mind boggling problem regarding trigonometric functions

[vivid magnet]

If sinx = cosy, would cosx = siny and therefore y" = 0

[vivid magnet]

Given that y is dependent on x

@frozen snow

This is a really vague question ik

I don't really understand the nature of it

@frozen snow

Hallo

What do u think of this

[frozen snow]

I see

[vivid magnet]

Scroll up it's explained there

Yep

I hate physics in English this is confusing

[frozen snow]

Well in such cases you have to use trigonometric identities and some sprinkle of differentiation
It's easy. You don't need to have "advance level" of mathematics to solve this

It's precalculus and introductory stuff to calculus

[vivid magnet]

I used Pythagorean identity

And normal differentiation

I got 0

But the problem is

I don't really understand the nature

Yeah I did smth similar

This is also a good way

Oh I thought u did cos2x = sin2y

Yep this as well

[frozen snow]

Well you won't get the exact preemptive answer you expect to get

[vivid magnet]

But do u understand the nature of the relation here

Kinda

Is there a way to graph it

I got the solutions

In fact I already did them

But

How would the graph be

Of y

The original expression

Cos-1(sinx) = y does this work

[vivid magnet]

That's what I meant by nature btw

[frozen snow]

(\cos(x)) is the same as (\sin(90^\circ - x)) (or (\sin(\frac{\pi}{2} - x)) in radians). So, the equation (\sin(x) = \cos(y)) can be rewritten as: [ \sin(x) = \sin\left(\frac{\pi}{2} - y\right) ]

• considering angles.

(\cos(x) = \sin(y)), then using the identity (\sin(y) = \cos(\frac{\pi}{2} - y)), this implies: [ \cos(x) = \cos\left(\frac{\pi}{2} - y\right) ] which gives the same scenarios:(x = \frac{\pi}{2} - y + 2m\pi), or(x = -\left(\frac{\pi}{2} - y\right) + 2m\pi), which simplifies to (x = \frac{\pi}{2} + y + 2m\pi).Notice the first scenario for (x) matches both conditions (\sin(x) = \cos(y)) and (\cos(x) = \sin(y)).

Considering (\cos(x) = \sin(y))

y'' = 0), we would need an explicit function for (y) as a function of (x). From the above relationship, one would need to differentiate (y) twice with respect to (x) to check if (y'') equals zero.

-considering derivatives

[vivid magnet]

Wait lemme show u my graph

[vivid magnet]

Ok lemme check this

[frozen snow]

Using Pythagorean identity is an excellent approach but the problem is you end up juggling with multiple identities in your head.

[vivid magnet]

True

But but

Here is the thing

Well

I also thought that as well

But then I tried this

And it was the same result

Ohhhh

Wait u are right

This is the correct graph I am tweaking

[frozen snow]

@frozen snow use graphing tools like desmos and geogabra

[vivid magnet]

Yes

[frozen snow]

Enter the equation (\sin(x) = \cos(y)).Plot (y = \frac{\pi}{2} - x) or (y = \frac{\pi}{2} + x).Observe the intersection points with the sine and cosine functions.

[vivid magnet]

It requires differential equations right

I've seen a similar problem before

Or actually nvm it's not similar

Pursuit problems

[frozen snow]

What do you want me to help with

Which topic

[merry hound]

I mean cos(y - pi/2) = sin(y)