He will respond to the prayer of the destitute; he will not despise their plea. Psalms 102:17 NIV
5-07 Product-to-Sum Formulas
VIDEO
Mr. Wright teaches the lesson.
Summary: In this section, you will:
Use product-to-sum formulas to evaluate trigonometric functions.
Use product-to-sum formulas to derive new trigonometric identities.
Use product-to-sum formulas to solve trigonometric equations.
SDA NAD Content Standards (2018): PC.5.1
Figure 1: Trumpets. credit (wikimedia/7trumpetsmusicband)
Adding some trigonometric functions can have the same result as multiplying other trigonometric functions. This can be used to analyze complex sound waves that are built by the addition of several sine sound waves.
Sometimes it is useful to rewrite addition of trigonometric functions as multiplying. This could be useful when solving equations so the zero product property can be used.
Product-to-Sum Formulas
Product-to-Sum Formulas
\(\sin u \sin v = \frac{1}{2}\left(\cos\left(u - v\right) - \cos\left(u + v\right)\right)\)
\(\cos u \cos v = \frac{1}{2}\left(\cos\left(u - v\right) + \cos\left(u + v\right)\right)\)
\(\sin u \cos v = \frac{1}{2}\left(\sin\left(u + v\right) + \sin\left(u - v\right)\right)\)
\(\cos u \sin v = \frac{1}{2}\left(\sin\left(u + v\right) - \sin\left(u - v\right)\right)\)
where typically u > v .
Example 1: Rewrite an Expression as a Sum or Difference
Rewrite cos 3α cos 2α as a sum or difference.
Solution
This is only cosines, so use the second formula with u = 3α and v = 2α.
$$\cos u \cos v = \frac{1}{2}\left(\cos\left(u - v\right) + \cos\left(u + v\right)\right)$$
$$\cos 3α \cos 2α = \frac{1}{2}\left(\cos\left(3α - 2α\right) + \cos\left(3α + 2α\right)\right)$$
$$= \frac{1}{2}\left(\cos α + \cos 5α\right)$$
Try It 1
Rewrite sin 4θ cos θ as a sum or difference.
Answer
\(\frac{1}{2}\left(\sin 5θ + \sin 3θ\right)\)
Sum-to-Product Formulas
Sum-to-Product Formulas
\(\sin u + \sin v = 2\sin\left(\frac{u + v}{2}\right)\cos\left(\frac{u - v}{2}\right)\)
\(\sin u - \sin v = 2\cos\left(\frac{u + v}{2}\right)\sin\left(\frac{u - v}{2}\right)\)
\(\cos u + \cos v = 2\cos\left(\frac{u + v}{2}\right)\cos\left(\frac{u - v}{2}\right)\)
\(\cos u - \cos v = -2\sin\left(\frac{u + v}{2}\right)\sin\left(\frac{u - v}{2}\right)\)
Example 2: Evaluate a Trigonometric Expression
Find the exact value of cos 75° − cos 15°.
Solution
This is a difference of cosines, so use the last formula with u = 75° and v = 15°.
$$\cos u - \cos v = -2\sin\left(\frac{u + v}{2}\right)\sin\left(\frac{u - v}{2}\right)$$
$$\cos 75° - \cos 15° = -2\sin\left(\frac{75° + 15°}{2}\right)\sin\left(\frac{75° - 15°}{2}\right)$$
$$= - \sin 45° \sin 30°$$
$$= -2\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$$
$$= \frac{\sqrt{2}}{2}$$
Try It 2
Find the exact value of sin 255° + sin 15°.
Answer
\(-\frac{\sqrt{2}}{2}\)
Example 3: Solve a Trigonometric Equation
Solve on the interval [0, 2π): sin 5x + sin x = 0.
