Given: sin a = 1/3, in which the angle a is in the range of (pi/2, pi). Find the value of cos (a - pi/6).

Solution:

The initial side of the angle alfa lies in the positive x-axis. The terminal side of the angle alfa lies in between pi/2 to pi, which is in Quadrant2. In Quadrant2, sin a > 0 and cos a < 0.

cos (a - pi/6)

= cos a cos pi/6 + sin a sin pi/6

We use formula: cos (a + b) = cos a cos b - sin a sin b. Note: cos pi/6 = cos 30^{o} = sin 60^{o} = (square root of 3)/2. sin pi/6 = sin 30^{o} = 1/2

cos (a - pi/6)

= [(square root of 3)/2] cos a + (1/2) sin a

Now we are going to find cos a

sin^{2}a + cos^{2}a = 1

cos^{2}a = 1 - sin^{2}a

Because in Quadrant2, cos a < 0

cos a = - square root of (1 - sin^{2}a)

given: sin a = 1/3

cos a = - square root of [1 - (1/3)^{2}]

= - square root of (1 - 1/9)

= - square root of 8/9

= - 2 (square root of 2)/3

cos (a - pi/6)

= (square root of 3)/2 × (- 2 square root of 2/3) + (1/2)(1/3)

= 1/6 - square root of 6/3

= 1/6 - 2/2 × (square root of 6)/3

= 1/6 - 2 (square root of 6)/6

= (1/6)(1 - 2 square root of 6)

= - 0.65

Therefore, the value of cos (a - pi/6) is - 0.65. Watch the video for more details.