Eric Nicholas K's answer to Si En's Junior College 1 H2 Maths Singapore question.
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First part
Here the sin y in -1/sin y comes as an “expression”, not an “equation”. Hence, when we express sin y in terms of cos y, there will not be a +- in front of the square root.
Here the sin y in -1/sin y comes as an “expression”, not an “equation”. Hence, when we express sin y in terms of cos y, there will not be a +- in front of the square root.
Date Posted:
4 years ago
y = cos-1(x) has a restricted range of ［0,π ］
when -1 < x < 1, 0 < cos-1(x) < π
sin y = sin(cos-1(x))
Since 0 < cos-1(x) < π,
0 < siny < 1
So sin y is always positive for this range. No ± is needed
Update: if you would like to know why it's restricted, you can check out this link :
https://www.themathpage.com/atrig/inverseTrig.htm#abs
when -1 < x < 1, 0 < cos-1(x) < π
sin y = sin(cos-1(x))
Since 0 < cos-1(x) < π,
0 < siny < 1
So sin y is always positive for this range. No ± is needed
Update: if you would like to know why it's restricted, you can check out this link :
https://www.themathpage.com/atrig/inverseTrig.htm#abs
Hmmm...I did think of the [0, pi] limit but completely forgot that sin of the number between those is never negative
Actually back in JC the explanations for these were never covered... picked it up over the years
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