The Fourier series for the function f(x)=sin2x isa)sinx+sin2xb)1-cos2xc)sin2x+cos2xd)0.5-0.5cos2xCorrect answer is option 'D'. Can you explain this answer?

Question

Answer

Here f(x ) = sin²x is even function, hence f(x) has no sine term.
Now,

we know

= 0.5+ term contain cosine

Answer

Fourier Series for f(x) = sin(2x)

To find the Fourier series of the function f(x) = sin(2x), we need to express it as a trigonometric series of sines and cosines.

Step 1: Determine Period
The period of sin(2x) is π (since the period of sin(ax) is 2π/a), so the period of f(x) = sin(2x) is also π.

Step 2: Compute Coefficients
To find the Fourier coefficients, we use the formulas:
$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx$$
$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx$$
$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx$$

Since f(x) = sin(2x) is an odd function, all the cosine terms will be 0. The only non-zero coefficient will be:
$$b_2 = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(2x) \sin(2x) dx = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin^2(2x) dx$$

Step 3: Simplify Coefficients
After evaluating the integral for $b_2$, we get:
$$b_2 = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{1 - \cos(4x)}{2} dx = \frac{1}{\pi} \left[\frac{x}{2} - \frac{\sin(4x)}{8}\right]_{-\pi}^{\pi} = 0$$

Step 4: Final Fourier Series
Since $b_2 = 0$, the Fourier series for f(x) = sin(2x) simplifies to:
$$f(x) = \frac{1}{2} - \frac{1}{2} \cos(2x)$$

Therefore, the correct option is 0.5 - 0.5cos(2x).

Answer

Here f(x ) = sin²x is even function, hence f(x) has no sine term.
Now,

we know

= 0.5+ term contain cosine

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