Given-sin-cos-4-3-where-0-lt-lt-pi-4-Find-the-value-of-cos-sin-

Question Number 123160 by liberty last updated on 23/Nov/20

G i v e n \sin ρ + \cos ρ = \frac{4}{3}, w h e r e

0 < ρ < \frac{π}{4} . F i n d t h e v a l u e o f \cos ρ - \sin ρ .

Answered by benjo_mathlover last updated on 23/Nov/20

\Rightarrow {(\sin ρ + \cos ρ)}^{2} = \frac{16}{9}

\Rightarrow 1 + 2 \sin ρ \cos ρ = \frac{16}{9}

\Rightarrow 2 \sin ρ \cos ρ = \frac{7}{9}

s i n c e 0 < ρ < \frac{π}{4} t h e n \cos ρ > \sin ρ

s o \cos ρ - \sin ρ > 0

c o n s i d e r \cos ρ - \sin ρ = \sqrt{1 - 2 \sin ρ \cos ρ}

\cos ρ - \sin ρ = \sqrt{1 - \frac{7}{9}} = \frac{\sqrt{2}}{3} .

Answered by Dwaipayan Shikari last updated on 23/Nov/20

{(s i n p + c o s p)}^{2} + {(c o s p - s i n p)}^{2} = 2

(c o s p - s i n p) = \sqrt{2 - \frac{16}{9}} = \frac{\sqrt{2}}{3} (A s 0 < p < \frac{π}{4})