Cos theta+sin theta =√2cos theta show thatcos theta-sin theta =√2sin theta

Open in App

Solution

cosθ+sinθ = √2cosθ

on squaring both sides,
cos²θ + sin²θ +2cosθ.sinθ = 2cos²θ = 2(1-sin²θ)
1+2cosθ.sinθ = 2(1-sin²θ)
2cosθ.sinθ = 2(1-sin²θ) -1
2cosθ.sinθ = 1-2sin²θ ................eq1)

Now,
(cosθ-sinθ)² = cos²θ + sin²θ -2cosθ.sinθ
= 1 -2cosθ.sinθ
= 1 -(1-2sin²θ) [ as per eq1)
= 2sin²θ
cosθ-sinθ = √2sinθ proved

Suggest Corrections

293

Similar questions

Q. Show that

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ}

= - 2 sec 2 θ

Prove that following identities:
(i) $\frac{sin θ - cos θ}{s i n θ + cos θ} + \frac{sin θ + cos θ}{sin θ - cos θ} = \frac{2}{2 s i n^{2} θ - 1}$
(ii) $\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{2}{1 - 2 {cos}^{2} θ}$

$Solve the following equations : (i) c o s θ + c p s 2 θ + c o s 3 θ = 0 (i i) c o s θ + c o s 3 θ - c o s 2 θ = 0 (i i i) s i n θ + s i n 5 θ = s i n 3 θ (i v) c o s θ c o s 2 θ c o s 3 θ = \frac{1}{4} (v) c o s θ + s i n θ = c o s 2 θ + s i n 2 θ (v i) s i n θ + s i n 2 θ + s i n 3 θ = 0 (v i i) s i n θ + s i n 2 θ + s i n 3 θ + s i n 4 θ = 0 (v i i i) s i n 3 θ - s i n θ = 4 c o s^{2} θ - 2 (i x) s i n 2 θ - s i n 4 θ + s i n 6 θ = 0$

Join BYJU'S Learning Program