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About LinkBacksShow that cos(sin^(-1) v + cos^(-1) v) = 0.
LetOriginally Posted by magentarita
and
First, we can rewriteusing the identity
Its good to note thatand
[why do you think this is the case?]. So our expression can be written as
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Now let's construct two different triangles, one that has an angleand an angle
Starting withand that the hypotenuse has a value of
. Using the pythagorean theorem, we can find the value of the adjacent side:
^2+(\text{adj})^2=(\text{hyp})^2\impli es (\text{adj})^2=(\text{hyp})^2-(\text{opp})^2\implies (\text{adj})^2=1^2-v^2)
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Now, we construct at triangle for. We see that the side adjacent of the angle has a value of
and that the hypotenuse has a value of
^2+(\text{adj})^2=(\text{hyp})^2\impli es (\text{opp})^2=(\text{hyp})^2-(\text{adj})^2\implies (\text{opp})^2=1^2-v^2)
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Now what do you thinkequals?
--Chris
haha...wow...I'm surprised I didn't notice that! Good one!and
are complementary angles! What is cos(90 degrees)?
The way I ended up doing this is acceptable, given that they didn't realize thatand
were complementary angles [Like I did
]
--Chris
This question is very involved.haha...wow...I'm surprised I didn't notice that! Good one!
The way I ended up doing this is acceptable, given that they didn't realize that
--Chris
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