How would you transform f(x) = 3 sin(4x - π) + 4 into a cosine function in the form f(x) = a cos(bx - c) + d?
\[Sin(\frac{\pi}{2} - x) = Cos(x)\] But we have to be careful, as you have a 4 in front of your x. We'll also need: Sin(-x) = - Sin(x)
So what do we do with the 4?
Glad I checked on that, It actually doesn't cause us a problem :) So first, let's rewrite sin(4x-pi) by factoring out a negative from the argument.
For example: \[Sin(x-a) = Sin(-(a-x)) = - Sin(a-x)\] We want it to look something like the form: \[Sin(\frac{\pi}{2} - x)\]
Wait... but if I factor out a negative, I get f(x)= 3-sin(-4x + pi)...
Unless that's right...
We aren't done yet, but the 3 is multiplying the Sin, it looks like, so you'd get: \[-3 Sin(\pi - 4x)\]
There is another way to do this, if you'd rather use the angle addition formulas. Sin(a+b) = Sin(a)Cos(b)+Cos(a)Sin(b)
I don't care. I just need to know how to do it this one time.
@Vandreigan
Well, let's use that way. I think it'll be a little more clear. Use that formula on your original Sin(4x-pi). a = 4x, b = -pi
Sin(4x-pi) = Sin(4x)Cos(-pi) + Cos(4x)Sin(-pi) right?
Right! Now, we know what Cos(-pi) and Sin(-pi) are, so just evaluate those.
Sin(4x)(-1) + Cos(4x)(0)
Yep, which equals what?
-sin(4x) + 0 -sin(4x)
But I have to make it a cosine function... how do I do that?
Right, sorry. We know that Sin(x) = Cos(pi/2 - x) Using this, we can make it a cosine.
-sin(4x) = cos(pi/2-4x) Where would the negative sign go?
You keep the negative sign in front. -sin(4x) = -cos(pi/2 - 4x)
Now, notice we didn't do anything to the +4 to the right. It just tags along, as well.
So the answer would be f(x)= -cos(pi/2-4x) +4
That's the one :) There's another answer, as well: -sin(4x)+4 = sin(-4x)+4 = cos(pi/2 + 4x) + 4 in case they don't want to take that one (both are correct)
Okay! THANKS SO MUCH!!!! (:
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