Example 25 - If sin x = 3/5, cos y = -12/13, find sin (x + y)

Example 25 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Example 25 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Example 25 - Chapter 3 Class 11 Trigonometric Functions - Part 4
Example 25 - Chapter 3 Class 11 Trigonometric Functions - Part 5
Example 25 - Chapter 3 Class 11 Trigonometric Functions - Part 6

  1. Chapter 3 Class 11 Trigonometric Functions (Term 2)
  2. Serial order wise

Transcript

Example 25 If sin π‘₯ = 3/5 , cos y = βˆ’12/13 , where π‘₯ and y both lie in second quadrant, find the value of sin (π‘₯ + y). We know that sin (x + y) = sin x cos y + cos x sin y We know that value of sin x and cos y but we do not know of cos x and sin y Let us first find cos x We know that sin2x + cos2x = 1 (3/5)^2+ cos2x = 1 9/25 + cos2x = 1 9/25 + cos2x = 1 cos2x = 1 – 9/25 cos2x = (25 βˆ’ 9)/25 cos2x = 16/25 cos x = Β± √(16/25) cos x = Β± 4/5 Since x is in llnd Quadrant cos x is negative So, cos x = (βˆ’πŸ’)/πŸ“ Similarly, Finding sin y We know that sin2 y + cos2 y = 1 sin2 y = 1 – cos2 y sin2 y = 1 – ((βˆ’12)/13)^2 sin2 y = 1 – 144/169 ("Given cos y =" (βˆ’12)/13) sin2 y = (169 βˆ’ 144)/169 sin2 y = 25/169 sin y = Β± √(25/169) sin y = Β± √((5 Γ— 5)/(13 Γ—13)) sin y = Β± 5/13 sin y = Β± 5/13 Since y lies in llnd Quadrant So, sin y is positive ∴ sin y = 5/13 Now, Putting value of sin x , sin y, cos x, cos y in sin (x + y) = sin x cos y + cos x sin y = 3/5 Γ— ((βˆ’12)/13) + ((βˆ’4)/5) (5/13) = (βˆ’12 Γ— 3)/(5 Γ— 13) + ((βˆ’4 Γ— 5)/(5 Γ— 13)) = (βˆ’36)/65 + ((βˆ’20)/65) = (βˆ’36 βˆ’20)/65 = (βˆ’56)/65 Hence, sin (x + y) = (βˆ’πŸ“πŸ”)/πŸ”πŸ“

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.