Mirandaregina
  • Mirandaregina
If csc theta = √5, then what is cot theta?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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Nnesha
  • Nnesha
what is the reciprocal of csc ?
Mirandaregina
  • Mirandaregina
@Nnesha Sine
Nnesha
  • Nnesha
not just sin :=) 1/sin = csc so csc = ?

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anonymous
  • anonymous
In addition to sine, cosine, and tangent, there are three other trigonometric functions you need to know for the Math IIC: cosecant, secant, and cotangent. These functions are simply the reciprocals of sine, cosine, and tangent. Cotangent is the reciprocal of tangent.
Mirandaregina
  • Mirandaregina
@Nnesha so csc = 1/sine
Nnesha
  • Nnesha
yes right so if csc = sqrt{5} then sin = ???
Nnesha
  • Nnesha
remember thy are reciprocal of each other
Mirandaregina
  • Mirandaregina
@Nnesha then sin = 1/√5 = √5/5 ?
phi
  • phi
draw a triangle and label its opposite side 1 and its hypotenuse sqr(5)
Nnesha
  • Nnesha
hmm \[\csc = \frac{ \sqrt{5} }{ 1 } ~~~~~ \sin=\frac{ 1 }{ \sqrt{5} }\] so draw right triangle for now it's okay leave the radical sign at the denominator |dw:1440603990058:dw| \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
Mirandaregina
  • Mirandaregina
@Nnesha so if sin theta = opp/hyp, and opp is 1 and hyp is √5, is it √5/5?
Nnesha
  • Nnesha
yes right sin = opp .hyp|dw:1440604248339:dw| apply the Pythagorean theorem to find 3rd side of right triangle
phi
  • phi
yes you could use sqr(5) for the opposite side and 5 for the hypotenuse but 1 and sqr(5) seem easier (smaller numbers) but either set will give the same answer.
Mirandaregina
  • Mirandaregina
@Nnesha 1^2 + √5^2 = 6
phi
  • phi
c^2 = a^2 + b^2 the unknown is a leg.
Nnesha
  • Nnesha
c=hypotenuse(longest leg of right triangle )
Mirandaregina
  • Mirandaregina
@Nnesha @phi did I use the Pythagorean Theorem correctly? doesn't a = 1 and b = √5 ?
phi
  • phi
no a and b are the legs you know a and c (and you need to find b, the other leg)
phi
  • phi
you know the hypotenuse (because sin is opp/hyp) you know a side and the hypotenuse
Nnesha
  • Nnesha
|dw:1440604696536:dw|
Mirandaregina
  • Mirandaregina
@phi how do I find b?
anonymous
  • anonymous
quick question is 1/3 a rational number
phi
  • phi
a^2 + b^2 = c^2 put in numbers for a and c can you do that ?
Mirandaregina
  • Mirandaregina
@phi but does a = 1? because before I said 1^2 + √5^2 = 6
phi
  • phi
a=1 and c= \( \sqrt{5} \) put those in the formula \[a^2 + b^2 = c^2\]
phi
  • phi
notice c is on the other side of the = sign from a and b
Nnesha
  • Nnesha
i guess she is trying to say that sqrt of 5 +1 equal to 6
Nnesha
  • Nnesha
square of sqrt 5**
phi
  • phi
when you write 1^2 + √5^2 = c^2 a^2 + b^2 = c^2 that means a is 1 and b is sqr(5). but that is not what we want
Mirandaregina
  • Mirandaregina
@phi I understand, thanks :) so I'm not sure if I did this right but I have 1^2 + b^2 = √5 b + 1^2 = √5 b^2 + 1 - 1 = √5 - 1 b^2 = √1.236068
phi
  • phi
almost, but the formula says c^2 so you want (sqrt(5))^2
Mirandaregina
  • Mirandaregina
@phi as is √1.111786 = 1.0544126
Mirandaregina
  • Mirandaregina
*as in
phi
  • phi
you wrote 1^2 + b^2 = √5 but what it should be is 1^2 + b^2 = (√5)^2 now find b
Mirandaregina
  • Mirandaregina
@phi oh my mistake 1^2 + b^2 = (√5)^2 b = 2
phi
  • phi
yes. now you know all 3 sides of the triangle |dw:1440605865953:dw|
phi
  • phi
the want the cotangent of theta (I would write the tangent of theta, then "flip it" to make it the cotangent) or you can remember cotangent is adj/opp
Mirandaregina
  • Mirandaregina
@phi cot theta = adj/opp = 2/1 = 2
phi
  • phi
yes
Mirandaregina
  • Mirandaregina
@phi so that's the overall answer? I got that answer when I first tried it but I thought it was wrong! lol
phi
  • phi
now you know how to do it so you believe it
Mirandaregina
  • Mirandaregina
@phi @Nnesha thank you for all your help and patience everyone :)
Nnesha
  • Nnesha
my pleasure o^_^o

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