anonymous
  • anonymous
given the sin a=-2/3 and a is in quadrant what are the exact values of the following trigonometric functions of a? cos a= tan a= csc a=
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
what quadrant is a in?
anonymous
  • anonymous
3
anonymous
  • anonymous
Sin=y/r Cos=x/r Tan=y/x Csc=r/r

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More answers

anonymous
  • anonymous
|dw:1433564957553:dw|
anonymous
  • anonymous
And remember that in quadrant 3 only tang is positive
anonymous
  • anonymous
Sinx is negative in the 3rd quadrant
anonymous
  • anonymous
and also in the fourth quadrant
anonymous
  • anonymous
sin=-2/3
anonymous
  • anonymous
Do you know how to approach this question ?
anonymous
  • anonymous
has anyone suggested drawing a triangle yet? it would really make things easy
anonymous
  • anonymous
no not really
anonymous
  • anonymous
the do it label the "opposite side " 2, the hypotenuse 3, and use pythagoras to find the length of the adjacent side
anonymous
  • anonymous
|dw:1433565255841:dw|
anonymous
  • anonymous
5
anonymous
  • anonymous
oh no!
anonymous
  • anonymous
\[2^2+?^2=3^2\\ ?=\sqrt{3^2-2^2}\]
anonymous
  • anonymous
1^2
anonymous
  • anonymous
is it right @satellite73
anonymous
  • anonymous
no not at all
anonymous
  • anonymous
what is \(3^2\)?
anonymous
  • anonymous
9
anonymous
  • anonymous
ok and \(2^2\)?
anonymous
  • anonymous
4
anonymous
  • anonymous
and \[9-4\]?
anonymous
  • anonymous
5
anonymous
  • anonymous
so what is \[\sqrt{3^2-2^2}\]?
anonymous
  • anonymous
5^2?
anonymous
  • anonymous
no dear not \(5^2\) but rather \(\sqrt 5\)
anonymous
  • anonymous
my bad
anonymous
  • anonymous
then what next
anonymous
  • anonymous
Then you use the sides to find the rest of the trig functions
anonymous
  • anonymous
|dw:1433566350237:dw|
anonymous
  • anonymous
now you can take any trig ratio you want, just remember what quadrant you are in to make sure to know if it is positive or negative
anonymous
  • anonymous
so for cos a
anonymous
  • anonymous
so for cos a its 5/3 @satellite73
anonymous
  • anonymous
and for tan a 2/sqrt5
anonymous
  • anonymous
You can't leave a radical in the denominator. You would have to multiply the 2/sqrt5 by sqrt5/sqrt5
anonymous
  • anonymous
so like 3/2sqrt5

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