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Am I able to evaluate sin(cos^1(7/11)) without a calculator? If yes, how?
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Is that supposed to be cosine inverse?
Yes
\[\text{Sin}\left[\text{ArcCos}\left[\frac{7}{11}\right]\right]\]
You want to draw the right triangle associated with that. Im trying to think of a better way to explain it >.<
\[\cos^{-1}(\frac{7}{11}) \Rightarrow \cos(\theta) = \frac{7}{11}\] and in a right triangle: \[\cos(\theta) = \frac{adj}{hyp}\] You want to draw that right triangle.
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\[cos(\theta) = \frac{7}{11} = \frac{adj}{hyp}\] \[adj^2 + opp^2 = hyp^2\]
\[sin(\theta) = \frac{opp}{hyp}\]
so find opp using the Pythagorean triangle and then put it over the hypotenuse.
6(rad2)/11
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