangle gives us three different formulas to choose from to find \(\cos (2\theta )\). Using the first formula, we get: \(\cos (2\theta ) = \cos ^{2}(\theta ) - \sin ^{2}(\theta ) = \left (-\frac {3}{5}\right )^2 - \left (\frac {4}{5}\right )^2 = -\frac {7}{25}\). For \(\sin (2\theta )\), we get \(\sin (2\theta ) = 2 \sin (\theta ) \cos (\theta ) = 2 \left (\frac {4}{5}\right )\left (-\frac {3}{5}\right ) = -\frac {24}{25}\).
Since both cosine and sine of \(2\theta \) are negative, the terminal side of \(2\theta \), when plotted in standard position, lies in Quadrant III. To see this more clearly, we plot the terminal side of \(2\theta \), along with the terminal side of \(\theta \) below on the right.
Note that in order to find the point \(Q(x,y)\) on the terminal side of \(2\theta \) of a circle of radius \(5\), we use Theorem cosinesinecircle again and find \(x = r \cos (2\theta ) = 5 \left (-\frac {7}{25} \right ) = -\frac {7}{5}\) and \(y = r \sin (2\theta ) = 5 \left (-\frac {24}{25}\right ) = -\frac {24}{5}\).
If your first reaction to ‘\(\sin (\theta ) = x\)’ is ‘No it’s not, \(\cos (\theta ) = x\)!’ then you have indeed learned something, and we take comfort in that.
While we have mostly used ‘\(x\)’ to represent the \(x\)-coordinate of the point the terminal side of an angle \(\theta \), here, ‘\(x\)’ represents the quantity \(\sin (\theta )\) and our task is to express \(\sin (2\theta )\) in terms of \(x\).
Since \(\sin (2\theta ) = 2 \sin (\theta ) \cos (\theta ) = 2 x \cos (\theta )\), what remains is to express \(\cos (\theta )\) in terms of \(x\).
Substituting \(\sin (\theta ) = x\) into the Pythagorean Identity, we get \(\cos ^{2}(\theta ) = 1- \sin ^{2}(\theta ) = 1 - x^2\), or \(\cos (\theta ) = \pm \sqrt {1-x^2}\). Since \(-\frac {\pi }{2} \leq \theta \leq \frac {\pi }{2}\), \(\cos (\theta ) \geq 0\), and thus \(\cos (\theta ) = \sqrt {1-x^2}\).
Our final answer is \(\sin (2\theta ) = 2 \sin (\theta ) \cos (\theta ) = 2x\sqrt {1-x^2}\).
We start with the right hand side of the identity and note that \(1 + \tan ^{2}(\theta ) = \sec ^{2}(\theta )\). Next, we use the Reciprocal and Quotient Identities to rewrite \(\tan (\theta )\) and \(\sec (\theta )\) in terms of \(\sin (\theta )\) and \(\cos (\theta )\):
In Theorem doubleangle, one of the formulas for \(\cos (2\theta )\), namely \(\cos (2\theta ) = 2\cos ^{2}(\theta ) - 1\), expresses \(\cos (2\theta )\) as a polynomial in terms of \(\cos (\theta )\). We are now asked to find such an identity for \(\cos (3\theta )\).
Using the sum formula for cosine, we begin with
Our ultimate goal is to express the right hand side in terms of \(\cos (\theta )\) only. To that end, we substitute \(\cos (2\theta ) = 2\cos ^{2}(\theta ) -1\) and \(\sin (2\theta ) = 2\sin (\theta )\cos (\theta )\) which yields:
Finally, we exchange \(\sin ^{2}(\theta ) = 1 - \cos ^{2}(\theta )\) courtesy of the Pythagorean Identity, and get
Hence, \(\cos (3\theta ) = 4\cos ^{3}(\theta )- 3\cos (\theta )\). □
In the last problem in Example doubleangleex, we saw how we could rewrite \(\cos (3\theta )\) as sums of powers of \(\cos (\theta )\). In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine.
Solving the identity \(\cos (2\theta ) = 2\cos ^{2}(\theta ) -1\) for \(\cos ^{2}(\theta )\) and the identity \(\cos (2\theta ) = 1 - 2\sin ^{2}(\theta )\) for \(\sin ^{2}(\theta )\) results in the aptly-named ‘Power Reduction’ formulas below.
Our next example is a typical application of Theorem powerreduction that you’ll likely see in Calculus.
Solution. We begin with a straightforward application of Theorem powerreduction
Next, we apply the power reduction formula to \(\cos ^{2}(2\theta )\) to finish the reduction
□
Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to \(\cos ^{2}\left (\frac {\theta }{2}\right )\)
We can obtain a formula for \(\cos \left (\frac {\theta }{2}\right )\) 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.
where the choice of \(\pm \) depends on the quadrant in which the terminal side of \( \frac {\theta }{2}\) lies.
