Question

Without using tables , find the value of sin600cos600

A 3√
B 3
C 1
D4√

Answer 1

Question #51816

Without using tables, find the value of $\sin 600{}^{\circ} \cos 600{}^{\circ}$.

Solution.

Using double angle formula for $\sin$, we have:

$$\sin 600{}^{\circ} \cos 600{}^{\circ} = \frac{1}{2} \cdot 2 \cdot \sin 600{}^{\circ} \cos 600{}^{\circ} = \frac{1}{2} \cdot \sin 2 \cdot 600{}^{\circ} = \frac{1}{2} \sin 1200{}^{\circ}$$ $$\frac{1}{2} \sin 1200{}^{\circ} = \frac{1}{2} \sin (1080{}^{\circ} + 120{}^{\circ})$$

Using, that sine has the smallest positive period $360{}^{\circ}$, we can rewrite:

$$\frac{1}{2} \sin (1080{}^{\circ} + 120{}^{\circ}) = \frac{1}{2} \sin (3 \cdot 360{}^{\circ} + 120{}^{\circ}) = \frac{1}{2} \sin 120{}^{\circ} = \frac{1}{2} \sin (180{}^{\circ} - 60{}^{\circ}) = \frac{1}{2} \sin 60{}^{\circ}$$ $$\frac{1}{2} \sin 60{}^{\circ} = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$$

Answer: $\frac{\sqrt{3}}{4}$

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