Solution
Since the equation equals zero, the zero product property could be used if the equation were a product and not a sum. So, use a sum-to-product formula with u = 5x and v = x .
$$\sin 5x + \sin x = 0$$
$$2 \cos\left(\frac{5x + x}{2}\right)\sin\left(\frac{5x - x}{2}\right) = 0$$
$$2\cos 3x \sin 2x = 0$$
\(\cos 3x = 0\)
\(\sin 2x = 0\)
\(3x = \frac{π}{2} + πn\)
\(2x = 0 + πn\)
\(x = \frac{π}{6} + \frac{πn}{3}\)
\(x = \frac{πn}{2}\)
$$x = 0, \frac{π}{6}, \frac{π}{2}, \frac{5π}{6}, π, \frac{7π}{6}, \frac{3π}{2}, \frac{11π}{6}$$
Try It 3
Solve on the interval [0, 2π): cos 3x + cos x = 0.
Answer
\(\frac{π}{4}\), \(\frac{π}{2}\), \(\frac{3π}{4}\), \(\frac{5π}{4}\), \(\frac{3π}{2}\), \(\frac{7π}{4}\)
Example 4: Verify a Trigonometric Identity
Verify \(\frac{\sin 3x + \sin x}{\cos 3x − \cos x} = −\cot x\)
Solution
If the fraction used terms that were multiplied together instead of added together, then they might cancel out. So, use a sum-to-product formula for the numerator and denominator with u = 3x and v = x .
$$\frac{\sin 3x + \sin x}{\cos 3x − \cos x} = −\cot x$$
$$\frac{2 \sin\left(\frac{3x + x}{2}\right)\cos\left(\frac{3x - x}{2}\right)}{−2\sin\left(\frac{3x + x}{2}\right)\sin\left(\frac{3x - x}{2}\right)} = −\cot x$$
$$\frac{2 \sin 2x \cos x}{−2 \sin 2x \sin x} = −\cot x$$
$$−\frac{\cos x}{\sin x} = −\cot x$$
$$−\cot x = −\cot x$$
Try It 4
Verify \(\frac{\cos 6x + \cos 2x}{\cos 6x - \cos 2x} = -\cot 4x \cot 2x\).
Answer
\(\frac{\cos 6x + \cos 2x}{\cos 6x - \cos 2x}\) \(= \frac{2\cos\left(\frac{6x + 2x}{2}\right)\cos\left(\frac{6x - 2x}{2}\right)}{-2\sin\left(\frac{6x+2x}{2}\right)\sin\left(\frac{6x - 2x}{2}\right)}\) \(= \frac{2\cos 4x \cos 2x}{-2\sin 4x \sin 2x}\) \(= -\cot 4x \cot 2x\)
Product-to-Sum Formulas
\(\sin u \sin v = \frac{1}{2}\left(\cos\left(u - v\right) - \cos\left(u + v\right)\right)\)
\(\cos u \cos v = \frac{1}{2}\left(\cos\left(u - v\right) + \cos\left(u + v\right)\right)\)
\(\sin u \cos v = \frac{1}{2}\left(\sin\left(u + v\right) + \sin\left(u - v\right)\right)\)
\(\cos u \sin v = \frac{1}{2}\left(\sin\left(u + v\right) - \sin\left(u - v\right)\right)\)
where typically u > v .
Sum-to-Product Formulas
\(\sin u + \sin v = 2\sin\left(\frac{u + v}{2}\right)\cos\left(\frac{u - v}{2}\right)\)
\(\sin u - \sin v = 2\cos\left(\frac{u + v}{2}\right)\sin\left(\frac{u - v}{2}\right)\)
\(\cos u + \cos v = 2\cos\left(\frac{u + v}{2}\right)\cos\left(\frac{u - v}{2}\right)\)
\(\cos u - \cos v = -2\sin\left(\frac{u + v}{2}\right)\sin\left(\frac{u - v}{2}\right)\)
Helpful videos about this lesson.
Practice Exercises
Rewrite the expression as a sum or difference.
sin 7α cos 2α
cos 7α sin 2α
sin 2x sin x
cos 3x cos 2x
Find the exact value of the expression.
sin 105° + sin 15°
cos 165° − cos 105°
sin 105° cos 15°
cos 285° cos 75°
Solve the equation on the interval [0, 2π).
cos 3x − cos 2x = 0
sin 4x = sin x
sin 2x cos x = 0
cos 2x + cos x = 0
Verify the identity.
