(x + y) formula

Chapter 3 Class 11 Trigonometric Functions
Concept wise

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Example 25 If sin π₯ = 3/5 , cos y = β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 β 9/25 cos2x = (25 β 9)/25 cos2x = 16/25 cos x = Β± β(16/25) cos x = Β± 4/5 Since x is in llnd Quadrant cos x is negative So, cos x = (βπ)/π Similarly, Finding sin y We know that sin2 y + cos2 y = 1 sin2 y = 1 β cos2 y sin2 y = 1 β ((β12)/13)^2 sin2 y = 1 β 144/169 ("Given cos y =" (β12)/13) sin2 y = (169 β 144)/169 sin2 y = 25/169 sin y = Β± β(25/169) sin y = Β± 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 β΄ 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 ((β12)/13) + ((β4)/5) (5/13) = (β12 Γ 3)/(5 Γ 13) + ((β4 Γ 5)/(5 Γ 13)) = (β36)/65 + ((β20)/65) = (β36 β20)/65 = (β56)/65 Hence, sin (x + y) = (βππ)/ππ