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nikdorinn [45]
3 years ago
8

Reference angle for -19pi/6? 

Mathematics
1 answer:
kow [346]3 years ago
7 0
\bf \cfrac{19}{6}\implies 3+\cfrac{1}{6}
\\ \quad \\\\
\cfrac{-19\pi }{6}\implies -\left(3\pi +\cfrac{1\pi }{6}  \right)\impliedby clockwise
\\ \quad \\\\
or\implies -\left(2\pi +\pi +\cfrac{\pi }{6}  \right)

go that much "clockwise", you'll end up at the 2nd quadrant,
the reference angle, is its reflective angle on the 1st quadrant

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Answer:

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3 years ago
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The result of expanding the trigonometry expression \sin^2(\theta) * (1 + \cos(\theta)) is cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

<h3>How to evaluate the expression?</h3>

The expression is given as:

\sin^2(\theta) * (1 + \cos(\theta))

Express \sin^2(\theta) as 1 - \cos^2(\theta).

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Open the bracket

\sin^2(\theta) * (1 + \cos(\theta)) =  1 + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Express 1 as cos°(Ф)

\sin^2(\theta) * (1 + \cos(\theta)) =  cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Hence, the result of expanding the trigonometry expression \sin^2(\theta) * (1 + \cos(\theta)) is cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)

Read more about trigonometry expressions at:

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3 years ago
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aleksandrvk [35]
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