This identity is a consequence of the sum of the angles of a triangle being pi radians. For a right angled triangle, the right angle itself accounts for pi/2 radians, and so the other two angles sum to pi/2 regardless of what they are. The main trick with that approach is proving that the sine of one of those angles is the cosine of the other. Or in other words, one angle's adjacent is the other's opposite and vice versa.
It is better to impose the conditions that when we let a= arsin(x) and b =arcos(x), then x is in [-1,1] and a in [-pi/2,pi/2] and b is in [0,pi] and note what happens when you add a to b. I think it is better to let x=cos(b) and use sin(pi/2-b)=cos(b) this will let us work with one angle only. YOU CAN ALSO DUFFERENTIATE AND OBTAIN A CONSTANT AND CINCLUDE THAT YOUR CONSTANT IS PI/2. THANK YOU SIR FOR ALL YOU EFFORT
I was thinking, why can’t it be like 5π/2? For completeness, we should say that the range of arcsin(x)+arccos(x) is [-π/2,3π/2] so the only possible angle value is π/2.
Let sin(a) = x So a = arcsin x ..... (1) Also since sin(a) = x cos(pi/2 - a) = x pi/2 - a = arccos x ..... (2) Add (1) and (2) to get pi/2 = arcsinx + arccosx Note that even if a>pi/2 cos is an even function so it will reduce to cos |(pi/2 -a)|
I have a first quadrant proof. Construct a right triangle with one side equal to x and the hypotenuse equal to 1. Let x be opposite of angle α and adjacent to angle ß. So...sin(α) = x/1 = x. We also see that cos(ß) = x/1 = x. We can also say that sin-1(x) = α and cos-1(x) = ß. Adding the arcsine and arccosine together we get α+ß. And in our right triangle, we know α +ß = 90° (π/2).
@@PrimeNewtons An equivalent approach would be to consider a rectangle with parallel sides length x, and diagonals, length 1. 1. Considering one of the diagonals, an angle Alpha between the diagonal and the rectangle side, with the side opposite length x, and the diagonal length 1, the hypotenuse, the sine value would be x. 2. Consider the complement of Angle alpha, Beta, in the same corner as Alpha. The cosine value would be x. Each corner of a rectangle is Pi /2 radians, or 90 degrees.
@@raivogrunbaum4801 Choose 1 of the corners of the Rectangle above, and place it on the origin of the sine/cosine graph. Put 1 rectangle, as above, in each quadrant. Make the angles considered in the corner with its point on the origin.
@@michaeledwards2251 Look at the graph of arccosine. It intersects the y-axis at y = π/2 and x = 0. The average of arccos(x) and arccos(-x) is π/2. Applying the formula for an average, you can now say arccos(-x) = π-arccos(x). Similarly arcsin(-x) = -arcsin(x). Now apply x = -0.5. Arccos(-0.5) = π-arccos(0.5). Arcsin(-0.5) = -arcsin(0.5). So we get π-arccos(0.5)-arcsin(0.5) = π/2 as a replacement equation and the triangle can then be constructed. You get π-π/3-π = π/2, which is true.
Simpler solution: Construct right triangle whose hypotenuse is 1, one side is x , and by pythagorus the other side is sqrt(1-x^2) The angle opposite x plus the angle opposite sqrt(1-x^2) = pi (sum of 3 angles of triangle) - pi/2 (right angle of the triangle) = pi/2
Interesting video. I would do it like this. (arcsinx+arccosx)’=(arcsinx)’+( arccosx)’ =1/√(1-x²)+(-1/√(1-x²))=0 It is then concluded that arcsinx+arccosx=constant, (k)'=0 So, for all x in [-1,1], it will always be the same If x= 0 arcsin 0+arccos 0=π/2, which is demonstrated.
Here's my solution: Let u be such that x = sin(u): Therefore arcsin(x) = arcsin(sin(u)) = u cos(π/2 - u) = sin(u) = x -> arccos(x) = π/2 - u -> arcsin(x) + arccos(x) = u + π/2 - u = π/2
É fácil provar essa identidade desenhando um circulo trigonométrico e um triângulo retângulo inscrito tocando o ponto até onde vai o arco como uma definição.
How Id solve this is take the derivative on the left, i’ll get zero then I’ve proven that the function on the left is a constant. I can then substitute a value to get pi/2, how’s that?
![int[sin^4(x)]](http://i.ytimg.com/vi/Lwmn-ZTCBMI/mqdefault.jpg)
![lim [x-tanx]/[x-sinx] as x approaches 0](/img/1.gif)
This identity is a consequence of the sum of the angles of a triangle being pi radians. For a right angled triangle, the right angle itself accounts for pi/2 radians, and so the other two angles sum to pi/2 regardless of what they are. The main trick with that approach is proving that the sine of one of those angles is the cosine of the other. Or in other words, one angle's adjacent is the other's opposite and vice versa.
I like the way you manage to cover different math topics on your channel. This is really helpful for self-studying student. Definitely appreciate it.
