Question

If sin(alpha)+ sin(beta)= m and cos(alpha) + cos(beta)= n then prove that cos (alpha - beta) = 1÷2{(m^2 + n^2 - 2).​

Answer 1

Solution :-

As we know that

$\bullet\sf\ \ Cos(\alpha-\beta)= Cos\alpha Cos\beta+Sin\alpha Sin\beta$

Proof :-

By taking RHS -

$\sf\ sin\alpha+ sin\beta=m\\ \\ \sf\ cos\alpha+cos\beta= n$

Solution In attachment!

Answer 2

Gɪᴠᴇɴ :

✯ $\bf{\sin{\alpha}\:+\:\sin{\beta}\:=\:m}$

Aɴᴅ,

✯ $\bf{\cos{\alpha}\:+\:\cos{\beta}\:=\:n}$

Tᴏ Pʀᴏᴏғ :

• $\bf{\cos(\alpha\:-\:\beta)\:=\:\dfrac{1}{2}\:(m^2\:+\:n^2\:-\:2)}$

Sᴏᴍᴇ Pʀᴏᴘᴇʀᴛɪᴇs :

$\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{\cos{(\alpha\:-\:\beta)}\:=\:\cos\alpha\:\cos\beta\:+\:\sin\alpha\:\sin\beta}}}}}}$

$\pink\bigstar\:\:{\underline{\blue{\boxed{\bf{\purple{\sin^2\theta\:+\:\cos^2\theta\:=\:1\:}}}}}}$

Pʀᴏᴠᴇ :

☆ We take R.H.S.

➙ $\bf{\dfrac{1}{2}\:(m^2\:+\:n^2\:-\:2)}$

☆ Now we calculate the value of 'm² + n²'.

$:\rightarrow\:\bf{(\sin{\alpha}\:+\:\sin{\beta})^2\:+\:(\cos{\alpha}\:+\:\cos{\beta})^2}$

$:\rightarrow\:\bf{\Big(\sin^2{\alpha}\:+\:2\:{\sin\alpha}\:{\sin\beta}\:+\:\sin^2{\beta}\Big)\:+\:\Big(\cos^2{\alpha}\:+\:2\:{\cos\alpha}\:{\cos\beta}\:+\:\cos^2{\beta}\Big)}$

$:\rightarrow\:\bf{(\sin^2{\alpha}\:+\:\cos^2{\alpha})\:+\:(\sin^2{\beta}\:+\:\cos^2{\beta})\:+\:(2\:{\sin\alpha}\:{\sin\beta}\:+\:2\:{\cos\alpha}\:{\cos\beta})\:}$

$:\rightarrow\:\bf{1\:+\:1\:+\:2\:(\sin\alpha\:\sin\beta\:+\:\cos\alpha\:\cos\beta)\:}$

$:\rightarrow\bf{2\:+\:2\:\cos(\alpha\:-\:\beta)\:}$

☆ Now putting the value of 'm² + n²' in R.H.S, we get

➙ $\bf{\dfrac{1}{2}\:\Big\{2\:+\:2\:\cos(\alpha\:-\:\beta)\:-\:2\Big\}}$

➙ $\bf{\dfrac{1}{2}\times{2\:\cos(\alpha\:-\:\beta)}}$

➙ $\bf{\underline{\green{\boxed{\pink{\cos(\alpha\:-\:\beta)}}}}}\:(L.H.S)$

$\Large\bf{Therefore,}$

★ L.H.S = R.H.S [Hence Proved]