Math, asked by Arceus02, 10 months ago

cosθ + sinθ = √2 cosθ
prove that cosθ - sinθ = √2 sinθ​

Answers

Answered by amitkumar44481
21

SolutioN :

 \tt \dagger \:  \:  \:  \:  \:  Cos\, \theta+ Sin\,\theta= \sqrt{2} Cos\,\theta.

Squaring Both sides, We get.

 \tt  : \implies  \Big(Cos\, \theta+ Sin\,\theta \Big)^2=\Big( \sqrt{2} Cos\,\theta \Big)^2

We know, that.

  • ( a + b )² = a² + b² + 2ab.

 \tt  : \implies  Cos^2\, \theta+ Sin^2\,\theta +2Sin\,\theta.Cos\,\theta=2 Cos^2\,\theta.

 \tt  : \implies  2Sin\,\theta.Cos\,\theta=2 Cos^2\,\theta - Cos^2\, \theta -  Sin^2\,\theta

 \tt  : \implies  2Sin\,\theta.Cos\,\theta=Cos^2\,\theta  -  Sin^2\,\theta

Taking Common both sides ( negative )

 \tt  : \implies   - (2Sin\,\theta.Cos\,\theta)= - (Cos^2\,\theta -  Sin^2\,\theta)

 \tt  : \implies   - 2Sin\,\theta.Cos\,\theta= - Cos^2\,\theta +  Sin^2\,\theta

 \tt  : \implies   - 2Sin\,\theta.Cos\,\theta=Sin^2\,\theta - Cos^2\,\theta

 \tt  : \implies Sin^2\,\theta - Cos^2\,\theta= - 2Sin\,\theta.Cos\,\theta.

Taking SinA both sides.

 \tt  : \implies Sin^2\,\theta +  Sin^2\,\theta - Cos^2\,\theta= Sin^2\,\theta  - 2Sin\,\theta.Cos\,\theta.

 \tt  : \implies 2Sin^2\,\theta= Sin^2\,\theta+ Cos^2\,\theta - 2Sin\,\theta.Cos\,\theta.

 \tt  : \implies  Sin^2\,\theta+ Cos^2\,\theta - 2Sin\,\theta.Cos\,\theta = 2Sin^2\,\theta

We know,

  • ( a - b )² = a² + b² - 2ab.

 \tt  : \implies  \bigg(Cos\,\theta -  Sin\,\theta \bigg)^2 = 2Sin^2\,\theta

 \tt  : \implies  Cos\,\theta -  Sin\,\theta =  \sqrt{2Sin^2\,\theta}

 \tt  : \implies  Cos\,\theta -  Sin\,\theta =  \sqrt2Sin\,\theta.

Hence Proved.

\rule{200}3

Other MethodS :

 \tt \dagger \:  \:  \:  \:  \:  Cos\, \theta+ Sin\,\theta= \sqrt{2} Cos\,\theta.

Squaring Both sides.

 \tt  : \implies  \Big(Cos\, \theta+ Sin\,\theta \Big)^2=\Big( \sqrt{2} Cos\,\theta \Big)^2

 \tt  : \implies  Cos^2\, \theta+ Sin^2\,\theta +2Sin\,\theta.Cos\,\theta=2 Cos^2\,\theta.

We know,

  • Sin²A + Cos²A = 1.
  • Cos²A = 1 - Sin²A.

 \tt  : \implies  Cos^2\, \theta+ Sin^2\,\theta +2Sin\,\theta.Cos\,\theta=2(1 -Sin^2\,\theta )

 \tt  : \implies  Cos^2\, \theta+ Sin^2\,\theta +2Sin\,\theta.Cos\,\theta=2 -  2Sin^2\,\theta

 \tt  : \implies  1 +2Sin\,\theta.Cos\,\theta=2 -  2Sin^2\,\theta

 \tt  : \implies 2Sin^2\,\theta =  2-  1  - 2Sin\,\theta.Cos\,\theta

 \tt  : \implies 2Sin^2\,\theta =  1 - 2Sin\,\theta.Cos\,\theta

 \tt  : \implies 2Sin^2\,\theta =   Cos^2\, \theta+ Sin^2\,\theta -  2Sin\,\theta.Cos\,\theta

 \tt  : \implies 2Sin^2\,\theta =   (Cos\, \theta -  Sin\,\theta )^2

 \tt  : \implies 2Sin^2\,\theta =   Cos^2\, \theta+ Sin^2\,\theta -  2Sin\,\theta.Cos\,\theta

 \tt:\implies Cos\,\theta - Sin\,\theta=  \sqrt{2}Sin\,\theta.

Hence Proved.

Answered by GreatBhavya2009
8

Cosθ+Sinθ=

2

Cosθ.

Squaring Both sides, We get.

\tt : \implies \Big(Cos\, \theta+ Sin\,\theta \Big)^2=\Big( \sqrt{2} Cos\,\theta \Big)^2:⟹(Cosθ+Sinθ)

2

=(

2

Cosθ)

2

We know, that.

( a + b )² = a² + b² + 2ab.

\tt : \implies Cos^2\, \theta+ Sin^2\,\theta +2Sin\,\theta.Cos\,\theta=2 Cos^2\,\theta.:⟹Cos

2

θ+Sin

2

θ+2Sinθ.Cosθ=2Cos

2

θ.

\tt : \implies 2Sin\,\theta.Cos\,\theta=2 Cos^2\,\theta - Cos^2\, \theta - Sin^2\,\theta:⟹2Sinθ.Cosθ=2Cos

2

θ−Cos

2

θ−Sin

2

θ

\tt : \implies 2Sin\,\theta.Cos\,\theta=Cos^2\,\theta - Sin^2\,\theta:⟹2Sinθ.Cosθ=Cos

2

θ−Sin

2

θ

Taking Common both sides ( negative )

\tt : \implies - (2Sin\,\theta.Cos\,\theta)= - (Cos^2\,\theta - Sin^2\,\theta):⟹−(2Sinθ.Cosθ)=−(Cos

2

θ−Sin

2

θ)

\tt : \implies - 2Sin\,\theta.Cos\,\theta= - Cos^2\,\theta + Sin^2\,\theta:⟹−2Sinθ.Cosθ=−Cos

2

θ+Sin

2

θ

\tt : \implies - 2Sin\,\theta.Cos\,\theta=Sin^2\,\theta - Cos^2\,\theta:⟹−2Sinθ.Cosθ=Sin

2

θ−Cos

2

θ

\tt : \implies Sin^2\,\theta - Cos^2\,\theta= - 2Sin\,\theta.Cos\,\theta.:⟹Sin

2

θ−Cos

2

θ=−2Sinθ.Cosθ.

Taking SinA both sides.

\tt : \implies Sin^2\,\theta + Sin^2\,\theta - Cos^2\,\theta= Sin^2\,\theta - 2Sin\,\theta.Cos\,\theta.:⟹Sin

2

θ+Sin

2

θ−Cos

2

θ=Sin

2

θ−2Sinθ.Cosθ.

\tt : \implies 2Sin^2\,\theta= Sin^2\,\theta+ Cos^2\,\theta - 2Sin\,\theta.Cos\,\theta.:⟹2Sin

2

θ=Sin

2

θ+Cos

2

θ−2Sinθ.Cosθ.

\tt : \implies Sin^2\,\theta+ Cos^2\,\theta - 2Sin\,\theta.Cos\,\theta = 2Sin^2\,\theta:⟹Sin

2

θ+Cos

2

Hope it helps..

Similar questions