Prove the trigonometric identity. cosA - cosB = -2sin((A+B)/(2))sin((A-B)/(2))

Apply Sine Angle Subtraction Formula: Use the sine angle subtraction formula: sin(x)sin(y) = (1/2)[cos(x-y) - cos(x+y)]. So, -2sin((A+B)/2)sin((A-B)/2) = -2 * (1/2)[cos((A+B)/2 - (A-B)/2) - cos((A+B)/2 + (A-B)/2)].Simplify Cosines: Simplify the cosines inside the brackets: cos((A+B)/2 - (A-B)/2) = cos(B/2 + B/2) = cos(B), and cos((A+B)/2 + (A-B)/2) = cos(A/2 + A/2) = cos(A). So, -2 * (1/2)[cos(B) - cos(A)].Eliminate Negative Sign: Multiply through by -1 to get rid of the negative sign: -1 * (1/2)[cos(B) - cos(A)] = -(1/2)cos(B) + (1/2)cos(A).Final Multiplication: Now multiply by -2: -2 * (-(1/2)cos(B) + (1/2)cos(A)) = cos(B) - cos(A).

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Q. Prove the trigonometric identity.

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\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)

Apply Sine Angle Subtraction Formula: Use the sine angle subtraction formula: $\sin(x)\sin(y) = \frac{1}{2}[\cos(x-y) - \cos(x+y)]$ . So, $-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) = -2 \times \frac{1}{2}[\cos\left(\frac{A+B}{2} - \frac{A-B}{2}\right) - \cos\left(\frac{A+B}{2} + \frac{A-B}{2}\right)]$ .
Simplify Cosines: Simplify the cosines inside the brackets: $\cos\left(\frac{A+B}{2} - \frac{A-B}{2}\right) = \cos\left(\frac{B}{2} + \frac{B}{2}\right) = \cos(B)$ , and $\cos\left(\frac{A+B}{2} + \frac{A-B}{2}\right) = \cos\left(\frac{A}{2} + \frac{A}{2}\right) = \cos(A)$ . $\newline$ So, $-2 \times \left(\frac{1}{2}\right)[\cos(B) - \cos(A)]$ .
Eliminate Negative Sign: Multiply through by $-1$ to get rid of the negative sign: $-1 \times (\frac{1}{2})[\cos(B) - \cos(A)] = -(\frac{1}{2})\cos(B) + (\frac{1}{2})\cos(A)$ .
Final Multiplication: Now multiply by $-2$ : $-2 \times (-(\frac{1}{2})\cos(B) + (\frac{1}{2})\cos(A)) = \cos(B) - \cos(A)$ .