Given: sin a × cos a = -1/4, in which 3pi/2 < a < 2pi. What is the value of sin a - cos a?

Solution:

For any angle trigonometry, the initial side of the angle a is overlap with the positive x-axis. The terminal side of the angle a lies in between 3pi/2 and 2pi which is in Quadrant4. In Quadrant4, sin a < 0 and cos a > 0.

We are given the product of sin a and cos a and we are asking for the difference between sin a and cos a. Now we make the product of sin a and cos a and difference of sin a and cos a.

(sin a - cos a)^{2} = sin^{2}a = 2 sin a cos a + cos^{2}a

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

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

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

Because in Quadrant4, sin a < 0 and cos a > 0. So, sin a - cos a < 0. The value of sin a - cos a is a negative number. So,

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

= - square root of [1 - 2 × (-1/4)]

= - square root of [1 + 2 × (1/4)]

= - square root of [1 + 1/2]

= - square root of (3/2)

= - (square root of 3)/(square root of 2)

To remove the square root from the denominator, both numerator and denominator times square root of 2.

sin a - cos a = - (square root of 6)/2

Therefore, the value of sin a - cos a = - (square root of 6)/2. Watch the video for more details.