Question

If, sin theta+sin phi=a, cos theta+cos phi=b, then show that, tan (theta - phi)/2 = plus/minus sqrt((4 - a ^ 2 - b ^ 2)/(a ^ 2 + b ^ 2))​

Answer 1

$\huge\mathfrak\green{answer}$

sinθ+sinϕ=a ________ (1) cosθ+cosϕ=b _______ (2)

(1)

2 +(2) 2

⇒a 2 +b 2

=sin 2 θ+sin

2 ϕ+2sinθsinϕ+cos 2 θ+cos 2

ϕ+2cosθcosϕ

(a 2 +b 2 )=1+1+2[sinθsinϕ+cosθcosϕ](a 2 +b 2 )=2[1+cos(θ−ϕ)]2a 2 +b 2

=2cos 2 ( 2θ−ϕ )

⇒cos 2 ( 2θ−ϕ )= 4a 2 +b

2 ______ (2)tan( 2θ−ϕ )= cos 2 ( 2θ−ϕ )

1−cos 2 ( 2θ−ϕ )

= ( 4a 2 +b 2 )1−( 4a 2 +b 2 )

= a 2 +b 2 4−a 2 −b 2

∴tan( 2θ−ϕ )= a 2 +b 2 4−a 2 −b 2

Answer 2

Answer:

$answer$

Step-by-step explanation:

sinθ+sinϕ=a ________ (1) cosθ+cosϕ=b _______ (2)

(1)

2 +(2) 2

⇒a 2 +b 2

=sin 2 θ+sin

2 ϕ+2sinθsinϕ+cos 2 θ+cos 2

ϕ+2cosθcosϕ

(a 2 +b 2 )=1+1+2[sinθsinϕ+cosθcosϕ](a 2 +b 2 )=2[1+cos(θ−ϕ)]2a 2 +b 2

=2cos 2 ( 2θ−ϕ )

⇒cos 2 ( 2θ−ϕ )= 4a 2 +b

2 ______ (2)tan( 2θ−ϕ )= cos 2 ( 2θ−ϕ )

1−cos 2 ( 2θ−ϕ )

= ( 4a 2 +b 2 )1−( 4a 2 +b 2 )

= a 2 +b 2 4−a 2 −b 2

∴tan( 2θ−ϕ )= a 2 +b 2 4−a

$.$