Use the identity given in number doubleanglesinewtan of Example doubleangleex to derive the identity
Solution.
To use the half angle formula, we note that \(15^{\circ } = \frac {30^{\circ }}{2}\) and since \(15^{\circ }\) is a Quadrant I angle, its cosine is positive. Thus we have
Back in Example cosinesumdiffex, we found \(\cos \left (15^{\circ }\right ) = \frac {\sqrt {6}+ \sqrt {2}}{4}\) by using the difference formula for cosine. The reader is encouraged to prove that these two expressions are equal algebraically.
If \(-\pi \leq t \leq 0\), then \(-\frac {\pi }{2} \leq \frac {t}{2} \leq 0\), which means \( \frac {t}{2}\) corresponds to a Quadrant IV angle. Hence, \(\sin \left (\frac {t}{2}\right ) < 0\), so we choose the negative root formula from Theorem halfangle:
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 doubleanglesinewtan of Example doubleangleex and manipulate it into the identity we are asked to prove.
The identity we are asked to start with is \(\; \sin (2\theta ) = \frac {2\tan (\theta )}{1 + \tan ^{2}(\theta )}\). If we are to use this to derive an identity for \(\tan \left (\frac {\theta }{2}\right )\), it seems reasonable to proceed by replacing each occurrence of \(\theta \) with \(\frac {\theta }{2}\)
We now have the \(\sin (\theta )\) we need, but we somehow need to get a factor of \(1+\cos (\theta )\) involved. We substitute \(1 + \tan ^{2}\left (\frac {\theta }{2}\right ) = \sec ^{2}\left (\frac {\theta }{2}\right )\), and continue to manipulate our given identity by converting secants to cosines.
Finally, we apply a power reduction formula, and then solve for \(\tan \left ( \frac {\theta }{2} \right )\)
□
Our next batch of identities, the Product to Sum Formulas, are easily verified by expanding each of the right hand sides in accordance with Theorem circularsumdifference 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.
Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section TrigonometricEquationsandInequalities. These are essentially restatements of the Product to Sum Formulas (by re-labeling the arguments of the sine and cosine functions) and as such, their proofs are left as exercises.
Solution.
Identifying \(\alpha = 2\theta \) and \(\beta = 6\theta \), we find
Identifying \(\alpha = \theta \) and \(\beta = 3\theta \) yields
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.
We first studied sinusoids in Section ??. Using the sum formulas for sine and cosine, we can expand the forms given to us in Theorem ??:
and
As we’ll see in the next example, recognizing these ‘expanded’ forms of sinusoids allows us to graph functions as sinusoids which, at first glance, don’t appear to fit the forms of either \(C(t)\) or \(S(t)\).
Check your answers analytically using identities and using a graphing utility.
Solution.
The key to this problem is to use the expanded forms of the sinusoid formulas and match up corresponding coefficients. We start by equating \(f(t) = \cos (2t) - \sqrt {3} \sin (2t)\) with the expanded form of \(C(t) = A \cos (\omega t + \phi ) + B\): \(\cos (2t) - \sqrt {3} \sin (2t) = A\cos (\omega t) \cos (\phi ) - A \sin (\omega t) \sin (\phi ) + B\).
If we take \(\omega = 2\) and \(B = 0\), we get: \(\cos (2t) - \sqrt {3} \sin (2t) = A\cos (2t) \cos (\phi ) - A \sin (2t)\sin (\phi )\).
To determine \(A\) and \(\phi \), a bit more work is involved. We get started by equating the coefficients of the trigonometric functions on either side of the equation.
On the left hand side, the coefficient of \(\cos (2t)\) is \(1\), while on the right hand side, it is \(A \cos (\phi )\). Since this equation is to hold for all real numbers, we must have that \(A \cos (\phi ) = 1\).
Similarly, we find by equating the coefficients of \(\sin (2t)\) that \(A \sin (\phi ) = \sqrt {3}\). In conjunction with \(A \cos (\phi ) = 1\), we have a system of two (nonlinear) equations and two unknowns.
As usual, our first task is to reduce this system of two equations and two unknowns to one equation and one unknown. We can temporarily eliminate the dependence on \(\phi \) by using a Pythagorean Identity. From \(\cos ^{2}(\phi ) + \sin ^{2}(\phi ) = 1\), we multiply through by \(A^2\) to get \(A^2\cos ^{2}(\phi ) + A^2\sin ^{2}(\phi ) = A^2\).
