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find the values of y in the following equqtions
(1) siny=cos48degree
(2) cos(90-y)=sin56degree47
(3) find the value of sin 0,and cos0,if tan0 equal to 43
A useful thing to know here is that siny = cos(90-y).
So in the first question you can say that cos(90-y) = cos48 and work from there.
I'm not sure exactly what the second question is saying, but I'm guessing that you can turn the cos(90-y) into siny to make things simpler.
The third question is a little harder. tanθ = sinθ/cosθ, so if tanθ = 43, then sinθ = 43cosθ.
Also, sin²θ + cos²θ = 1, for any value of θ.
Hence, (43cosθ)² + (cosθ)² = 1.
1850cos²θ = 1
cosθ = √(1/1850).
You can find sinθ by using that result and the bolded equation.
Why did the vector cross the road?
It wanted to be normal.
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And in case you didn't know, -1 ≤ sinθ ≤ 1 and -1 ≤ cosθ ≤ 1.
"Unit Circle" is good to know.
http://www.sparknotes.com/testprep/book … ion5.rhtml
igloo myrtilles fourmis
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