# Determine sina and cosa if cota=3. a belongs to (0;pi)

hala718 | Certified Educator

cota = 3

But cota = cosa/ sina = 3

==> cos a = 3sin a

But sin^2 a + cos^2 a = 1

==> cos a = sqrt(1-sin^2 a)

==> sqrt(1-sin^2 a ) = 3sin a

Square both sides:

==> 1- sin^2 a = 9sin^2 a

==> 10 sin^2 a = 1

==> sin^2 a = 1/10

==> sin a = 1/sqrt10 = sqrt10/10

==> sin a = sqrt10/10

==> cos a = sqrt(1-sin^2 a)

= sqrt(1- 1/10)

= sqrt(9/10) = 3/sqrt10

But in second quadrant cosine is negative:

==> cos a = +-3/sqrt10

neela | Student

cota = 3 in (0, pi). To find sina and cos a.

Solution:

Cota = 3 Or tana = 1/3. a is in 1st quadrant as it is tana = 1/3 and tana is -ve in 2nd quadrant.

Consider the triangle  with angle  B = 90 and AB = 3, BC =1 and hytemuse CA = sqrt(1^2+3^2) = sqrt10.

tanA = 1/3.

Sina = sinA = BC/AC =  1/10^(1/2)

cosa = AB/AC= 3/sqrt10.

giorgiana1976 | Student

The interval (0,pi) covers the first and the second quadrant where the values of the sine function are positive. The cosine function is positive over the first quadrant, but it's negative in the second quadrant.

We'll use the fundamental formula of trigonometry:

(sin a)^2 + (cos a)^2=1

We'll divide the formula with the value  (sin a)^2:

(sin a)^2/ (sin a)^2 + (cos a)^2/(sin a)^2 = 1 / (sin a)^2

But the ratio cos a/sin a= cot a = 3

The formula will become:

1 + (cot a)^2 = 1/(sin a)^2

sin a = 1/sqrt[1+(cot a)^2]

sin a = 1/sqrt[1+(3)^2]

sin a = 1/sqrt[1+(3)^2]

sin a = 1/sqrt[1+9]

sin a = 1/sqrt 10

sin a = sqrt 10/10

cos a = +/- sqrt (1 - 1/10)

cos a = +/- sqrt (9/10)

cos a = +/- 3/sqrt 10

cos a = +/- (3*sqrt 10)/10