Question

If $\alpha $ and $\beta $ be two distinct real numbers such that \[\left( \alpha -\beta \right)\ne 2n\pi \] for any integer n satisfying the equations \[a\text{ }cos\text{ }\theta +b\text{ }sin\text{ }\theta =c\] then prove that $\cos \left( \alpha +\beta \right)=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$.

Answer 1

Hint: In this question, $\alpha $ and $\beta $ are two different roots satisfying the equation \[a\text{ }cos\text{ }\theta +b\text{ }sin\text{ }\theta =c\].You can use factorization formulas or sum to product formulas$\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right) $, \[\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)\].

Complete step by step answer:
As $\alpha $ and $\beta $ are two roots of the equation\[a\text{ }cos\text{ }\theta +b\text{ }sin\text{ }\theta =c\], we get
\[a\text{ }cos\text{ }\alpha +b\text{ }sin\text{ }\alpha =c......................(1)\]
\[a\text{ }cos\text{ }\beta +b\text{ }sin\text{ }\beta =c......................(2)\]
Subtracted equation (2) from the equation (1), we get
 \[a\text{ }\left( cos\text{ }\alpha -\cos \beta \right)+b\text{ }\left( \sin \alpha -\sin \beta \right)=0\]
Applying the factorization or sum to product formula $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$ , we get
\[a\left[ 2\sin \left( \dfrac{\alpha +\beta }{2} \right)\sin \left( \dfrac{\beta -\alpha }{2} \right) \right]+b\text{ }\left( \sin \alpha -\sin \beta \right)=0\]
Also, applying the factorization or sum to product formula\[\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)\], we get
\[a\left[ 2\sin \left( \dfrac{\alpha +\beta }{2} \right)\sin \left( \dfrac{\beta -\alpha }{2} \right) \right]+b\left[ 2\cos \left( \dfrac{\alpha +\beta }{2} \right)\sin \left( \dfrac{\alpha -\beta }{2} \right) \right]=0..........(3)\]
We know that, $\sin (-\theta )=-\sin \theta $
$\sin \left( \dfrac{\alpha -\beta }{2} \right)=\sin \left[ -\left( \dfrac{\beta -\alpha }{2} \right) \right]=-\sin \left[ \left( \dfrac{\beta -\alpha }{2} \right) \right]$
Now put this value in the equation (3), we get
\[a\left[ 2\sin \left( \dfrac{\alpha +\beta }{2} \right)\sin \left( \dfrac{\beta -\alpha }{2} \right) \right]-b\left[ 2\cos \left( \dfrac{\alpha +\beta }{2} \right)\sin \left( \dfrac{\beta -\alpha }{2} \right) \right]=0..........(3)\]
Dividing both sides by $2\sin \left( \dfrac{\beta -\alpha }{2} \right)$ , we get
\[a\sin \left( \dfrac{\alpha +\beta }{2} \right)-b\cos \left( \dfrac{\alpha +\beta }{2} \right)=0..................(4)\]
It is given that $\alpha -\beta \ne 2n\pi $ and dividing both sides by 2, we have
$\dfrac{\alpha -\beta }{2}\ne n\pi $
Taking sine on the both sides, we get
$\sin \left( \dfrac{\alpha -\beta }{2} \right)\ne \sin n\pi $
We know that
$\sin n\pi =0$ for all values of n
$\sin \left( \dfrac{\alpha -\beta }{2} \right)\ne \sin n\pi =0$
Dividing equation (4) by $\cos \left( \dfrac{\alpha +\beta }{2} \right)$ , we get
$\tan \left( \dfrac{\alpha +\beta }{2} \right)=\dfrac{b}{a}..............(5)$
The cosine double angle formula tells us that $\cos \theta $ is always equal to, \[\cos \theta =\dfrac{1-{{\tan }^{2}}\left( \dfrac{\theta }{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{\theta }{2} \right)}\]
Now put $\theta =\alpha +\beta $ in the above formula, we get
\[\cos (\alpha +\beta )=\dfrac{1-{{\tan }^{2}}\left( \dfrac{\alpha +\beta }{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{\alpha +\beta }{2} \right)}...........(6)\]
Put the value of $\tan \left( \dfrac{\alpha +\beta }{2} \right)$ is $\dfrac{b}{a}$ in the equation (6), we get
\[\cos (\alpha +\beta )=\dfrac{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
\[\cos (\alpha +\beta )=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
Hence proved.
Note: The sine double angle formula tells us that $\sin \theta $ is always equal to $\dfrac{2\tan \dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}$ . In this you can put $\theta =\alpha +\beta $ and we have
\[\sin \left( \alpha +\beta \right)=\dfrac{2\tan \left( \dfrac{\alpha +\beta }{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{\alpha +\beta }{2} \right)}=\dfrac{2\dfrac{b}{a}}{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}=\dfrac{2b}{\dfrac{{{a}^{2}}+{{b}^{2}}}{a}}=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}\]