Question

If $\sin \alpha =\frac{336}{625}$ and ${450}^0<\alpha <{540}^0$ , then find the value of $\sin \frac{\alpha }{4}$ .

Answer 1

Putting $\sin \left(\alpha /4\right)=x$, we get $\cos \left(\alpha /2\right)=1-2x^2$ and $\cos \alpha =2(1-2x^2{)}^2-1=1-8x^2+8x^4$ ..... $\left(1\right)$

Futhenmore, $\sin \alpha =336/625$ implies

$\cos \alpha =\pm \sqrt{1-{\left(\frac{336}{625}\right)}^2}$

Choosing the minus sign because $\alpha$ lies in the second quadrant we get

$\cos \alpha =-\sqrt{\frac{(625+336)(625-336)}{(625{)}^2}}$

$=-\sqrt{\frac{\left(961\right)\left(289\right)}{(625{)}^2}}=\frac{\left(31\right)\left(17\right)}{625}=-\frac{527}{625}$

Putting this in $\left(1\right)$ gives $1-8x^2+8x^4=-527/625$

$\Rightarrow x^4-x^2+144/625=0$

$\Rightarrow x^2=\frac{1}{2}[1\pm \sqrt{1-\frac{\left(4\right)\left(144\right)}{625}}]=\frac{1}{2}(1\pm \frac{7}{25})$

i.e $x^2=16/25$ or $9/25$.Since $x=\sin \left(\alpha /4\right)$ is positive for the specified values of $\alpha$. we get $\sin \left(\alpha /4\right)=4/5$ or $3/5$.

Now, ${450}^o<\alpha <{540}^o$ means the angle $\alpha /4$ lies in the second quadrant with ${112.5}^o<\alpha /4<{135}^o$

Therefore we must have

$\sin {135}^o<\sin \alpha /4<\sin {112.5}^o$

$\Rightarrow \sin \alpha /4>1/\surd 2\Rightarrow \sin \alpha /4>0.7$.

This excludes the values $\sin \alpha =3/5=0.6$, and the answer is $\sin \left(\alpha /4\right)=4/5$