\(\frac{\sin 2x + \sin x}{\cos 2x + \cos x} = \tan \frac{3x}{2}\)
\(\frac{\sin 3x \cos 2x}{\cos 3x \sin 2x} = \frac{\sin 5x + \sin x}{\sin 5x - \sin x}\)
\(\frac{\cos 4x \cos x}{\sin 4x \sin x} = \frac{\cos 3x + \cos 5x}{\cos 3x - \cos 5x}\)
Mixed Review
(5-06) Rewrite the expression as a sum of 1st powers of cosine: sin2 x cos2 x .
(5-06) If \(\tan θ = \frac{3}{4}\), find (a) \(\sin \frac{θ}{2}\), (b) \(\cos \frac{θ}{2}\), (c) \(\tan \frac{θ}{2}\).
(5-05) Verify tan(x + π) cot x = 1.
(5-03) Verify \(\frac{1}{\sec x \csc x} = \tan x - \sin^2 x \tan x\).
(5-02) Simplify \(\frac{\sin x}{1 - \cos x} - \frac{\cos x}{\sin x}\).
Answers
\(\frac{1}{2}\left(\sin 9α + \sin 5α\right)\)
\(\frac{1}{2}\left(\sin 9α - \sin 5α\right)\)
\(\frac{1}{2}\left(\cos x - \cos 3x\right)\)
\(\frac{1}{2}\left(\cos x + \cos 5x\right)\)
\(\frac{\sqrt{6}}{2}\)
\(-\frac{\sqrt{2}}{2}\)
\(\frac{\sqrt{3} + 2}{4}\)
\(\frac{-\sqrt{3} + 2}{4}\)
0, \(\frac{2π}{5}\), \(\frac{4π}{5}\), \(\frac{6π}{5}\), \(\frac{8π}{5}\)
0, \(\frac{π}{5}\), \(\frac{3π}{5}\), \(\frac{2π}{3}\), π, \(\frac{4π}{3}\), \(\frac{7π}{5}\), \(\frac{9π}{5}\)
0, \(\frac{π}{2}\), π, \(\frac{3π}{2}\)
\(\frac{π}{3}\), π, \(\frac{5π}{3}\)
\(\frac{\sin 2x - \sin x}{\cos 2x + \cos x}\) \(= \frac{2 \sin\left(\frac{2x + x}{2}\right)\cos\left(\frac{2x - x}{2}\right)}{2 \cos\left(\frac{2x + x}{2}\right) \cos\left(\frac{2x - x}{2}\right)}\) \(= \frac{\sin \frac{3x}{2}}{\cos{3x}{2}}\) \(= \tan \frac{3x}{2}\)
\(\frac{\sin 3x \cos 2x}{\cos 3x \sin 2x}\) \(= \frac{\frac{1}{2}\left(\sin\left(3x + 2x\right) + \sin\left(3x - 2x\right)\right)}{\frac{1}{2}\left(\sin\left(3x + 2x\right) - \sin\left(3x - 2x\right)\right)}\) \(= \frac{\sin 5x + \sin x}{\sin 5x - \sin x}\)
\(\frac{\cos 4x \cos x}{\sin 4x \sin x}\) \(= \frac{\frac{1}{2}\left(\cos\left(4x - x\right) + \cos\left(4x + x\right)\right)}{\frac{1}{2}\left(\cos\left(4x - x\right) - \cos\left(4x + x\right)\right)}\) \(= \frac{\cos 3x + \cos 5x}{\cos 3x - \cos 5x}\)
\(\frac{1-\cos 4x}{8}\)
(a) \(\frac{\sqrt{10}}{10}\), (b) \(\frac{3\sqrt{10}}{10}\), (c) \(\frac{1}{3}\)
\(\tan\left(x + π\right) \cot x\) \(= \left(\frac{\tan x + \tan π}{1 - \tan x \tan π}\right)\cot x\) \(= \left(\frac{\tan x + 0}{1 - \left(\tan x\right)\left(0\right)}\right)\cot x\) \(= \tan x \cot x = 1\)
\(\frac{1}{\sec x \csc x}\) \(= \frac{\sec x}{\sec^2 x \csc x}\) \(= \cos^2 x \frac{\sin x}{\cos x}\) \(= \left(1 - \sin^2 x\right) \tan x\) \(= \tan x - \sin^2 x \tan x\)
csc x
