Set $alpha=arccos x$. Since, by definition, $alphain<0,pi>$, you know that $sinalphage0$, so$$sinarccos x=sinalpha=sqrt1-cos^2alpha=sqrt1-(cosarccos x)^2=sqrt1-x^2$$
Let $arccos(x)= heta$
$x=cos( heta)$
You require $y=sin(arccos(x))=sin( heta)$
Identity: $cos^2( heta)+sin^2( heta)=1$
$x^2+y^2=1$
What is $arccos(x)$? it is the angle given by the ratio of sides of length $x$ và $1$. That is, on a triangle with adjacent side length $x$ and hypotenuse length $1$ we will find the angle $arccos(x)$.
(There is a picture for this which I hope you can get from what I describe).
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Now what is $sin(y)$? It is the ratio of the opposite side to the hypotenuse of a triangle with angle $y$ in the respective part. So if we use $y=arccos(x)$, we have that $sin(y)$ is the ratio of the sides with degree given by a triangle whose adjacent and hypotenuse sides are length $x$ và $1$ respectively.
Since we have the adjacent & hypotenuse lengths, we can calculate the opposite sides length by the Pythagorean theorem. This means, if we use $z$ for the opposite side, that $z^2+x^2=1^2=1$. Solving for $z$ gives $z=sqrt1-x^2$.
Then $sin(y)$ is the ratio of the opposite size, $z=sqrt1-x^2$ & the hypotenuse, $1$. We may now say $sin(arcos(x))=sqrt1-x^2$.
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edited Apr 26, năm ngoái at 6:30
answered Apr 26, 2015 at 6:21
EoinEoin
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Draw a right triangle with one of the angles being $cos^-1 x$, then fill out the lengths of the sides of the triangle using the definition of cosine for two of the sides, & Pythagoras"s theorem for the third side. Then, use the definition of sine lớn find $sin(cos^-1 x)$. For questions lượt thích these, always try drawing a triangle first.
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answered Apr 26, 2015 at 6:15
Kevin HsuKevin Hsu
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$$y=sinarccos x$$$$x=cosarccos x$$
Therefore $x$ và $y$ are the sin and cos of the same angle and
$$x^2+y^2=1$$
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answered Apr 26, năm ngoái at 6:21
ajotatxeajotatxe
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