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If we know that sin(t) = 5/7 , how to find other identies like cos(t) and tan(t) ?

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Tushar Chandra eNotes educator | Certified Educator

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We know that sin t = 5/7. We use the relation (sin x)^2 + (cos x)^2 = 1, to find cos t

(cos t)^2 + (5/ 7)^2 = 1

=> (cos t)^2 = 1 - 25 / 49

=> (cos t)^2 = 24 / 49

=> cos t = (sqrt 24) / 7 or -(sqrt 24) / 7

tan t = sin t / cos t

=> (5/7)/ [(sqrt 24)/7] or (5/7)/ [-(sqrt 24)/7]

=> 5/ (sqrt 24) or -5/ (sqrt 24)

The required value of...

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hala718 eNotes educator | Certified Educator

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giorgiana1976 | Student

We'll start from the identity:

1 + (cot t)^2 = 1/(sin t)^2

(cot t)^2 = 1/(sin t)^2 - 1

(cot t)^2 = 49/25 - 1

(cot t)^2 = (49-25)/25

(cot t)^2 = 24/25

cot t = 2sqrt6/5 or cot t = -2sqrt6/5

We know that tan t = 1/cot t

tan t = 5/2sqrt6 => tan t = 5sqrt6/12 or tan t = -5sqrt6/12

We also know that tan t = sin t/cos t

cos t = sin t/tan t

cos t = 5/7/5sqrt6/12 => cos t = 12/7sqrt6

cos t = 12sqrt6/42 = 6sqrt6/21

cos t = -6sqrt6/21

The values for cos t and tan t are: cos t = +6sqrt6/21; cos t = - 6sqrt6/21 ;  tan t = 5sqrt6/12 ; tan t = -5sqrt6/12.