For example, we're given a problem in which sin $\theta = \sqrt3/2$ and cos $\theta = -1/2$. To find out the angle $\theta$, I look at the unit circle and I get the answer. However, I was just curious whether there's an alternative to this, any idea? Because when I tried using cos(-$\theta$) = cos$\theta$, I get the wrong value of $\theta$ as we've been provided with the value of sin $\theta$ as well...
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asked Sep 13 '14 at 8:15
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$\begingroup$ Why duplicating your question of yesterday: math.stackexchange.com/questions/929151/… ? $\endgroup$ – Christian Blatter Sep 13 '14 at 12:08
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$\begingroup$ @ChristianBlatter Sorry, no idea why I did that! $\endgroup$ – Always Learning Forever Sep 13 '14 at 12:10
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$\begingroup$ any help here, if you can, will be really appreciated: chemistry.stackexchange.com/questions/16259/… $\endgroup$ – Always Learning Forever Sep 13 '14 at 12:11
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$$\sin\theta=\frac{\sqrt3}2=\sin\frac\pi3\implies\theta=n\pi+(-1)^n\frac\pi3$$ where $n$ is any integer
Set $n=2s+1,$(odd) $=2s$(even) one by one
Again, $$\cos\theta=-\frac12=-\cos\frac\pi3=\cos\left(\pi-\frac\pi3\right)$$ $$\implies\theta=2m\pi\pm\left(\pi-\frac\pi3\right)$$ where $m$ is any integer
Check for '+','-' one by one
Observe that the intersection of the above two solutions is $$\theta=2r\pi+\frac{2\pi}3$$ where $r$ is any integer
answered Sep 13 '14 at 8:19
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The given data determine $\theta$ up to an additive multiple of $2\pi$. When $\cos\theta\ne-1$ the principal value $\theta\in\ ]{-\pi},\pi[\ $ can be found using the formula $$\tan{\theta\over2}={\sin\theta\over 1+\cos\theta}\ ,$$ which leads to $$\theta=2\arctan{\sin\theta\over 1+\cos\theta}\ .$$ In your example we get $$\theta=2\arctan{{\sqrt{3}\over2}\over 1-{1\over2}}=2\arctan\sqrt{3}={2\pi\over3}\ .$$
answered Sep 13 '14 at 9:56
