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anonymous
 one year ago
If sin θ = 12 over 37, use the Pythagorean Identity to find cos θ.
anonymous
 one year ago
If sin θ = 12 over 37, use the Pythagorean Identity to find cos θ.

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1Identity is \[\huge\rm sin^2 \theta +\cos^2 \theta =1\] substitute sin for 12/37 solve for cos

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1or you can draw a right triangle ^^^\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440269641978:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alright thx. wasn't sure i was solving it right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Refer to the attachment.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos θ = ± 35 over 37 cos θ = ± 23 over 37 cos θ = ± 35 over 12 cos θ = ± 20 over 35 are my answer choices... little confused now.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you need cos theta for right triangle you hve to apply Pythagorean theorem \[\huge\rm a^2+b^2=c^2\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1cos = adjacent over hypotenuse

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm \sin \theta =\frac{ opposite }{ hyp } =\frac{ 12 }{ 37 }\]dw:1440270658205:dw find 3rd side of right triangle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats where i get lost. do i just substitute each of those in for the variables

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1remember c is the longest side of right triangle (hypotenuse )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a^2+12^2=37^2... right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and if i rearrange it to solve for a i get 35^2?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you would subtract 12^2 both sides or you can take square of 12 first and the subtract it doesn't matter \[c^2= \sqrt{37^212^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the answer is +35/37?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you help with one more?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What are the sine, cosine, and tangent of 5 pi over 4 radians?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i can't find the tan

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm tan \theta =\frac{ \sin \theta }{ \cos \theta }\] :=)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1on the unit circle there are cos and sin (x,y)(cos,sin) xcoordinate represent cos ycoordinate represents sin so tan is sin over cos

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sin Θ = negative square root 2 over 2; cos Θ = negative square root 2 over 2; tan Θ = 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you, would give 2 medals.. but one at a time :]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1ahha that fine one medal is enough :) & my pleasure
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