I agree 💯👍
It is better to impose the conditions that when we let a= arsin(x) and b =arcos(x), then x is in [-1,1] and a in [-pi/2,pi/2] and b is in [0,pi] and note what happens when you add a to b. I think it is better to let x=cos(b) and use sin(pi/2-b)=cos(b) this will let us work with one angle only. YOU CAN ALSO DUFFERENTIATE AND OBTAIN A CONSTANT AND CINCLUDE THAT YOUR CONSTANT IS PI/2. THANK YOU SIR FOR ALL YOU EFFORT
I was thinking, why can’t it be like 5π/2? For completeness, we should say that the range of arcsin(x)+arccos(x) is [-π/2,3π/2] so the only possible angle value is π/2.
his way of explaining things are like an asmr to me
I love how you simplify everything
Nice calculation sir👍
I love your enthusiasm. I'm sure your students love you
Let sin(a) = x
So a = arcsin x ..... (1)
Also since sin(a) = x
cos(pi/2 - a) = x
pi/2 - a = arccos x ..... (2)
Add (1) and (2) to get pi/2 = arcsinx + arccosx
Note that even if a>pi/2 cos is an even function so it will reduce to cos |(pi/2 -a)|
I have a first quadrant proof. Construct a right triangle with one side equal to x and the hypotenuse equal to 1. Let x be opposite of angle α and adjacent to angle ß. So...sin(α) = x/1 = x. We also see that cos(ß) = x/1 = x. We can also say that sin-1(x) = α and cos-1(x) = ß. Adding the arcsine and arccosine together we get α+ß. And in our right triangle, we know α +ß = 90° (π/2).
Nice
@@PrimeNewtons
An equivalent approach would be to consider a rectangle with parallel sides length x, and diagonals, length 1.
1. Considering one of the diagonals, an angle Alpha between the diagonal and the rectangle side, with the side opposite length x, and the diagonal length 1, the hypotenuse, the sine value would be x.
2. Consider the complement of Angle alpha, Beta, in the same corner as Alpha. The cosine value would be x.
Each corner of a rectangle is Pi /2 radians, or 90 degrees.
But when x is negative x=-1/2 proof for that case???
@@raivogrunbaum4801
Choose 1 of the corners of the Rectangle above, and place it on the origin of the sine/cosine graph. Put 1 rectangle, as above, in each quadrant.
Make the angles considered in the corner with its point on the origin.
@@michaeledwards2251 Look at the graph of arccosine. It intersects the y-axis at y = π/2 and x = 0. The average of arccos(x) and arccos(-x) is π/2. Applying the formula for an average, you can now say arccos(-x) = π-arccos(x). Similarly arcsin(-x) = -arcsin(x). Now apply x = -0.5. Arccos(-0.5) = π-arccos(0.5). Arcsin(-0.5) = -arcsin(0.5).
So we get π-arccos(0.5)-arcsin(0.5) = π/2 as a replacement equation and the triangle can then be constructed. You get π-π/3-π = π/2, which is true.
nice, we started trig identities in my class recently and this is much related to what we learn
My solution, given the basic identity that cos(x) = sin(pi/2 - x) for all x:
arcsin(x) + arccos(x) = pi/2
arcsin(x) = pi/2 - arccos(x)
x = sin(pi/2 - arccos(x))
Substitute x=cos(y)
cos(y) = sin(pi/2 - y)
did we just prove that the an right triangle other 2 angels sums up to 90° (½π)?
3:35 The domain of arc sin is -1 to 1. -π/2 to π/2 is the range. This is what he means there.
Simpler solution:
Construct right triangle whose hypotenuse is 1, one side is x , and by pythagorus the other side is sqrt(1-x^2)
The angle opposite x plus the angle opposite sqrt(1-x^2) = pi (sum of 3 angles of triangle) - pi/2 (right angle of the triangle) = pi/2
This identity is correct also if x is negative take x=-1/2 proof for that case ??
If I had u as my maths teacher back in high school. I would hv turned into an influential mathematician
The geometric proof in the right triangle is so drastically more simple - just the two angles beside the right angle.
Hey, hey, @PrimeNewtons
Your preview for this video is wrong a bit 😅
sin^-1(x) is not arcsin(x)
Interesting video.
I would do it like this.
(arcsinx+arccosx)’=(arcsinx)’+( arccosx)’
=1/√(1-x²)+(-1/√(1-x²))=0
It is then concluded that
arcsinx+arccosx=constant, (k)'=0
So, for all x in [-1,1], it will always be the same
If x= 0
arcsin 0+arccos 0=π/2, which is demonstrated.
Here's my solution:
Let u be such that x = sin(u):
Therefore
arcsin(x) = arcsin(sin(u)) = u
cos(π/2 - u) = sin(u) = x ->
arccos(x) = π/2 - u ->
arcsin(x) + arccos(x) = u + π/2 - u = π/2
Thank you so much you just saved me from a big problem❤❤
Shouldn't x be restricted to the first quadrant?
That was a nifty proof
a very enjoyable video! :)
É fácil provar essa identidade desenhando um circulo trigonométrico e um triângulo retângulo inscrito tocando o ponto até onde vai o arco como uma definição.
❤
How Id solve this is take the derivative on the left, i’ll get zero then I’ve proven that the function on the left is a constant. I can then substitute a value to get pi/2, how’s that?
Very good. Thanks Sir
Could this be shown by using the definitions of arcsin and arccos only?
Yes
I was thinking of a geometric where u draw a line with 2 angles that add up to 90 degrees and form a triangle with respects to the 2 axis
The prime reason to learn math is Prime Newtons! 🎉😊
كان ينبغي ان يشترط كونcosbأكبر من الصفر
❤❤❤
This problem was in my boards math question paper 😅
You can also use the fact that the derivative is zero
Sir I have send you a problem in your mail. Please solve that