Question

$\alpha ,\beta ,\gamma$ and $\delta$ are the smallest positive angle in ascending order of magnitude which have their sines equal to the positive quantity $x$. The value of $4\sin \frac{\alpha }{2}+\sin \frac{\beta }{2}+2\sin \frac{\gamma }{2}+\sin \frac{\delta }{2}$ is equal to

• A. $2\sqrt{1+x}$
• B. $2\sqrt{1-x}$
• C. $2\sqrt{x}$
• D. $2x$

Answer 1

Answer: (A), (B)

The correct options are
A $2\sqrt{1+x}$
B $2\sqrt{1-x}$

$sin\alpha =sin\beta =sin\gamma =sin\delta =k$

$\beta =\pi -a\Rightarrow \frac{\beta }{2}=\frac{\pi }{2}-\frac{a}{2}$

$r=2\pi +\alpha \Rightarrow \frac{r}{2}+\pi +\frac{\alpha }{2}$

$\delta =3\pi -\alpha \Rightarrow \frac{\delta }{2}=\frac{3\pi }{2}-\frac{\alpha }{2}$

$4sin\frac{\alpha }{2}+3sin\frac{\beta }{2}+2sin\frac{\gamma }{2}+sin\frac{\delta }{2}=4sin\frac{\alpha }{2}+3cos\frac{\alpha }{2}-2sin\frac{\alpha }{2}$

$-cos\frac{\alpha }{2}=2sin\frac{\alpha }{2}+2cos\frac{\alpha }{2}$

Let $y=2[sin\frac{\alpha }{2}+cos\frac{\alpha }{2}]$

$=\sqrt[2]{2(sin\frac{a}{2}+cos\frac{a}{2}{)}^2}=2\sqrt{1+sina}=2\sqrt{1+x}$