Question

If the sides of a right-angled triangle are $\{\cos 2\alpha +\cos 2\beta +2\cos (\alpha +\beta )\}$ and $\{\sin 2\alpha +\sin 2b+2\sin (\alpha +\beta \left)\right\},$ then the length of the hypotenuse is:

• A. $4[1+\cos (\alpha -\beta \left)\right]$
• B. $4[1-\cos (\alpha +\beta \left)\right]$
• C. $4{\cos }^2\frac{\alpha -\beta }{2}$
• D. $4{\sin }^2\frac{\alpha +\beta }{2}$

Answer 1

Answer: (C)

$4{\cos }^2\frac{\alpha -\beta }{2}$

Let, $a,b,c$ be the sides of a right angled triangle. $c$ is the Hypotenuse.

Let,

$a=\cos 2\alpha +\cos 2\beta +2\cos (\alpha +\beta )$

$b=\sin 2\alpha +\sin 2\beta +2\sin (\alpha +\beta )$

We know that,

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

$\sin 2\theta =2\sin \theta \cos \theta$

$\cos (a+b)=\cos a\cos b-\sin a\sin b$

$\sin (a+b)=\sin a\cos b+\cos a\sin b$

Consider,

$a=\cos 2\alpha +\cos 2\beta +2\cos (\alpha +\beta )$

$a={\cos }^2\alpha -{\sin }^2\alpha +{\cos }^2\beta -{\sin }^2\beta +2(\cos \alpha \cos \beta -\sin \alpha \sin \beta )$

$a=(\cos \alpha +\cos \beta {)}^2-(\sin \alpha +\sin \beta {)}^2$

Consider,

$b=\sin 2\alpha +\sin 2\beta +2\sin (\alpha +\beta )$

$b=2\sin \alpha \cos \alpha +2\sin \beta \cos \beta +2(\sin \alpha \cos \beta +\cos \alpha \sin \beta )$

$b=2(\sin \alpha +\sin \beta )(\cos \alpha +\cos \beta )$

From Pythagoras theorem,

$c^2=a^2+b^2$

$c^2=\left(\right(\cos \alpha +\cos \beta {)}^2-(\sin \alpha +\sin \beta {)}^2{)}^2+(2(\sin \alpha +\sin \beta )(\cos \alpha +\cos \beta ){)}^2$

$c^2=(\cos \alpha +\cos \beta {)}^4+(\sin \alpha +\sin \beta {)}^4-2(\cos \alpha +\cos \beta {)}^2(\sin \alpha +\sin \beta {)}^2+4(\sin \alpha +\sin \beta {)}^2(\cos \alpha +\cos \beta {)}^2$

$c^2=(\cos \alpha +\cos \beta {)}^4+(\sin \alpha +\sin \beta {)}^4+2(\sin \alpha +\sin \beta {)}^2(\cos \alpha +\cos \beta {)}^2$

$c^2=(\cos \alpha +\cos \beta {)}^4+(\sin \alpha +\sin \beta {)}^4+2(\sin \alpha +\sin \beta {)}^2(\cos \alpha +\cos \beta {)}^2$

$c^2=(\cos \alpha +\cos \beta {)}^4+(\sin \alpha +\sin \beta {)}^4+2(\sin \alpha +\sin \beta {)}^2(\cos \alpha +\cos \beta {)}^2$

$c^2=\left(\right(\cos \alpha +\cos \beta {)}^2+(\sin \alpha +\sin \beta {)}^2{)}^2$

Taking square root on both sides, we get:

$c=(\cos \alpha +\cos \beta {)}^2+(\sin \alpha +\sin \beta {)}^2$

$c={\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta +{\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta$

$c=2+2\cos \alpha \cos \beta +2\sin \alpha \sin \beta$ --------( ${\sin }^2\theta +{\cos }^2\theta =1$ )

$c=2(1+\cos \alpha \cos \beta +\sin \alpha \sin \beta )$

$c=2(1+\cos (\alpha -\beta \left)\right)$

$c=4{\cos }^2\frac{\alpha -\beta }{2}$

This is the required length of the hypotenuse.