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

Asked on by totoo

### 3 Answers |Add Yours

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given that sin(t) = 5/7, we need to find cos(x) and tan(x)

We will use trigonometric identities to find cos(x)

We know that sin^2 x + cos^2 x = 1

==> cosx = +-sqrt(1-sin^2 x)

Let us substitute with sinx = 5/7

==> cosx = +-sqrt(1- (5/7)^2]

= +- sqrt(1-25/49)

= +-sqrt(24/49)

= +-2sqrt6/ 7

Then cosx = +- 2sqrt6 / 7

Now we know that tanx = sinx/cosx

==> tanx = (5/7) / (+-2sqrt6 / 7)

= +- 5/2sqrt6

= +-5sqrt6/ 12

==> tanx= +-5sqrt6 / 12

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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 cos t is (sqrt 24)/ 7 or -(sqrt 24) / 7 and tan t = 5/(sqrt 24) or -5/ (sqrt 24)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 317,681 questions. We can answer yours, too.