FIND THE VALUE OF COS X MINUS Y WHEN QUADRANTS ARE MENTIONED

Question 1 :

Find cos(x − y), given that cos x = −4/5 with π < x < 3π/2 and sin y = −24/25 with π < y < 3π/2.

Solution :

Given that, cos x = −4/5 and sin y = −24/25 both are lying in the 3rd quadrant.

So all trigonometric ratios other than tan and cot, we will have negative sign. 

cos (x - y)  =  cos x cos y - sin x sin y 

cos x = -4/5 

sin x  =  √(1 - cos2x)  

  =  √(1 - (-4/5)2

  =  √(1 - (16/25))

  =  √(25 - 16)/25 

  =  √(9/25) 

sin x  =  -3/5

sin y = −24/25

cos y  =  √(1 - sin2y)

  =  √(1 - (-24/25)2

  =  √(1 - (576/625)

  =  √(625 - 576)/625 

  =  √49/625 

cos y  =  -7/25

cos (x - y)  =  (-4/5)(-7/25) + (-3/5) (-24/25)

  =  (28/125) + (72/125)

  =  (28 + 72)/125

  =  100/125

  =  4/5

Question 2 :

Find sin(x − y), given that sin x = 8/17 with 0 < x < π/and cos y = −24/25 with π < y < 3π/2

Solution :

Given that, sin x = 8/17 and cos y = −24/25

Here x lies in the 1st quadrant and y lies in the 3rd quadrant.

sin (x - y)  =  sin x cos y - cos x sin y 

sin x = 8/17 

cos x  =  √(1 - sin2x)  

  =  √(1 - (8/17)2

  =  √(1 - (64/289))

  =  √(289 - 64)/289 

  =  √(225/289) 

cos x  =  15/17 (x lies in 1st quadrant)

cos y = −24/25

sin y  =  √(1 - cos2y)

  =  √(1 - (-24/25)2

  =  √(1 - (576/625)

  =  √(625 - 576)/625 

  =  √49/625 

sin  y  =  -7/25 (y lies in the 3rd quadrant)

sin (x - y)  =  (8/17) (-24/25) - (15/17) (-7/25)

  =  (-192/425) + (105/425)

  =  (-192 + 105)/425

  =  -87/425

Hence the value of sin (x - y) is -87/425.

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