Question

Prove that tan22.5=√2-1 given sin22.5=√[2-√2]/2

Answer 1

Hey there !

tan22.5= sin22.5/cos22.5

find the value of cos 22.5

cos 22.5= √ 1- sin²22.5

= √ 1- {√[2-√2]/2}²

=√ [1- (2-√2)/4 ]

= √ [4-2+√2 / 4 ]

= √ [2+√2 /4 ]

= √ 2+√2 / 2

Now tan22.5 = sin22.5/cos22.5

= √ (2-√2) /2 ÷ √(2+√2)/2

= √( 2-√2 / 2+√2)

= √ {(√2) (√2-1)/√2 (√2+1) }

= √ [ √2-1/√2+1 ]

Rationalise the denominator .

= √ (√2-1 /√2+1)(√2-1)/√2+1

= √ (√2-1)² / √2²-1²

= √2-1

hope helped !

Answer 2

sin(22.5°) = √(2 - √2)/2
take square both sides,
sin²(22.5°) = (2 - √2)/2²
1 - cos²(22.5°) = (2 - √2)/4
1 - (2 - √2)/4 = cos²(22.5°)
(4 - 2 + √2)/4 = cos²(22.5°)
{√(2 + √2)/2}² = cos²(22.5°)
cos(22.5°) = ±√(2 - √2)/2
but cos(22.5°) > 0 so, [ 1st quadrant ]
cos(22.5°) = √(2 + √2)/2

now,
tan(22.5°) = sin(22.5°)/cos(22.5°)
= √(2 - √2)/√(2 + √2)
= √{((2 - √2)(2 - √2)/(2 +√2)(2 -√3)}
= √{(2 - √2)²/(2² -√2²)
= (2 - √2)/√2
= √2 - 1
hence, tan(22.5°) = √2 - 1