How to turn 9 sin120 into (9*(squareroot3))/2

**anees7112** · February 12th 2011, 05:44 AM

Hello

How would I go about converting 9 sin120 into the form with a surd:
9*(squareroot of 3)
-------------------
............2
?
Thank you

**Prove It** · February 12th 2011, 05:47 AM

You should know that $\displaystyle \sin{\left(180^{\circ} - \theta^{\circ}\right)} \equiv \sin{\theta^{\circ}}$ for $\displaystyle 0^{\circ} < \theta^{\circ} < 90^{\circ}$ .

**skeeter** · February 12th 2011, 05:52 AM

Originally Posted by anees7112

Hello

How would I go about converting 9 sin120 into the form with a surd:
9*(squareroot of 3)
-------------------
............2
?
Thank you

value from the unit circle ...

$\sin(120^\circ) = \dfrac{\sqrt{3}}{2}$

**e^(i*pi)** · February 12th 2011, 06:45 AM

Although the other two methods are easier you can use the fact that $\sin(A+B) = \sin A \cos B + \cos A \sin B \text{ where } A = 90^{\circ} \text{ and } B = 30^{\circ}$

30 and 90 are, of course, found on a 30-60-90 triangle

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