Question

If $\alpha ,\beta$ are the different values of $\Theta$ satisfying the equation $3cos\Theta +4sin\Theta =\frac{9}{2}$ then $sin(\alpha +\beta )=....(if0<\alpha ,\beta <\frac{\pi }{2})$

Answer 1

We have,

$3\cos \theta +4\sin \theta =\frac{9}{2}$

$6\cos \theta +8\sin \theta =9$

Let $\alpha and\beta$ are the different value$\theta$

Then,

$3\cos \alpha +4\sin \alpha =\frac{9}{2}$

$6\cos \alpha +8\sin \alpha =9......\left(1\right)$

$6\cos \beta +8\sin \beta =9......\left(2\right)$

Subtract equation (2) from (1) and we get,

$\Rightarrow 6(\cos \alpha -\cos \beta )+8(\sin \alpha -\sin \beta )=0$

$\Rightarrow 6(\cos \alpha -\cos \beta )=-8(\sin \alpha -\sin \beta )$

$\Rightarrow 3(\cos \beta -\cos \alpha )=4(\sin \alpha -\sin \beta )$

$\Rightarrow \frac{(\sin \alpha -\sin \beta )}{(\cos \beta -\cos \alpha )}=\frac{3}{4}$

$\Rightarrow \frac{2\cos \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)}{2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)}=\frac{3}{4}$

$\Rightarrow \frac{\cos \left(\frac{\alpha +\beta }{2}\right)}{\sin \left(\frac{\alpha +\beta }{2}\right)}=\frac{3}{4}$

$\Rightarrow \cot \left(\frac{\alpha +\beta }{2}\right)=\frac{3}{4}$

Squaring both side and we get,

$\Rightarrow {\cot }^2\left(\frac{\alpha +\beta }{2}\right)={\left(\frac{3}{4}\right)}^2$

$\Rightarrow {\csc }^2\left(\frac{\alpha +\beta }{2}\right)-1={\left(\frac{3}{4}\right)}^2\therefore {\csc }^2\theta -{\cot }^2\theta =1$

$\Rightarrow {\csc }^2\left(\frac{\alpha +\beta }{2}\right)=\frac{9}{16}+1$

$\Rightarrow {\csc }^2\left(\frac{\alpha +\beta }{2}\right)=\frac{25}{16}$

$\Rightarrow {\sin }^2\left(\frac{\alpha +\beta }{2}\right)=\frac{16}{25}$

$\Rightarrow \sin \left(\frac{\alpha +\beta }{2}\right)=\frac{4}{5}$

But we know that,

${\sin }^2\theta +{\cos }^2\theta =1$

${\cos }^2\theta =1-{\sin }^2\theta$

$\cos \theta =\sqrt{1-{\sin }^2\theta }$

$\cos \left(\frac{\alpha +\beta }{2}\right)=\sqrt{1-{\sin }^2\left(\frac{\alpha +\beta }{2}\right)}$

$=\sqrt{1-\frac{16}{25}}$

$=\sqrt{\frac{9}{25}}$

$\cos \left(\frac{\alpha +\beta }{2}\right)=\frac{3}{5}$

Now,

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

$=2\times \frac{4}{5}\times \frac{3}{5}$

$=\frac{24}{25}$

$\sin (\alpha +\beta )=\frac{24}{25}$

Hence, this is the answer.