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@elusive
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\[\cos(sin^{-1}(1/2)) = ?\]
thats easy
i think
Haha its made easy on purpose I want you use the triangle method
\[\frac{ \sqrt{3} }{ 2 }\]
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Yep, but no prize to you. You haven't showed steps
I think I got sin, I'll have to study the other functions, now!
|dw:1467919730161:dw|
Perfect ! Here is your medal
thank you :D
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Wana try another problem ?
Yeah!
\[\sin(sin^{-1}(2/3)) = ?\]
you can use "arcsin" instead of sin^-1
I like arcsin too
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|dw:1467920041360:dw| I think that was a trick question! it's just 2/3
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