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

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

Mathematics
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Identity is \[\huge\rm sin^2 \theta +\cos^2 \theta =1\] substitute sin for 12/37 solve for cos
|dw:1440269530637:dw|
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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|dw:1440269641978:dw|
alright thx. wasn't sure i was solving it right.
np :)
Refer to the attachment.
1 Attachment
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.
you need cos theta for right triangle you hve to apply Pythagorean theorem \[\huge\rm a^2+b^2=c^2\]
cos = adjacent over hypotenuse
\[\huge\rm \sin \theta =\frac{ opposite }{ hyp } =\frac{ 12 }{ 37 }\]|dw:1440270658205:dw| find 3rd side of right triangle
thats where i get lost. do i just substitute each of those in for the variables
remember c is the longest side of right triangle (hypotenuse )
a^2+12^2=37^2... right?
yes right!!
and if i rearrange it to solve for a i get 35^2?
hmm not square
you would subtract 12^2 both sides or you can take square of 12 first and the subtract it doesn't matter \[c^2= \sqrt{37^2-12^2}\]
35
yes right
so the answer is +-35/37?
yes right :=)
thx much
can you help with one more?
i'll try
What are the sine, cosine, and tangent of 5 pi over 4 radians?
i can't find the tan
\[\huge\rm tan \theta =\frac{ \sin \theta }{ \cos \theta }\] :=)
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
so then its -1?
sin Θ = negative square root 2 over 2; cos Θ = negative square root 2 over 2; tan Θ = -1?
yes right :=)
thank you, would give 2 medals.. but one at a time :]
ahha that fine one medal is enough :) & my pleasure

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