solve f(x)=sin(2arccos(1/4))
This is a function of x? There is no x :o That's strange
So anyway we can ummm...
\[\Large\rm \sin\left[2\color{orangered}{\arccos\left(\frac{1}{4}\right)}\right]\]Work from the inside out. This inverse cosine is equal to some angle, let's call it theta.\[\Large\rm \color{orangered}{\arccos\left(\frac{1}{4}\right)=\theta}\]
\[\Large\rm \implies \qquad \cos \theta=\frac{1}{4}\]
We'll draw a triangle which satisfies this relationship.
|dw:1415857655849:dw|
Applying our Pythagorean Theorem gives us the missing side:|dw:1415857683397:dw|
Then we can think of our problem:\[\Large\rm \sin\left[2\color{orangered}{\arccos\left(\frac{1}{4}\right)}\right]\]As:\[\Large\rm \sin\left[2\color{orangered}{\theta}\right]\]Applying our Sine Double Angle Identity:\[\Large\rm =2\sin \theta \cos \theta\]And use the triangle to finish it up.
I still don't see what this has to do with x. Is it a constant function? Or were you plugging in some value for x?
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