Question

2. If cos 0 + sin 0 = 12 cos e, prove that cos 0 - sin 0 =
√2 sine.
0​

Answer 1

we know that,

sin²x + cos²x = 1

sin²x = 1 - cos²x

$cosx + sinx = \sqrt{2} cosx \\ \\ squaring \: both \: sides \\ \\ {cos}^{2} x + {sin}^{2} x + 2sinxcos = 2 {cos}^{2} x \\ 1 + 2sinxcos = 2 {cos}^{2} x \\ 2sinxcosx = 2 {cos}^{2} x - 1 \: \: \: \: \: \: \: \: ...(i) \\ \\ {(cosx - sinx)}^{2} \\ = {cos}^{2} x + {sin}^{2}x - 2sinxcosx \\ = 1 - (2 {cos}^{2} x - 1) \: \: \: \: \: ...(from \: i) \\ = 1 - 2 {cos}^{2}x + 1 = 2 - 2 {cos}^{2} x \\ = 2(1 - {cos}^{2} x) \\ = 2 {sin}^{2} x \\ \\ {(cosx - sinx)}^{2} = 2 {sin}^{2} x \\ \\ taking \: square \: roots \: on \: both \: sides \\ \\ cosx - sinx = \sqrt{2} sinx$

Hence proved!