The curve is given by `x = cos 2t, y = sin t` represents

Options

A. A parabola
B. Circle
C. Part of a parabola
D. A pair of straight lines

Answer

To determine the nature of the curve given by the parametric equations \( x = \cos(2t) \) and \( y = \sin(t) \), we will follow these steps:

### Step 1: Rewrite the equation for \( x \)
We know that \( x = \cos(2t) \). Using the double angle formula for cosine, we can express \( \cos(2t) \) as:
\[ \cos(2t) = \cos^2(t) - \sin^2(t) \]
Thus, we have:
\[ x = \cos^2(t) - \sin^2(t) \]

### Step 2: Rewrite the equation for \( y \)
Given \( y = \sin(t) \), we can express \( \sin^2(t) \) in terms of \( y \):
\[ \sin^2(t) = y^2 \]

### Step 3: Substitute \( \sin^2(t) \) into the equation for \( x \)
Now, we can substitute \( \sin^2(t) \) into the equation for \( x \):
\[ x = \cos^2(t) - y^2 \]

### Step 4: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \cos^2(t) + \sin^2(t) = 1 \]
Substituting \( \sin^2(t) = y^2 \) into this identity gives us:
\[ \cos^2(t) + y^2 = 1 \implies \cos^2(t) = 1 - y^2 \]

### Step 5: Substitute \( \cos^2(t) \) back into the equation for \( x \)
Now we can substitute \( \cos^2(t) \) back into the equation for \( x \):
\[ x = (1 - y^2) - y^2 \]
This simplifies to:
\[ x = 1 - 2y^2 \]

### Step 6: Rearranging the equation
Rearranging the equation gives us:
\[ 2y^2 = 1 - x \implies y^2 = \frac{1 - x}{2} \]

### Conclusion
The equation \( y^2 = \frac{1 - x}{2} \) is in the standard form of a parabola that opens to the left. Therefore, the curve represented by the parametric equations \( x = \cos(2t) \) and \( y = \sin(t) \) is a part of a parabola.

### Final Answer
The curve represented by the equations \( x = \cos(2t) \) and \( y = \sin(t) \) is a part of a parabola.

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