Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Certainly! Let's solve the equation step by step.
Given the equation:
[tex]\[ (\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \][/tex]
We can start by using trigonometric identities. Recall the product-to-sum identities:
[tex]\[ \cos A \cos B = \frac{1}{2} [\cos (A + B) + \cos (A - B)] \][/tex]
and
[tex]\[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \][/tex]
Using [tex]\(A = 3\theta\)[/tex] and [tex]\(B = \theta\)[/tex], we have:
[tex]\[ \cos 3\theta \cos \theta = \frac{1}{2} [\cos (3\theta + \theta) + \cos (3\theta - \theta)] = \frac{1}{2} [\cos 4\theta + \cos 2\theta] \][/tex]
and
[tex]\[ \sin 3\theta \sin \theta = \frac{1}{2} [\cos (3\theta - \theta) - \cos (3\theta + \theta)] = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Substitute these into the given equation:
[tex]\[ \frac{1}{2} [\cos 4\theta + \cos 2\theta] + 1 = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Multiply through by 2 to clear the fractions:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Rearrange terms to combine like terms:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Combine the [tex]\(\cos 4\theta\)[/tex] terms:
[tex]\[ 2\cos 4\theta + 2 = 0 \][/tex]
Subtract 2 from both sides:
[tex]\[ 2\cos 4\theta = -2 \][/tex]
Divide by 2:
[tex]\[ \cos 4\theta = -1 \][/tex]
The general solution to [tex]\(\cos (x) = -1\)[/tex] is:
[tex]\[ x = \pi + 2k\pi \quad \text{for integer } k \][/tex]
Set [tex]\(x = 4\theta\)[/tex]:
[tex]\[ 4\theta = \pi + 2k\pi \][/tex]
Solve for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{k\pi}{2} \][/tex]
Now, let's find the specific values of [tex]\(\theta\)[/tex] that satisfy the original equation within one full cycle (0 to [tex]\(2\pi\)[/tex]). We consider the first few values of [tex]\(k\)[/tex]:
For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} \][/tex]
For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \][/tex]
For [tex]\(k = -1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4} \][/tex]
For [tex]\(k = -2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \pi = -\frac{3\pi}{4} \][/tex]
Thus, the solutions to the equation [tex]\((\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \)[/tex] are:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
Given the equation:
[tex]\[ (\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \][/tex]
We can start by using trigonometric identities. Recall the product-to-sum identities:
[tex]\[ \cos A \cos B = \frac{1}{2} [\cos (A + B) + \cos (A - B)] \][/tex]
and
[tex]\[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \][/tex]
Using [tex]\(A = 3\theta\)[/tex] and [tex]\(B = \theta\)[/tex], we have:
[tex]\[ \cos 3\theta \cos \theta = \frac{1}{2} [\cos (3\theta + \theta) + \cos (3\theta - \theta)] = \frac{1}{2} [\cos 4\theta + \cos 2\theta] \][/tex]
and
[tex]\[ \sin 3\theta \sin \theta = \frac{1}{2} [\cos (3\theta - \theta) - \cos (3\theta + \theta)] = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Substitute these into the given equation:
[tex]\[ \frac{1}{2} [\cos 4\theta + \cos 2\theta] + 1 = \frac{1}{2} [\cos 2\theta - \cos 4\theta] \][/tex]
Multiply through by 2 to clear the fractions:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Rearrange terms to combine like terms:
[tex]\[ \cos 4\theta + \cos 2\theta + 2 = \cos 2\theta - \cos 4\theta \][/tex]
Combine the [tex]\(\cos 4\theta\)[/tex] terms:
[tex]\[ 2\cos 4\theta + 2 = 0 \][/tex]
Subtract 2 from both sides:
[tex]\[ 2\cos 4\theta = -2 \][/tex]
Divide by 2:
[tex]\[ \cos 4\theta = -1 \][/tex]
The general solution to [tex]\(\cos (x) = -1\)[/tex] is:
[tex]\[ x = \pi + 2k\pi \quad \text{for integer } k \][/tex]
Set [tex]\(x = 4\theta\)[/tex]:
[tex]\[ 4\theta = \pi + 2k\pi \][/tex]
Solve for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{k\pi}{2} \][/tex]
Now, let's find the specific values of [tex]\(\theta\)[/tex] that satisfy the original equation within one full cycle (0 to [tex]\(2\pi\)[/tex]). We consider the first few values of [tex]\(k\)[/tex]:
For [tex]\(k = 0\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} \][/tex]
For [tex]\(k = 1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \][/tex]
For [tex]\(k = -1\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4} \][/tex]
For [tex]\(k = -2\)[/tex]:
[tex]\[ \theta = \frac{\pi}{4} - \pi = -\frac{3\pi}{4} \][/tex]
Thus, the solutions to the equation [tex]\((\cos 3\theta)(\cos \theta) + 1 = (\sin 3\theta)(\sin \theta) \)[/tex] are:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
