Need help solving this cosine equation from a graph?!

Adorable98

I have no idea how to solve parts a and b? :frown:

Reply 1

1420787

a) What do values P and Q mean for a trig function?

Spoiler

For b) Set the function equal to 0 and solve. This we do whenever we want to find roots.

Reply 2

Adorable98

Original post by offhegoes

a) What do values P and Q mean for a trig function?

Spoiler

For b) Set the function equal to 0 and solve. This we do whenever we want to find roots.

For Q why do we find the midpoint? :s-smilie:

Reply 3

1420787

Original post by Adorable98

For Q why do we find the midpoint? :s-smilie:

It works out the same, so the midpoint of the maximum value of 5 and the minimum value of -3 is 1, right?

Just a different way if doing it.

Reply 4

Adorable98

Original post by offhegoes

It works out the same, so the midpoint of the maximum value of 5 and the minimum value of -3 is 1, right?

Just a different way if doing it.

No, i mean't why do we find the midpoint? :smile:

And I also don't understand how you found P? :redface:

Reply 5

1420787

Original post by Adorable98

No, i mean't why do we find the midpoint? :smile:

And I also don't understand how you found P? :redface:

Sorry, let me correct myself from before. For Q we do not find the midpoint. For Q we need to know that Q, because it multiplies the trig function, gives the amplitude, which is equal to half the distance from the top to the bottom. The distance from top to bottom is 8, so the amplitude is 4, which means Q is 4.

P, added to the trig function as it is, moves the function up (or down if negative). A normal trig function with nothing added will oscillate up and down about the x-axis, and the middle point will be 0. Here the middle point (midpoint between top and bottom) is 1, so the function has been moved up 1. So P is 1.

Reply 6

Adorable98

Original post by offhegoes

Thaank you for explaining :smile:

, but I seriously for some reason don't seem to understand it. :redface:

Reply 7

Adorable98

@notnek Could you please help me with this question :redface:

Reply 8

Adorable98

@Zacken

Could you please explain this for me?

Reply 9

felixastejada

just thinking out of the box.
doesn't the equation look similar to y = mx + c ?
so since you have the y intercept -3, would that mean c = P = -3 ?
And then just substitute the values for y and x into the equation.
bring everything to one side by using inverse of cos and so on to get Q on its own?

ps. I'm probably very very wrong, just makes sense to me.
please somebody correct me

(edited 7 years ago)

Reply 10

username1732491

Original post by felixastejada

It's not a linear graph, so you can't use y = mx + c

Reply 11

felixastejada

Original post by TimGB

It's not a linear graph, so you can't use y = mx + c

thought so
thanks

Reply 12

username1732491

Original post by Adorable98

Thaank you for explaining :smile:

, but I seriously for some reason don't seem to understand it. :redface:

It's a lot simpler than it seems. You know the graph of cos(x) oscillates between -1 and 1, as does cos(2x) and cos(3x) and so on. The graph in question oscillates between -3 and 5, so you can use this to work out how much the graph has been stretched by (the value of Q) and how much it has been shifted by (the value of P).

Reply 13

Adorable98

Original post by TimGB

I finally understood part (a) :biggrin:

but it's part (b) that I still don't understand :redface:

Reply 14

username1732491

Original post by Adorable98

I finally understood part (a) :biggrin:

but it's part (b) that I still don't understand :redface:

Probably should have also mentioned that you should check whether Q is negative by seeing if the graph is upside down or not - if it is, then Q must be negative.

For part b), set y = 0 and solve for x. You're looking for the first 6 positive x values.