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pychu [463]
3 years ago
6

I am currently working on Trigonometry, but unfortunately I am having a hard time. Here are some of my questions I've answered s

o far. Could someone please confirm if it is correct, and if not, please help me find the correct answer. Thank you very much.

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
3 0
\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
-------------------------------\\\\
cos(\theta)=\cfrac{\sqrt{3}}{3}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}
\\\\\\
\textit{using the pythagorean theorem}\implies c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b
\\\\\\
\pm\sqrt{3^2-(\sqrt{3})^2}=b\implies \pm\sqrt{6}=b

now the root gives us the +/-, which is it? well, we know the sine is negative, so the "b" value must be negative, so it has to be -√6 = b

thus  \bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(\theta)=-\cfrac{\sqrt{6}}{3}
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