Trigonometric Identities and Equations

The integral of cos(x) dx is sin(x) + C. Since the derivative of sin(x) is cos(x), the antiderivative (integral) of cos(x) must be sin(x). A common mistake is to suggest -sin(x), which is actually the derivative of cos(x).

Use the double-angle identity: 2 sin x cos x = sin(2x). The period of sin(kx) is 2π/k, so the period of sin(2x) is 2π/2 = π.

Sin 90 is 1, Cos 90 is 0, but Tan 90 is undefined (approaching infinity). In basic mathematics, the presence of an undefined term in a sum makes the entire expression undefined or 'infinity'.

Using trigonometric identities: sin(x) * cosec(x) = 1 and sec(x) * cos(x) = 1. Therefore, 1 + 1 = 2.

Expanding (sinA+cosA)²+(sinA-cosA)² gives sin²A+2sinAcosA+cos²A+sin²A-2sinAcosA+cos²A = 2(sin²A+cos²A) = 2(1) = 2, using the fundamental identity sin²θ+cos²θ=1.

Substitute the given angles: sin45° = 1/√2, cos45° = 1/√2, tan45° = 1, sec45° = √2, sin90° = 1, cos90° = 0. The expression becomes (1/√2) × (1/√2) × 1 × √2 × 1 × 0 = 0 because cos90° = 0. Any expression multiplied by zero equals zero.

The expression is a product of cosine values from 0 to 99 degrees. Since cos(90°) = 0, and any number multiplied by zero is zero, the entire product equals 0.

MATLAB trigonometric functions use radians, not degrees. sin(90°) requires converting 90° to radians: 90° = pi/2 radians. Option B (sin(90)) is wrong because 90 is treated as 90 radians. Option C has wrong case (Sin vs sin). Option A has wrong case for pi (Pi vs pi).

For an even function Fourier series contains de term and cosine term (even and odd harmonics).

The Fourier series of a real periodic function has only cosine terms if it is even and sine terms if it is odd.

$sin x = x + \frac{x^3}{3!}+\frac{x^5}{5!}+ ... \\ cos x = 1 + \frac{x^2}{2!}+\frac{x^4}{4!}+ ... \\ \text{Thus only $(x^3)$ will have odd power of x.}$