Example 25 - Chapter 3 Class 11 Trigonometric Functions

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Example, 25 If sin 𝑥 = 3/5 , cos y = − 12/13 , where 𝑥 and y both lie in second quadrant, find the value of sin (𝑥 + 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 − 9)/25 cos2x = 16/25 cos x = ± √(16/25) cos x = ± 4/5 Since x is in llnd Quadrant cos x is negative So cos x = (−4)/5 Similarly, Finding sin y We know that sin2 y + cos2 y = 1 sin2 y = 1 – cos2 y sin2 y = 1 – ((−12)/13)^2 sin2 y = 1 – 144/169 sin2 y = (169 − 144)/169 sin2 y = 25/169 sin y = ± √(25/169) sin y = ± √((5 × 5)/(13 ×13)) sin y = ± 5/13 sin y = ± 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 × ( (− 12)/13 ) + ( (−4)/5 ) ( 5/13 ) = (− 12 × 13)/(5 × 13) + ((−4 × 5)/(5 × 13)) = (−36)/65 + ((−20)/65) = (−36 −20)/65 = (−56)/65 Hence sin (x + y) = (−56)/65