+118654 Hi asinus,
Your answer does not make sense.
x is not an angle it is a ratio. so you cannot have sin(x)
Certainly if the angles are both 45 degrees this will work
but one angle might be 30 degrees and the other 60degrees
or as I have said, one is x degrees and the other is 90-x degrees.
Think of this triangle and it might help you understand. :)
+14985 sin(arcsin x + arccos x)
\(sin(arcsin\ x+arccos\ x)\)
If
\(sinx=cosx\)
then
\(sinx=cosx=\frac{\sqrt 2}{2}\)
\(\large sin(arcsin\frac{\sqrt 2}{2}+arccos\frac{\sqrt 2}{2})=sin(\frac{\pi}{4}+\frac{\pi}{4})=sin\frac{\pi}{2}=1\)
!
+118654 Hi asinus,
Your answer does not make sense.
x is not an angle it is a ratio. so you cannot have sin(x)
Certainly if the angles are both 45 degrees this will work
but one angle might be 30 degrees and the other 60degrees
or as I have said, one is x degrees and the other is 90-x degrees.
Think of this triangle and it might help you understand. :)
+118654 sin(arcsin x + arccos x)
\(sin(arcsin x + arccos x)\\ let\;\;asin(x)=\theta\\ then\;\;acos(x)=90-\theta\\ so\\ sin(arcsin x + arccos x)=sin(\theta+90-\theta)=sin(90)=1\)
