can some1 quick recap me on how to do such qs (1 Viewer)

[ekjchale#1]

specifically the finding exact value, have become rusty so forgot that part

 

[cossine]

If the derivative is zero this means the function is constant. So once you plug in some point that is in the domain you will get the answer.

 

[5uckerberg]

Considering this is a range question we need to focus on the domain between [-1, 1] because otherwise, we will have complex numbers for both which will require complex analysis which is way too out of your reach.

Here, for at x=-1, 0, 1 you will see that they have the following values .

So therefore,

 

[Run hard@thehsc]

When you differentiate it, you get 0 right..... this implies that the original function must have been a constant for the derivative to be zero. Now for the working out, I am sure you let x = sin (theta). Hence , this means that x = cos (pi/2 - theta). Then we can re write these statements as theta = sin^-1 (x) and (pi/2 - theta) = cos^-1(x). And when you sub in the terms you get (pi/2 - theta + theta = pi/2) which proves as required. Sorry for the clumped up form

Edit: @5uckerberg and @cossine have explained it better than me... did not see their posts lol

 

[cossine]

Considering this is a range question we need to focus on the domain between [-1, 1] because otherwise, we will have complex numbers for both which will require complex analysis which is way too out of your reach.

Here, for at x=-1, 0, 1 you will see that they have the following values .

So therefore,

I never heard of sin^-1 being defined for complex numbers. Cool

 

[5uckerberg]

I never heard of sin^-1 being defined for complex numbers. Cool

I am talking about something like this. For example, . I guarantee you that complex analysis will have to be used.

 

[ekjchale#1]

If the derivative is zero this means the function is constant. So once you plug in some point that is in the domain you will get the answer.

Ah fair I tried that but it seemed way to simple lol, thanks for the clarification @5uckerberg @cossine @Run hard@thehsc

 

[cossine]

Ah fair I tried that but it seemed way to simple lol, thanks for the clarification @5uckerberg @cossine @Run hard@thehsc

So if you have y = sin^-1(x) + cos^-1(x)

Then dy/dx = 0. So you have a differential equation that you can solve. Another alternative would be draw triangles.

So draw triangle with hypotenuse 1 and opposite side x for sin^-1(x) and hypotenuse 1 and adjacent side x for cos^-1(x). You will see the two angles will need to add up to 90.

 

[ekjchale#1]

So if you have y = sin^-1(x) + cos^-1(x)

Then dy/dx = 0. So you have a differential equation that you can solve. Another alternative would be draw triangles.

So draw triangle with hypotenuse 1 and opposite side x for sin^-1(x) and hypotenuse 1 and adjacent side x for cos^-1(x). You will see the two angles will need to add up to 90.

Ah yep that makes sense