for the question at the end, I considered a right angled triangle with one leg of length x and hypotenuse of length 1. Then defined angles a and b such that cos(a)=x/1 and sin(b)=x/1 and a+b+pi/2=pi. From here it’s trivial and we can see that the answer should be pi/2. However we should be careful because this is only true if x>0 or x=0. The easier way is to obviously substitute values. x=-1 gives arccos(x)=pi and arcsin(x)=-pi/2 which sums to pi/2. At x=0 it is also equal to pi/2. Therefore there are no discontinuities and the sum of arccos and arcsin is equal to pi/2.
@jasimmathsandphysics Nice geometric proof! Although, yes, it mainly works for positive x. For the second method you stated, you would need to show that the derivative is always 0 to make it evident that the function is constant; only then can you sub a point to conclude the function is always pi/2
for the question at the end, I considered a right angled triangle with one leg of length x and hypotenuse of length 1. Then defined angles a and b such that cos(a)=x/1 and sin(b)=x/1 and a+b+pi/2=pi. From here it’s trivial and we can see that the answer should be pi/2. However we should be careful because this is only true if x>0 or x=0.
The easier way is to obviously substitute values. x=-1 gives arccos(x)=pi and arcsin(x)=-pi/2 which sums to pi/2. At x=0 it is also equal to pi/2. Therefore there are no discontinuities and the sum of arccos and arcsin is equal to pi/2.
@jasimmathsandphysics Nice geometric proof! Although, yes, it mainly works for positive x. For the second method you stated, you would need to show that the derivative is always 0 to make it evident that the function is constant; only then can you sub a point to conclude the function is always pi/2
@ I mean yeah. The only reason I didn’t differentiate was because I thought it would be obvious from the topic of the video 😂
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Glad to be here relatively early