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

Transcript

Example, 25 If sin ๐ฅ = 3/5 , cos y = โ 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 โ 9)/25 cos2x = 16/25 cos x = ยฑ โ(16/25) cos x = ยฑ 4/5 Since x is in llnd Quadrant cos x is negative So cos x = (โ4)/5 Similarly, Finding sin y We know that sin2 y + cos2 y = 1 sin2 y = 1 โ cos2 y sin2 y = 1 โ ((โ12)/13)^2 sin2 y = 1 โ 144/169 sin2 y = (169 โ 144)/169 sin2 y = 25/169 sin y = ยฑ โ(25/169) sin y = ยฑ โ((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 ( (โ 12)/13 ) + ( (โ4)/5 ) ( 5/13 ) = (โ 12 ร 13)/(5 ร 13) + ((โ4 ร 5)/(5 ร 13)) = (โ36)/65 + ((โ20)/65) = (โ36 โ20)/65 = (โ56)/65 Hence sin (x + y) = (โ56)/65