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

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Example, 25 If sin ๐‘ฅ = 3/5 , cos y = โˆ’ 12/13 , where ๐‘ฅ and y both lie in second quadrant, find the value of sin (๐‘ฅ + 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 โ€“ 9/25 cos2x = (25 โˆ’ 9)/25 cos2x = 16/25 cos x = ยฑ โˆš(16/25) cos x = ยฑ 4/5 Since x is in llnd Quadrant cos x is negative So cos x = (โˆ’4)/5 Similarly, Finding sin y We know that sin2 y + cos2 y = 1 sin2 y = 1 โ€“ cos2 y sin2 y = 1 โ€“ ((โˆ’12)/13)^2 sin2 y = 1 โ€“ 144/169 sin2 y = (169 โˆ’ 144)/169 sin2 y = 25/169 sin y = ยฑ โˆš(25/169) sin y = ยฑ โˆš((5 ร— 5)/(13 ร—13)) sin y = ยฑ 5/13 sin y = ยฑ 5/13 Since y lies in llnd Quadrant so, sin y is positive So sin y = 5/13 Only 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 ร— ( (โˆ’ 12)/13 ) + ( (โˆ’4)/5 ) ( 5/13 ) = (โˆ’ 12 ร— 13)/(5 ร— 13) + ((โˆ’4 ร— 5)/(5 ร— 13)) = (โˆ’36)/65 + ((โˆ’20)/65) = (โˆ’36 โˆ’20)/65 = (โˆ’56)/65 Hence sin (x + y) = (โˆ’56)/65

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