Question

Find a quadratic equation whose roots are sin 15° and cos 15°

Answer 1

sin 15° = sin(45° – 30°)
= sin 45°. cos 30° – cos 45°. sin 30°

= `1/sqrt(2) * sqrt(3)/2 - 1/sqrt(2) * 1/2`

sin 15° = `1/(2sqrt(2)) (sqrt(3) - 1)`  ......(1)

cos 15° = cos(45° – 30°)
= cos 45° . cos 30° + sin 45° . sin 30°

= `1/sqrt(2) * sqrt(3)/2 + 1/sqrt(2) * 1/2`

sin 15° = `1/(2sqrt(2)) (sqrt(3) + 1)`  ......(2)

The quadratic whose roots cos 15° and sin 15° is
x² – (cos 15° + sin 15°)x + (cos 15°) (sin 15°) = 0  ......(3)

cos 15° + sin 15° = `1/(2sqrt(2)) (sqrt(3) + 1) + 1/(2sqrt(2)) (sqrt(3) - 1)`

= `1/(2sqrt(2)) (sqrt(3) + 1 + sqrt(3) - 1)`

= `(2sqrt(3))/(2sqrt(2))`

= `sqrt(3)/sqrt(2)`

= `sqrt(3)/sqrt(2) xx sqrt(2)/sqrt(2)`

= `sqrt(6)/2`

(cos 15°) (sin 15°) = `1/(2sqrt(2)) (sqrt(3) + 1) * 1/(2sqrt(2)) (sqrt(3) - 1)`

= `1/(4(2)) * (3 - 1)`

= `2/8`

= `1/4`
Substituting in equation (3) we have

`x^2 - sqrt(6)/2x + 1/4` = 0

`4x^2 - 2sqrt(6)x + 1` = 0