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Precalculus
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OpenStudy (anonymous):
sin(π/4)=cos @dumbcow
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OpenStudy (dumbcow):
sin(pi/4) = sqrt2/2
cos x = sqrt2/2 ---> x = pi/4
OpenStudy (anonymous):
sin(π/6)=cos
OpenStudy (dumbcow):
let me explain it this way
|dw:1375638733282:dw|
sin(pi/6) = y
and
cos A = y
thus
sin(pi/6) = cosA
pi/6 +A = pi/2
--> A = pi/3
OpenStudy (anonymous):
I get it!
OpenStudy (anonymous):
thank you for sooo much help @dumbcow
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OpenStudy (dumbcow):
your welcome
OpenStudy (anonymous):
so I'm always going to set it up like this? sin(pi/#) = cosA @dumbcow
OpenStudy (dumbcow):
yes as long as its in the form sin = cos
in which case the 2 angles will always add up to pi/2
OpenStudy (anonymous):
what if its cos=sin @dumbcow ?
OpenStudy (dumbcow):
same thing, it doesnt matter if you flip everything
x=5 is same as 5=x
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OpenStudy (dumbcow):
cos(pi/#) = sin A
where A is still the other angle in triangle
pi/# + A = pi/2
OpenStudy (anonymous):
so its the same way but the cos and sin are in different spots?
OpenStudy (anonymous):
@dumbcow
OpenStudy (dumbcow):
yeah
OpenStudy (dumbcow):
fyi i need to log off now
if have more questions, just tag me and i can look at it later
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OpenStudy (anonymous):
ALRIGHT THANK YOU SOO MUCH!
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