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weeeeeb [17]
3 years ago
5

HELP!!! I don't know how to do this!

Mathematics
1 answer:
alina1380 [7]3 years ago
4 0

\bf sin(\theta )=\cfrac{1}{2}\implies \measuredangle \theta =sin^{-1}\left( \cfrac{1}{2} \right)\implies \theta \measuredangle = \begin{cases} \frac{\pi }{6}\\\\ \frac{5\pi }{6} \end{cases}\\\\ -------------------------------\\\\ cot(\theta )=-1\implies \measuredangle \theta =cot^{-1}(-1)\implies \measuredangle \theta = \begin{cases} \frac{3\pi }{4}\\\\ \frac{7\pi }{4} \end{cases}


\bf -------------------------------\\\\ cos(3\theta )=\cfrac{1}{2}\implies 3\theta =cos^{-1}\left( \cfrac{1}{2} \right)\implies 3\theta = \begin{cases} \frac{\pi }{3}\\\\ \frac{5\pi }{3} \end{cases} \\\\\\ 3\theta =\cfrac{\pi }{3}\implies \theta =\cfrac{\pi }{9}\qquad \qquad ,\qquad \qquad 3\theta =\cfrac{5\pi }{3}\implies \theta =\cfrac{5\pi }{9}



for the second one, cot(θ) = -1, bear in mind that cotangent is cosine/sine, so where both of them are the same values BUT different signs, you'd get a -1 from their fraction, and that happens at halfway in II and IV Quadrants, since on those quadrants both of them are of different signs.

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