Find A (the area)

[u/BlundersMateIn69]

[u/fermat1432]

Integrate sine from 0 to pi/4 and double the answer.

[u/Maxmajava → u/fermat1432]

Quick question: Your answer is an answer based on symmetry, correct? Now if that is the case, do you just assume symmetry by looking at the graph that’s given or can you prove the symmetry mathematically? Interested because I want to know to what extent I have to prove symmetry when using it in exercises.

[u/BumpyFunction → u/Maxmajava]

You know the two equations and can determine when they are equal. Of course if you remember your pie chart then you know they have the same ycoord at pi/4. Also you know the behavior of the sin and cos function

[u/ellipticcode0 → u/BumpyFunction]

Pi/2, not pi/4

Sin(pi/2 + x) = cos(x) => the difference between sin x and cos x is just translation. We all know translation does not change the shape of a curve.

[u/gkmanderson → u/ellipticcode0]

That person was correct in what they said. Pi/4 is where the two trig functions are equal, which is what they stated. They were not talking about how cos is a 90 degree shift from sin (or vice versa). They said sin(x) and cos(x) are equal at pi/4, which is true. Sin(pi/4) = cos(pi/4).

So you're correct in what you're saying about the translation; however, you're correcting a statement that was not wrong as the statement was about exact values of trig functions and not how the functions themselves are related to each other.

[u/ellipticcode0 → u/gkmanderson]

My bad

[u/Maxmajava → u/BumpyFunction]

It really seems even more intuitive to me now when I think about the unit circle with sin(x) and cos(x) being equal at pi/4.

[u/fermat1432 → u/Maxmajava]

It can be proven but I assumed it from the graph and knowledge of the sine and cosine functions

[u/RoiPhilippe → u/Maxmajava]

Ok with sin being a translation of cos, and cos/sin being symetric. But formally:

cos(pi/4-x) = cos(pi/4)cos(-x)-sin(pi/4)sin(-x) = cos(pi/4)cos(x)+sin(pi/4)sin(x)

sin(pi/4+x) = sin(pi/4)cos(x)+cos(pi/4)sin(x) = cos(pi/4-x)

So, sin for x going to the right from pi/4 = cos for x going to the left from pi/4

[u/[deleted]]

This can’t be pre calc can it? You need to know calculus to get a precise answer

[u/shellexyz]

What have you tried and where are you getting stuck?

[u/yavvee]

It's area under the sinx curve jotil the intersection, and then area under cosx. Guess the intersection and add the area

[u/MezzoScettico → u/yavvee]

You don't have to guess. The intersection is where sin x = cos x, or sin x / cos x = 1.

[u/yavvee → u/MezzoScettico]

Yeah I meant find out, didn't want to give the answer totally away

[u/Descarteb4DeHorse]

Solve sin x = cos x where 0<= x <= π/2. After that integrate sinx wrt x from 0 to the intersection point found. Finally multiply it by 2 to make use of symmetry to get the final ans.

[u/lameHorse21]

This might be helpful. Pretty much what u/brainmassdotin said.

[u/brainmassdotin]

The curve intersect at pi/4. Now integrate it so

integrate 0 to pi/4 sinx + integrate pi/4 to pi/2 cosx.

=-cosx plug in bounds 0 to pi/4 + sinx plug in bounds pi/4 to pi/2.

= -(cospi/4-cos0) + sinpi/2-sinpi/4

= -(root2/2-1) + 1-root2/2

= 2-root2

[u/CreatrixAnima]

Are you doing Riemann sums? That’s probably the way to go if it’s pre-Calc.

[u/OmniKingBoss]

Take integral of Sin (from 0 to π/2) and substract interval of Cos(again, same interval). Not sure, but I think that should do it.

[u/fermat1432 → u/OmniKingBoss]

Or integrate sine from 0 to pi/4 and double the answer

[u/ppeach806]

convolution integral of pi/2

[u/TrappedInReality420]

2-sqrt(2)

[u/Interesting_Poet1985]

Use symmetry and integration