Respuesta :
Answer:
[tex]\cos(\frac{\pi}{4})\cos(\frac{\pi}{6})=\frac{1}{2}(\cos(\frac{\pi}{12})+\cos(\frac{5\pi}{12}))[/tex]
So the blank is cos.
Step-by-step explanation:
There is an identity for this:
[tex]\cos(a)\cos(b)=\frac{1}{2}(\cos(a+b)+\cos(a-b))[/tex]
Let's see if this is fit by your left hand and right hand side:
So [tex]a=\frac{\pi}{4}[/tex] while [tex]b=\frac{pi}{6}[/tex].
Let's plug these in to the identity above:
[tex]\cos(\frac{\pi}{4})\cos(\frac{\pi}{6})=\frac{1}{2}(\cos(\frac{\pi}{4}+\frac{\pi}{6})+\cos(\frac{\pi}{4}-\frac{\pi}{6}))[/tex]
Ok, we definitely have the left hand sides are the same.
Let's see if the right hand sides are the same.
Before we move on let's see if we can find the sum and difference of [tex]\frac{\pi}{4}[/tex] and [tex]\frac{\pi}{6}[/tex].
We will need a common denominator. How about 12? 12 works because 4 and 6 go into 12. That is 4(3)=12 and 6(2)=12.
[tex]\frac{\pi}{4}+\frac{\pi}{6}=\frac{3\pi}{12}+\frac{2\pi}{12}=\frac{5\pi}{12}[/tex].
[tex]\frac{\pi}{4}-\frac{\pi}{6}=\frac{3\pi}{12}-\frac{2\pi}{12}=\frac{\pi}{12}[/tex].
Let's go back to our identity now:
[tex]\cos(\frac{\pi}{4})\cos(\frac{\pi}{6})=\frac{1}{2}(\cos(\frac{\pi}{4}+\frac{\pi}{6})+\cos(\frac{\pi}{4}-\frac{\pi}{6}))[/tex]
[tex]\cos(\frac{\pi}{4})\cos(\frac{\pi}{6})=\frac{1}{2}(\cos(\frac{5\pi}{12})+\cos(\frac{\pi}{12}))[/tex]
We can rearrange the right hand side inside the ( ) using commutative property of addition:
[tex]\cos(\frac{\pi}{4})\cos(\frac{\pi}{6})=\frac{1}{2}(\cos(\frac{\pi}{12})+\cos(\frac{5\pi}{12}))[/tex]
So comparing my left hand side to their left hand side we see that the blank should be cos.
