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Trigonometry question
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|dw:1453799042511:dw|
\[\cos \alpha = -\frac{4}5= 2 \cos^2 \alpha/2 - 1\]\[\pi/2 \le \alpha/2 \le 3\pi/4 \Rightarrow \cos \alpha/2 < 0\]
sin(37)=3/5 sin(180+37)=-3/5
\(\large \color{black}{\begin{align} &\cos \dfrac{\alpha}{2}=\dfrac{1}{\sqrt{10}} \text{}\hspace{.33em}\\~\\ \end{align}}\)
I told you that \(\cos\alpha/2\) is negative because \( \pi/2\le \alpha/2\le 3\pi/4\). On the unit circle, those angles have a negative x-coordinate.
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\(\large \color{black}{\begin{align} &\cos \dfrac{\alpha}{2}=-\dfrac{1}{\sqrt{10}} \text{}\hspace{.33em}\\~\\ \end{align}}\)
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