Trig Identity

**RosieLaird** · Aug 8th 2010, 10:41 AM

Verify:
Sin(α/3)Cos(α/3)=.5Sin(2α/3)

**eumyang** · Aug 8th 2010, 10:46 AM

Use the sine double-angle identity.
$\begin{aligned} \tfrac{1}{2}\sin \left(\tfrac{2\alpha}{3} \right) &= \tfrac{1}{2}\sin \left(2 \cdot \tfrac{\alpha}{3} \right) \\ &= \tfrac{1}{2} \cdot 2 \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right) \\ &= \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right) \end{aligned}$

**RosieLaird** · Aug 8th 2010, 10:54 AM

Originally Posted by eumyang

Use the sine double-angle identity.
$\begin{aligned} \tfrac{1}{2}\sin \left(\tfrac{2\alpha}{3} \right) &= \tfrac{1}{2}\sin \left(2 \cdot \tfrac{\alpha}{3} \right) \\ &= \tfrac{1}{2} \cdot 2 \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right) \\ &= \sin \left(\tfrac{\alpha}{3} \right) \cos \left(\tfrac{\alpha}{3} \right) \end{aligned}$

In the second step how did you get Cos to appear?

**eumyang** · Aug 8th 2010, 11:04 AM

Like I said, I used the sine double angle identity:
$\sin\,2\theta = 2\sin\,\theta\,\cos\,\theta$ .
You should have learned this. If not, you could still verify the problem using the sine of a sum identity instead:
$\sin\,(u + v) = \sin\,u\,\cos\,v + \cos\,u\,\sin\,v$ .

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