cosθ + sinθ = √2 cosθ
prove that cosθ - sinθ = √2 sinθ
Answers
SolutioN :
Squaring Both sides, We get.
We know, that.
- ( a + b )² = a² + b² + 2ab.
Taking Common both sides ( negative )
Taking SinA both sides.
We know,
- ( a - b )² = a² + b² - 2ab.
Hence Proved.
Other MethodS :
Squaring Both sides.
We know,
- Sin²A + Cos²A = 1.
- Cos²A = 1 - Sin²A.
Hence Proved.
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..