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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

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Transcript

Example 18 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.