If-2cos-sin-1-2-0-lt-lt-90-then-2sin-cos-

Question Number 115062 by Sudip last updated on 23/Sep/20

If 2 \cos θ - \sin θ = \frac{1}{\sqrt{2}} (0 ° < θ < 90 °)

then 2 \sin θ + \cos θ = ¿

Answered by PRITHWISH SEN 2 last updated on 23/Sep/20

let 2 \sin θ + \cos θ = k \dots . (ii)

then by solving eqn (i) (given) & (ii) we get

\sin θ = \frac{1}{5} (2 k - \frac{1}{\sqrt{2}}) & \cos θ = \frac{1}{5} (k + \sqrt{2})

from \sin^{2} θ + \cos^{2} θ = 1 we get

k = \pm \frac{3}{\sqrt{2}} as 0 < θ < 90

∴ 2 \sin θ + \cos θ = \frac{3}{\sqrt{2}}

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

4 {c o s}^{2} θ + {s i n}^{2} θ - 4 s i n θ c o s θ = \frac{{s i n}^{2} θ + {c o s}^{2} θ}{2}

8 {c o s}^{2} θ + 2 {s i n}^{2} θ - 8 s i n θ c o s θ = {s i n}^{2} θ + {c o s}^{2} θ

7 {c o s}^{2} θ + {s i n}^{2} θ - 8 s i n θ c o s θ = 0

7 {c o s}^{2} θ - 7 s i n θ c o s θ - c o s θ s i n θ + {s i n}^{2} θ = 0

7 c o s θ (c o s θ - s i n θ) - s i n θ (c o s θ - s i n θ) = 0

c o s θ = s i n θ (\frac{π}{4}, - \frac{π}{4})

2 s i n θ + c o s θ = \pm \frac{3}{\sqrt{2}}

o r

θ = {t a n}^{- 1} \frac{1}{7}