In this video , I proved that arcsin x plus arccos x = pi/2
Пікірлер: 41
@lagomoof5 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.
@ijkhugeplay5 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_mathematics25 ай бұрын
I agree 💯👍
@Archimedes_Notes5 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
@GreenMeansGOF5 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.
@offgame16545 ай бұрын
nice, we started trig identities in my class recently and this is much related to what we learn
@ReyazulislamReayal5 ай бұрын
Nice calculation sir👍
@admkaan5045 ай бұрын
his way of explaining things are like an asmr to me
@shahaabudinbanday71575 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)|
@JSSTyger5 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).
@PrimeNewtons5 ай бұрын
Nice
@michaeledwards22515 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.
@raivogrunbaum48015 ай бұрын
But when x is negative x=-1/2 proof for that case???
@michaeledwards22515 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.
@JSSTyger5 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.
@PugganBacklund5 ай бұрын
did we just prove that the an right triangle other 2 angels sums up to 90° (½π)?
@barryzeeberg36725 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
@raivogrunbaum48015 ай бұрын
This identity is correct also if x is negative take x=-1/2 proof for that case ??
@77Chester775 ай бұрын
I love your enthusiasm. I'm sure your students love you
@khairyalkhalidy13165 ай бұрын
I love how you simplify everything
@fisimath405 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.
@sphakamisozondi5 ай бұрын
If I had u as my maths teacher back in high school. I would hv turned into an influential mathematician
@WhiteGandalfs5 ай бұрын
The geometric proof in the right triangle is so drastically more simple - just the two angles beside the right angle.
@kappascopezz51225 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)
@vafasadrif125 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
@JourneyThroughMath5 ай бұрын
That was a nifty proof
@kevinmadden16455 ай бұрын
Shouldn't x be restricted to the first quadrant?
@grezhz5 ай бұрын
a very enjoyable video! :)
@TheLukeLsd5 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.
@maburwanemokoena711714 күн бұрын
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?
@mohamedoumar73645 ай бұрын
كان ينبغي ان يشترط كونcosbأكبر من الصفر
@surendrakverma5555 ай бұрын
Very good. Thanks Sir
@chengkaigoh51015 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
@fabayojackson53495 ай бұрын
❤❤❤
@punditgi5 ай бұрын
The prime reason to learn math is Prime Newtons! 🎉😊
@miraj50865 ай бұрын
This problem was in my boards math question paper 😅
@ericerpelding23485 ай бұрын
Could this be shown by using the definitions of arcsin and arccos only?
@PrimeNewtons5 ай бұрын
Yes
@dkg94815 ай бұрын
You can also use the fact that the derivative is zero
@DEYGAMEDU5 ай бұрын
Sir I have send you a problem in your mail. Please solve that