For what values of m does y=mx-1 provide one solution to y=cos (2x)? (2 Viewers)

[eternallyboreduser]

How to do?

 

[WeiWeiMan]

How to do?

y=cos2x can be rewritten as 2cos^2 x -1
mx-1=2cos^2x -1
mx=2cos^2x
if u sketch this you get something like this (desmos isn't required since you just need a rough idea of what to look for)

you can observe that when m is between some values, the line is above the curve of 2cos^2x
from this you can probably just use pythagoras' theorem to figure out the bounds for m

edit:nvm pythag might be bad and lead to bad things just use calculus and figure out when y=mx is tangent

 

[eternallyboreduser]

y=cos2x can be rewritten as 2cos^2 x -1
mx-1=2cos^2x -1
mx=2cos^2x
if u sketch this you get something like this (desmos isn't required since you just need a rough idea of what to look for)
View attachment 42089
you can observe that when m is between some values, the line is above the curve of 2cos^2x
from this you can probably just use pythagoras' theorem to figure out the bounds for m

edit:nvm pythag might be bad and lead to bad things just use calculus and figure out when y=mx is tangent

but what if you were in an exam and couldnt use desmos to geaoh lol, wouldbt ur geaph br much more inaccurate so its a bit hatder to tell

 

[eternallyboreduser]

i havent done calculus with trig yet but do u mind providinf a dteo by step w/o @WeiWeiMan

 

[WeiWeiMan]

but what if you were in an exam and couldnt use desmos to geaoh lol, wouldbt ur geaph br much more inaccurate so its a bit hatder to tell

nah it doesn't rly matter you just gotta get a rough idea of what you're looking for
if you wanna find the shape of the graph just sub in some points into your calculator and use the fact that the thing is an even function

i havent done calculus with trig yet but do u mind providinf a dteo by step w/o @WeiWeiMan

not exactly in the mood currently to do step by step but the rough idea should be differentiating it and plugging it into slope form, maybe sure that the y intercept is 0
calculus with trig shouldn't that that much different from calculus with anytyhing else
also i could be mistaken and my method could be completely wrong so please don't hold this against me

 

[eternallyboreduser]

nah it doesn't rly matter you just gotta get a rough idea of what you're looking for
if you wanna find the shape of the graph just sub in some points into your calculator and use the fact that the thing is an even function


not exactly in the mood currently to do step by step but the rough idea should be differentiating it and plugging it into slope form, maybe sure that the y intercept is 0
calculus with trig shouldn't that that much different from calculus with anytyhing else
also i could be mistaken and my method could be completely wrong so please don't hold this against me

how would u figure out the exact x value the tangent is at

 

[Average Boreduser]

y=cos2x can be rewritten as 2cos^2 x -1
mx-1=2cos^2x -1
mx=2cos^2x
if u sketch this you get something like this (desmos isn't required since you just need a rough idea of what to look for)
View attachment 42089
you can observe that when m is between some values, the line is above the curve of 2cos^2x
from this you can probably just use pythagoras' theorem to figure out the bounds for m

edit:nvm pythag might be bad and lead to bad things just use calculus and figure out when y=mx is tangent

idk if this would be the best approach, observing mx=2cos^2x is harder to graph as opposed to mx-1=cos2x

 

[eternallyboreduser]

idk if this would be the best approach, observing mx=2cos^2x is harder to graph as opposed to mx-1=cos2x

bro do yk how to do rhe q

 

[WeiWeiMan]

how would u figure out the exact x value the tangent is at

calculus ig idk

 

[Average Boreduser]

forget what I just said that was dumb asf

 

[eternallyboreduser]

forget what I just said that was dumb asf

do yk how to solve

 

[Luukas.2]

I would work with the equations as given.

There will be one solution when is large, so there will be two sets of possible values, and where the value correspond to the line being a tangent to the curve at some value and the line also crossing the curve at some point between and .

The values are about and .

 

[Average Boreduser]

do yk how to solve

Yeah idk atp. I feel like just answering generally using the graph is the only real method.

 

[eternallyboreduser]

Yeah idk atp. I feel like just answering generally using the graph is the only real method.

thats kinda hard to do during exam though cos ur geaph most definitely aint gonna be 100% accurate

 

[WeiWeiMan]

thats kinda hard to do during exam though cos ur geaph most definitely aint gonna be 100% accurate

it doesn't ahve to be u just need a general idea of what ur looking for

I would work with the equations as given.

There will be one solution when is large, so there will be two sets of possible values, and where the value correspond to the line being a tangent to the curve at some value and the line also crossing the curve at some point between and .

The values are about and .

may i ask how you're expected to find these values
also i'm not sure but i feel like the question might want some exact bounds

 

[Average Boreduser]

Is this even possible to do without some other hint? I can't grasp anything here. I've always been taught, once u get x variables with trig, jus graph and thats as far as u can get it for the hsc.

 

[eternallyboreduser]

it doesn't ahve to be u just need a general idea of what ur looking for


may i ask how you're expected to find these values
also i'm not sure but i feel like the question might want some exact bounds

but then how r u gonna find the x value

 

[Average Boreduser]

but then how r u gonna find the x value

u dont. ur finding values for m. x is a fnction

 

[eternallyboreduser]

u dont. ur finding values for m. x is a fnction

yeah but when u diff cos2x wouldnt u sub in a x value where the two functions touch each other once (aka the x value where y=x-1 is a tangent)

 

[Luukas.2]

looks like a typical cosine graph but with a period of . The line must have a y-intercept at (0, -1), and so as decreases from infinity, there must be only a single solution until the line becomes a tangent. This will happen for some x-value near but smaller than pi, from the diagram. In that case, the line has gradient and the point of tangency has .

So, you know that (for the line and curve to meet) and (for the line to have the correct slope to be a tangent).

Solving these simultaneously would give the exact values of and , but most equations like these are not solvable exactly by HSC methods (or at all, in many cases) and so we are left to approximate.

My approximate solutions give:

​

and

​

which are pretty decent results, and I found them using graphical methods.