In our case, \(A \cos (\phi ) = 1\) and \(A \sin (\phi ) = \sqrt {3}\), hence \(A^2 = A^2\cos ^{2}(\phi ) + A^2\sin ^{2}(\phi ) = 1^2 + (\sqrt {3})^2 = 4\) so \(A = \pm 2\). In much the same way we fit a sinusoid to a graph in Example ??, we choose \(A = 2\), and then find the phase angle \(\phi \) associated with this choice.
Substituting \(A=2\) into our two equations, \(A \cos (\phi ) = 1\) and \(A \sin (\phi ) = \sqrt {3}\), we get \(2\cos (\phi ) = 1\) and \(2 \sin (\phi ) = \sqrt {3}\). After some rearrangement, \(\cos (\phi ) = \frac {1}{2}\) and \(\sin (\phi ) = \frac {\sqrt {3}}{2}\). One such angle \(\phi \) which satisfies this criteria is \(\phi = \frac {\pi }{3}\).
Hence, one way to write \(f(t)\) as a sinusoid is \(f(t) = 2 \cos \left (2t + \frac {\pi }{3}\right )\). We can check our answer using the sum formula for cosine :
Proceeding as before, we equate \(f(t) = \cos (2t) - \sqrt {3} \sin (2t)\) with the expanded form of of the sinusoid \(S(t) = A \sin (\omega t + \phi ) + B\) to get: \(\cos (2t) - \sqrt {3} \sin (2t) = A\sin (\omega t) \cos (\phi ) + A \cos (\omega t)\sin (\phi ) + B\).
Taking \(\omega = 2\) and \(B = 0\), we get \(\cos (2t) - \sqrt {3} \sin (2t) = A\sin (2t) \cos (\phi ) + A \cos (2t)\sin (\phi )\). We equate the coefficients of \(\cos (2t)\) on either side and get \(A\sin (\phi ) = 1\) and \(A\cos (\phi ) = -\sqrt {3}\).
Using \(A^2\cos ^{2}(\phi ) + A^2\sin ^{2}(\phi ) = A^2\) as before, we get \(A = \pm 2\), and again we choose \(A = 2\).
This means \(2 \sin (\phi ) = 1\), or \(\sin (\phi ) = \frac {1}{2}\), and \(2\cos (\phi ) = -\sqrt {3}\), so \(\cos (\phi ) = -\frac {\sqrt {3}}{2}\). One such angle which meets these criteria is \(\phi = \frac {5\pi }{6}\).
Hence, we have \(f(t) = 2 \sin \left (2t + \frac {5\pi }{6}\right )\). Checking our work analytically, we have
A couple of remarks about Example 7 are in order. First, had we chosen \(A = -2\) instead of \(A = 2\) as we worked through Example 7, our final answers would have looked different. The reader is encouraged to rework Example 7 using \(A = -2\) to see what these differences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent.
It is important to note that in order for the technique presented in Example 7 to fit a function into one of the forms in Theorem ??, the frequencies of the sine and cosine terms much match. For example, in the Exercises, you’ll be asked to write \(f(t) = 3\sqrt {3}\sin (3t) - 3\cos (3t)\) in the form of \(S(t)\) and \(C(t)\) above, and since both the sine and cosine terms have frequency \(3\), this is possible.
However, a function such as \(f(t) = \sin (t) - \sin (3t)\) cannot be written in the form of \(S(t)\) or \(C(t)\). The quickest way to see this is to examine its graph below which is decidedly not a sinusoid. That being said, we can still analyze this curve using identities.
Using our result from number 2 Example 6, we may rewrite \(f(t) = \sin (t) - \sin (3t) = -2 \sin (t) \cos (2t)\). Grouping factors, we can view \(f(t) = [ -2 \sin (t) ] \cos (2t) = A(t) \cos (2t)\) as the curve \(y = \cos (2t)\) with a variable amplitude, \(A(t) = -2 \sin (t)\).
Overlaying the graphs of \(f(t)\) with the (dashed) graphs of \(A_{1}(t) = 2 \sin (t)\) and \(A_{2}(t) = -2 \sin (t)\), we can see the role these two curves play in the graph of \(y = f(t)\). They create a kind of ‘wave envelope’ for the graph of \(y = f(t)\). This is an example of the beats phenomenon. Note that when written as a product of sinusoids, it is always the lower frequency factor which creates the ‘wave-envelope’ of the curve.
Note that in order to rewrite a sum or difference of sine and cosine functions with different frequencies into a product using the sum to product identities, Theorem 10, we need the amplitudes of each term to be the same. We explore more examples of these functions and this behavior in the Exercises.