6sin2theta-8sincube2theta
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It's a question from trigonometry. We are going to evaluate the given expression by using identities.
We know that, sin3x = 3sinx 4sin'x.
Now,
6 sin 20° - 8 sin 20
= 2 (3 sin 20" - 4 sin 20°)
= 2( sin ( 320)")
= 2 ( sin 60") %3!
=2(3/2)
= 3.
Therefore, 6 sin 20" - 8 sin 20° = V3.
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