Find the exact value of arccos (sin (pi divided by six)). For full credit, explain your reasoning.
I need help explaining. I have the answer, but I need to get this question correct!
Lemme explain something really quick in regards to trig functions.
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If I were to ask you for \(\Large\rm \cos\theta\), you would know that it's 3/5. But more specifically, when taking the trig function of an angle, we're `solving for a numerical value`, the ratio of sides. If I asked for \(\Large\rm \arccos\dfrac{3}{5}\), This gives us back our angle \(\Large\rm \theta\). So when you take the inverse trig function of something, you're `solving for an angle`.
So with that in mind,\[\Large\rm \arccos\left(\sin \frac{\pi}{6}\right)\]What does sin(pi/6) give you?
1/2
\[\Large\rm \arccos\left(\frac{1}{2}\right)\]Mmmm ok good. Since we're taking the inverse trig function of this value, we're solving for an angle yes? \[\Large\rm \arccos\left(\frac{1}{2}\right)=\theta\]
Do you remember how to switch between ... like cosine and arccosine? You just switch the arguments.\[\Large\rm \cos \theta=(stuff) \qquad\to\qquad \arccos(stuff)=\theta\]
yeah
So we can write our inverse function back in terms of cosine. \[\Large\rm \cos \theta=\frac{1}{2}\] Do you remember which special angle gives you 1/2?
30
I made a tiny boo boo I guess. Since we `started with` the inverse cosine function, we have to remember that our output is restricted to quadrants 1 and 2. Ummm I think it's 60 isn't it? It turns out that 300 also gives us 1/2 but that is in quadrant 4 so that one doesn't work for us.
Are you talking about in the Unit Circle? Cause the sin (π/6) gives us 1/2 which is the y value of 30 degrees the Unit Circle
Correct, but now we're dealing with cosine, not sine.
The inner function was sin(30) which gave us 1/2. Now we're trying to figure out what angle when we take the cosine of it, gives us 1/2. It's not 30 again. :o
It's 60. :p now that I check my Unit Circle
|dw:1402890063191:dw|Ah yes :) good!
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