anonymous
  • anonymous
If sin θ = 12 over 37, use the Pythagorean Identity to find cos θ.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Nnesha
  • Nnesha
Identity is \[\huge\rm sin^2 \theta +\cos^2 \theta =1\] substitute sin for 12/37 solve for cos
Nnesha
  • Nnesha
|dw:1440269530637:dw|
Nnesha
  • Nnesha
or you can draw a right triangle ^^^\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

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anonymous
  • anonymous
|dw:1440269641978:dw|
anonymous
  • anonymous
alright thx. wasn't sure i was solving it right.
Nnesha
  • Nnesha
np :)
anonymous
  • anonymous
Refer to the attachment.
1 Attachment
anonymous
  • anonymous
cos 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
  • Nnesha
you need cos theta for right triangle you hve to apply Pythagorean theorem \[\huge\rm a^2+b^2=c^2\]
Nnesha
  • Nnesha
cos = adjacent over hypotenuse
Nnesha
  • Nnesha
\[\huge\rm \sin \theta =\frac{ opposite }{ hyp } =\frac{ 12 }{ 37 }\]|dw:1440270658205:dw| find 3rd side of right triangle
anonymous
  • anonymous
thats where i get lost. do i just substitute each of those in for the variables
Nnesha
  • Nnesha
remember c is the longest side of right triangle (hypotenuse )
anonymous
  • anonymous
a^2+12^2=37^2... right?
Nnesha
  • Nnesha
yes right!!
anonymous
  • anonymous
and if i rearrange it to solve for a i get 35^2?
Nnesha
  • Nnesha
hmm not square
Nnesha
  • Nnesha
you would subtract 12^2 both sides or you can take square of 12 first and the subtract it doesn't matter \[c^2= \sqrt{37^2-12^2}\]
anonymous
  • anonymous
35
Nnesha
  • Nnesha
yes right
anonymous
  • anonymous
so the answer is +-35/37?
Nnesha
  • Nnesha
yes right :=)
anonymous
  • anonymous
thx much
anonymous
  • anonymous
can you help with one more?
Nnesha
  • Nnesha
i'll try
anonymous
  • anonymous
What are the sine, cosine, and tangent of 5 pi over 4 radians?
anonymous
  • anonymous
i can't find the tan
Nnesha
  • Nnesha
\[\huge\rm tan \theta =\frac{ \sin \theta }{ \cos \theta }\] :=)
Nnesha
  • Nnesha
on the unit circle there are cos and sin (x,y)(cos,sin) x-coordinate represent cos y-coordinate represents sin so tan is sin over cos
anonymous
  • anonymous
so then its -1?
anonymous
  • anonymous
sin Θ = negative square root 2 over 2; cos Θ = negative square root 2 over 2; tan Θ = -1?
Nnesha
  • Nnesha
yes right :=)
anonymous
  • anonymous
thank you, would give 2 medals.. but one at a time :]
Nnesha
  • Nnesha
ahha that fine one medal is enough :) & my pleasure

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