Please help me with this trig question
- Thread starter ryan882
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Using a product to sum identity, we may write:
\(\displaystyle 4\sin\left(x+\dfrac{\pi}{6} \right)\sin\left(x-\dfrac{\pi}{6} \right)=2\left(\cos\left(\dfrac{\pi}{3} \right)-\cos\left(2x \right) \right)\)
Or, using the angle sum identity for sine, we may write:
\(\displaystyle 4\sin\left(x+\dfrac{\pi}{6} \right)\sin\left(x-\dfrac{\pi}{6} \right)=4\left(\sin^2(x)\cos^2\left(\dfrac{\pi}{6} \right)-\cos^2(x)\sin^2\left(\dfrac{\pi}{6} \right) \right)\)
\(\displaystyle 4\sin\left(x+\dfrac{\pi}{6} \right)\sin\left(x-\dfrac{\pi}{6} \right)=2\left(\cos\left(\dfrac{\pi}{3} \right)-\cos\left(2x \right) \right)\)
Or, using the angle sum identity for sine, we may write:
\(\displaystyle 4\sin\left(x+\dfrac{\pi}{6} \right)\sin\left(x-\dfrac{\pi}{6} \right)=4\left(\sin^2(x)\cos^2\left(\dfrac{\pi}{6} \right)-\cos^2(x)\sin^2\left(\dfrac{\pi}{6} \right) \right)\)
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No; "pi" stands for \(\displaystyle \pi\). "Pie" stands for a tasty pastry treat.
