Question on: JAMB Mathematics - 2009
Find the value of sin 45o - cos 30o
A
\(\frac{2 + \sqrt{6}}{4}\)
B
\(\frac{ \sqrt{2} + \sqrt{3}}{4}\)
C
\(\frac{ \sqrt{2} + \sqrt{3}}{6}\)
D
\(\frac{ \sqrt{2} - \sqrt{3}}{2}\)
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Correct Option: D
sin 45o - cos 30o
\(\frac{1}{\sqrt{2}}\) - \(\frac{\sqrt{3}}{2}\)
= \(\frac{2 - \sqrt{2} \times \sqrt{3}}{2\sqrt{2}})\)
= \(\frac{2 - \sqrt{6} \times \sqrt{2}}{2\sqrt{2} \sqrt{2}}\)
= \(\frac{2\sqrt{2} - \sqrt{12}}{4}\)
= \(\frac{2\sqrt{2} - 2\sqrt{3}}{4}\)
= \(\frac{2(\sqrt{2} - \sqrt{3})}{4}\)
= \(\frac{\sqrt{2} - \sqrt{3}}{2}\)
\(\frac{1}{\sqrt{2}}\) - \(\frac{\sqrt{3}}{2}\)
= \(\frac{2 - \sqrt{2} \times \sqrt{3}}{2\sqrt{2}})\)
= \(\frac{2 - \sqrt{6} \times \sqrt{2}}{2\sqrt{2} \sqrt{2}}\)
= \(\frac{2\sqrt{2} - \sqrt{12}}{4}\)
= \(\frac{2\sqrt{2} - 2\sqrt{3}}{4}\)
= \(\frac{2(\sqrt{2} - \sqrt{3})}{4}\)
= \(\frac{\sqrt{2} - \sqrt{3}}{2}\)
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