The value of sin230°÷sin245°is

A
2
B
$\frac{1}{2}$
C
$\sqrt{3}$
D
$\frac{1}{\sqrt{3}}$

Solution:

$\begin{array}{l} \sin^{2} 30 ° = {(\frac{1}{2})}^{2} = \frac{1}{4} \\ \sin^{2} 45 ° = {(\frac{1}{\sqrt{2}})}^{2} = \frac{1}{2} \\ ∴ \sin^{2} 30 ° \div \sin^{2} 45 ° = \frac{1}{4} \div \frac{1}{2} \\ = \frac{(\frac{1}{4})}{(\frac{1}{2})} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2} \end{array}$

The value of $\sin^{2} 30 ° \div \sin^{2} 45 °$ is

Solution:

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