Question 34

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Jambmaths question: 

Find the value of sin45o – cos 30o

Jamb Maths Solution: 

$\begin{align}  & \sin {{45}^{\circ }}-\cos {{30}^{\circ }}=\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{6}}{2\sqrt{2}}=\frac{2-\sqrt{6}}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ & \sin {{45}^{\circ }}-\cos 30=\frac{2\sqrt{2}-\sqrt{12}}{4}=\frac{2\sqrt{2}-2\sqrt{3}}{4}=\frac{2(\sqrt{2}-\sqrt{3})}{4} \\ & \sin {{45}^{\circ }}-\cos 30=\frac{\sqrt{2}-\sqrt{3}}{2} \\\end{align}$

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