Question

Evaluate sin² $82\frac{1}{2}^{\textdegree}$ - sin² $22\frac{1}{2}^{\textdegree}$

Answer 1

we have to find value of sin²82 1/2° - sin²22 1/2°

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

so, sin²82 1/2° - sin²22 1/2° = [sin82 1/2° + sin22 1/2° ][sin82 1/2° - sin22 1/2° ]

use formula, sinC + sinD = 2sin(C + D)/2.cos(C - D)/2
sinC - sinD = 2cos(C + D)/2. sin(C - D)/2

so, sin82 1/2° + sin22 1/2° = 2sin52 1/2° . cos30°

= 2 sin52 1/2° (√3/2) = √3sin52 1/2°

similarly, sin82 1/2° - sin22 1/2° = 2cos52 1/2°sin30°

= 2cos52 1/2° (1/2) = cos52 1/2°

now, [sin82 1/2° + sin22 1/2° ][sin82 1/2° - sin22 1/2° ] = √3sin52 1/2° cos52 1/2°

= √3/2(2sin52 1/2° cos52 1/2°)

we know, 2sinx. cosx = sin2x

so, 2sin52 1/2° cos52 1/2° = sin105°

= √3/2 sin105°

= √3/2 sin(60° + 45°)

= √3/2 [ sin60°.cos45° + cos60° sin45°]

= √3/2 × (√3 + 1)/2√2

= (3 + √3)/4√2

Answer 2

Answer:

$\frac{\sqrt{3}+3}{4\sqrt{2}}$

Step-by-step explanation:

Formula used:

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

First we calculate

sin105

=sin(60+45)

=sin60 cos45+cos60 sin45

$=\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}+\frac{1}{2}.\frac{1}{\sqrt{2}}\\=\frac{\sqrt{3}+1}{2\sqrt{2}}$

${sin}^2(82.5)-{sin}^2(22.5)\\=sin(82.5+22.5).sin(82.5-22.5)\\=sin105. sin60$

$=\frac{\sqrt{3}+1}{2\sqrt{2}}.\frac{\sqrt{3}}{2}\\=\frac{\sqrt{3}+3}{4\sqrt{2}}$