Consider the following for the next items that follow: Let \(\rm p=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \) and \(\rm q=\cos \left(\frac{4 \pi}{5}\right) \cos \left(\frac{8 \pi}{5}\right)\).
What is the value of p + q?
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The problem asks us to find the value of \(p+q\), where \(p\) and \(q\) are defined as products of cosine terms with specific angles.
The given expressions are:
Let's evaluate the expression for \(p\):
\( \rm p=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \)
We can use the trigonometric identity \(2 \sin A \cos A = \sin 2A\). To apply this, we can multiply and divide the expression by \(2 \sin \left(\frac{\pi}{5}\right)\):
\( \rm p = \frac{1}{2 \sin \left(\frac{\pi}{5}\right)} \left(2 \sin \left(\frac{\pi}{5}\right) \cos \left(\frac{\pi}{5}\right)\right) \cos \left(\frac{2 \pi}{5}\right) \)
Applying the identity to the term in the parenthesis:
\( \rm p = \frac{1}{2 \sin \left(\frac{\pi}{5}\right)} \sin \left(2 \times \frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \)
\( \rm p = \frac{1}{2 \sin \left(\frac{\pi}{5}\right)} \sin \left(\frac{2 \pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \)
Now, we have \( \sin \left(\frac{2 \pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \). We can apply the same identity again by multiplying and dividing by 2:
\( \rm p = \frac{1}{2 \sin \left(\frac{\pi}{5}\right)} \times \frac{1}{2} \left(2 \sin \left(\frac{2 \pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right)\right) \)
\( \rm p = \frac{1}{4 \sin \left(\frac{\pi}{5}\right)} \sin \left(2 \times \frac{2 \pi}{5}\right) \)
\( \rm p = \frac{1}{4 \sin \left(\frac{\pi}{5}\right)} \sin \left(\frac{4 \pi}{5}\right) \)
We know that \( \sin(\pi - x) = \sin x \). So, \( \sin \left(\frac{4 \pi}{5}\right) = \sin \left(\pi - \frac{\pi}{5}\right) = \sin \left(\frac{\pi}{5}\right) \).
Substitute this into the expression for \(p\):
\( \rm p = \frac{1}{4 \sin \left(\frac{\pi}{5}\right)} \sin \left(\frac{\pi}{5}\right) \)
Assuming \( \sin \left(\frac{\pi}{5}\right) \neq 0 \), which is true since \(0 < \frac{\pi}{5} < \pi\), we can cancel the \( \sin \left(\frac{\pi}{5}\right) \) terms:
\( \rm p = \frac{1}{4} \)
Next, let's evaluate the expression for \(q\):
\( \rm q=\cos \left(\frac{4 \pi}{5}\right) \cos \left(\frac{8 \pi}{5}\right) \)
We can simplify the angles using properties of cosine:
For the first term:
\( \cos \left(\frac{4 \pi}{5}\right) = \cos \left(\pi - \frac{\pi}{5}\right) = -\cos \left(\frac{\pi}{5}\right) \)
For the second term:
\( \cos \left(\frac{8 \pi}{5}\right) = \cos \left(2\pi - \frac{2 \pi}{5}\right) = \cos \left(\frac{2 \pi}{5}\right) \)
Substitute these simplified terms back into the expression for \(q\):
\( \rm q = \left(-\cos \left(\frac{\pi}{5}\right)\right) \left(\cos \left(\frac{2 \pi}{5}\right)\right) \)
\( \rm q = -\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \)
Notice that the expression \( \cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \) is exactly the expression for \(p\).
So, \( \rm q = -p \).
Since we found \( \rm p = \frac{1}{4} \), it follows that \( \rm q = -\frac{1}{4} \).
Now we need to find the sum \(p + q\):
\( \rm p + q = \frac{1}{4} + \left(-\frac{1}{4}\right) \)
\( \rm p + q = \frac{1}{4} - \frac{1}{4} \)
\( \rm p + q = 0 \)
The value of \(p + q\) is 0.
| Angle (in radians) | Relationship | Identity |
|---|---|---|
| \( \frac{\pi}{5} \) | Base angle | N/A |
| \( \frac{2\pi}{5} \) | \( 2 \times \frac{\pi}{5} \) | Used in \(2 \sin A \cos A = \sin 2A\) |
| \( \frac{4\pi}{5} \) | \( \pi - \frac{\pi}{5} \) | \( \sin(\pi - x) = \sin x \) \( \cos(\pi - x) = -\cos x \) |
| \( \frac{8\pi}{5} \) | \( 2\pi - \frac{2\pi}{5} \) | \( \cos(2\pi - x) = \cos x \) |
This problem utilizes several fundamental concepts in trigonometry, essential for evaluating expressions involving angles and trigonometric functions.
Mastering these concepts is key to solving a wide range of trigonometric problems.
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