Question

Evaluate sin² $(\frac{\pi}{8} + \frac{A}{2})$ - sin² $(\frac{\pi}{8} - \frac{A}{2})$

Answer 1

sin²(π/8 + A/2) - sin²(π/8 - A/2)

use algebraic formula, a² - b² = (a - b)(a + b)

then, sin²(π/8 + A/2) - sin²(π/8 - A/2) = [sin(π/8 + A/2) + sin(π/8 - A/2) ] [ sin(π/8 + A/2) - sin(π/8 - A/2) ]

we know, sin(A + B) + sin(A - B) = 2sinA.cosB
sin(A + B) - sin(A - B) = 2cosA.sinB

so, [sin(π/8 + A/2) + sin(π/8 - A/2) ] = 2sin(π/8). cos(A/2)
[sin(π/8 + A/2) - sin(π/8 - A/2) ] = 2cos(π/8). sin(A/2)]

now, [sin(π/8 + A/2) + sin(π/8 - A/2) ] [ sin(π/8 + A/2) - sin(π/8 - A/2) ] = [2sin(π/8)cos(A/2)][2cos(π/8)sin(A/2)]

= [2sin(π/8)cos(π/8)] [ 2sin(A/2) cos(A/2)]

we know, sin2x = 2sinx. cosx

= sin2(π/8) . sin2(A/2)

= sin(π/4) . sinA

= sinA/√2

Answer 2

Answer:

$\frac{sinA}{\sqrt{2}}$

Step-by-step explanation:

Formula used:

${sin}^2A-{sin}^2B=sin(A+B).sin(A-B)$

${sin}^2(\frac{\pi}{8}+\frac{A}{2})-{sin}^2(\frac{\pi}{8}-\frac{A}{2})$

$=sin(\frac{\pi}{8}+\frac{A}{2}+\frac{\pi}{8}-\frac{A}{2}).sin(\frac{\pi}{8}+\frac{A}{2}-\frac{\pi}{8}+\frac{A}{2})$

$=sin(\frac{\pi}{8}+\frac{\pi}{8}).sin(\frac{A}{2}+\frac{A}{2})$

$=sin(\frac{\pi}{4}).sinA$

$=\frac{1}{\sqrt(2)}.sinA$

$= \frac{sinA}{\sqrt{2}}$