arcsin x + arccos x = pi/2

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Prime Newtons

4 ай бұрын

In this video , I proved that arcsin x plus arccos x = pi/2

Пікірлер: 40

@lagomoof 3 ай бұрын

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.

@ijkhugeplay 3 ай бұрын

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.

@dydx_mathematics2 3 ай бұрын

I agree 💯👍

@GreenMeansGOF 3 ай бұрын

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.

@Archimedes_Notes 3 ай бұрын

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

@ReyazulislamReayal 3 ай бұрын

Nice calculation sir👍

@admkaan504 3 ай бұрын

his way of explaining things are like an asmr to me

@offgame1654 3 ай бұрын

nice, we started trig identities in my class recently and this is much related to what we learn

@77Chester77 3 ай бұрын

I love your enthusiasm. I'm sure your students love you

@khairyalkhalidy1316 3 ай бұрын

I love how you simplify everything

@sphakamisozondi 3 ай бұрын

If I had u as my maths teacher back in high school. I would hv turned into an influential mathematician

@JSSTyger 3 ай бұрын

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 3 ай бұрын

Nice

@michaeledwards2251 3 ай бұрын

@@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 3 ай бұрын

But when x is negative x=-1/2 proof for that case???

@michaeledwards2251 3 ай бұрын

@@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.

@JSSTyger 3 ай бұрын

@@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.

@fisimath40 3 ай бұрын

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.

@barryzeeberg3672 3 ай бұрын

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

@raivogrunbaum4801 3 ай бұрын

This identity is correct also if x is negative take x=-1/2 proof for that case ??

@PugganBacklund 3 ай бұрын

did we just prove that the an right triangle other 2 angels sums up to 90° (½π)?

@WhiteGandalfs 3 ай бұрын

The geometric proof in the right triangle is so drastically more simple - just the two angles beside the right angle.

@shahaabudinbanday7157 3 ай бұрын

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)|

@kappascopezz5122 3 ай бұрын

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)

@JourneyThroughMath 3 ай бұрын

That was a nifty proof

@grezhz 3 ай бұрын

a very enjoyable video! :)

@kevinmadden1645 3 ай бұрын

Shouldn't x be restricted to the first quadrant?

@surendrakverma555 3 ай бұрын

Very good. Thanks Sir

@TheLukeLsd 3 ай бұрын

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.

@chengkaigoh5101 3 ай бұрын

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

@vafasadrif12 3 ай бұрын

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