Question

Trigonometry:
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?

Answer 1

@Michele_Laino

Answer 2

@Hero @dan815 @paki

Answer 3

From my understanding, I'd just have to change sine to cosine and add -pi/2

Answer 4

So would it just be \[f(x)=3\cos(4x- \pi/2)+4\] ???

Answer 5

@perl

Answer 6

hint:
\[\begin{gathered}
\sin \left( {4x - \pi } \right) = \sin \left( {4x} \right)\cos \pi - \cos \left( {4x} \right)\sin \pi = \hfill \\
= - \sin \left( {4x} \right) \hfill \\
\end{gathered} \]

Answer 7

i don't get it

Answer 8

That doesn't put it into the form f(x) = a cos(bx-c)+d

Answer 9

that is the first step

Answer 10

oh okay thats a trig identity right?

Answer 11

yes! It is the subtraction formula for sin function

Answer 12

sin(α - β) = sinα cos β - cos α sin β ?

Answer 13

ok!

Answer 14

then what goes after?

Answer 15

I'm trying to write the subsequent step, please wait

Answer 16

okay

Answer 17

here is the next step:
\[\Large \sin \left( {4x} \right) = \cos \left( {4x - \frac{\pi }{2}} \right)\]
@genesis98

Answer 18

so we can write:
\[\Large f\left( x \right) = - 3\cos \left( {4x - \frac{\pi }{2}} \right) + 4\]
@genesis98

Answer 19

oh okay! thank you!

Answer 20

thank you!