Question
Express 1/2cos&+√3/2sin& in terns of Cos& only
Answers
GPT-4o mini
To express \( \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \) in terms of \( \cos \theta \) only, we can use the sine function's representation in terms of cosine. Notice that we can rewrite \( \sin \theta \) using the identity:
\[
\sin \theta = \sqrt{1 - \cos^2 \theta}
\]
However, to express the original function completely in terms of \( \cos \theta \), it is often more convenient to rewrite it in a form compatible with cosine using a phase shift. We know that we can express a linear combination of sine and cosine as a single cosine function:
\[
a \cos \theta + b \sin \theta = R \cos(\theta - \phi)
\]
where:
- \( R = \sqrt{a^2 + b^2} \)
- \( \cos \phi = \frac{a}{R} \)
- \( \sin \phi = \frac{b}{R} \)
In this case, \( a = \frac{1}{2} \) and \( b = \frac{\sqrt{3}}{2} \).
First, compute \( R \):
\[
R = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1
\]
Next, calculate \( \phi \):
\[
\cos \phi = \frac{\frac{1}{2}}{1} = \frac{1}{2}, \quad \sin \phi = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2}
\]
From these values, we find that:
\[
\phi = \frac{\pi}{3}
\]
Thus, we can rewrite the original expression as:
\[
\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = \cos\left(\theta - \frac{\pi}{3}\right)
\]
If we substitute \( \theta - \frac{\pi}{3} \) back into terms of \( \cos \theta \), we don't obtain a form in terms of \( \cos \theta \) only. However, we can say:
\[
\cos\left(\theta - \frac{\pi}{3}\right) = \cos(\theta) \cos\left(\frac{\pi}{3}\right) + \sin(\theta) \sin\left(\frac{\pi}{3}\right)
\]
Calculating, we get:
\[
\cos\left(\theta - \frac{\pi}{3}\right) = \cos\theta \cdot \frac{1}{2} + \sin\theta \cdot \frac{\sqrt{3}}{2}
\]
This implies directly that:
\[
\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = \cos(\theta - \frac{\pi}{3})
\]
Thus, there isn’t a direct way to express it exclusively in terms of \( \cos \theta \) without involving \( \sin \theta \) or recalling the phase shift nature of combining these functions.
However, if you're specifically looking for a transformation to express with just \( \cos \), you need to substitute or relate angles according to a definite range or context, typically \( \sin \theta \) will still be involved indirectly.
In conclusion, you can keep it in this cos/sin format, or elaborate on its specific transformations based on your requirement for further analysis or calculations.
\[
\sin \theta = \sqrt{1 - \cos^2 \theta}
\]
However, to express the original function completely in terms of \( \cos \theta \), it is often more convenient to rewrite it in a form compatible with cosine using a phase shift. We know that we can express a linear combination of sine and cosine as a single cosine function:
\[
a \cos \theta + b \sin \theta = R \cos(\theta - \phi)
\]
where:
- \( R = \sqrt{a^2 + b^2} \)
- \( \cos \phi = \frac{a}{R} \)
- \( \sin \phi = \frac{b}{R} \)
In this case, \( a = \frac{1}{2} \) and \( b = \frac{\sqrt{3}}{2} \).
First, compute \( R \):
\[
R = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1
\]
Next, calculate \( \phi \):
\[
\cos \phi = \frac{\frac{1}{2}}{1} = \frac{1}{2}, \quad \sin \phi = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2}
\]
From these values, we find that:
\[
\phi = \frac{\pi}{3}
\]
Thus, we can rewrite the original expression as:
\[
\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = \cos\left(\theta - \frac{\pi}{3}\right)
\]
If we substitute \( \theta - \frac{\pi}{3} \) back into terms of \( \cos \theta \), we don't obtain a form in terms of \( \cos \theta \) only. However, we can say:
\[
\cos\left(\theta - \frac{\pi}{3}\right) = \cos(\theta) \cos\left(\frac{\pi}{3}\right) + \sin(\theta) \sin\left(\frac{\pi}{3}\right)
\]
Calculating, we get:
\[
\cos\left(\theta - \frac{\pi}{3}\right) = \cos\theta \cdot \frac{1}{2} + \sin\theta \cdot \frac{\sqrt{3}}{2}
\]
This implies directly that:
\[
\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = \cos(\theta - \frac{\pi}{3})
\]
Thus, there isn’t a direct way to express it exclusively in terms of \( \cos \theta \) without involving \( \sin \theta \) or recalling the phase shift nature of combining these functions.
However, if you're specifically looking for a transformation to express with just \( \cos \), you need to substitute or relate angles according to a definite range or context, typically \( \sin \theta \) will still be involved indirectly.
In conclusion, you can keep it in this cos/sin format, or elaborate on its specific transformations based on your requirement for further analysis or calculations.