Ex 5.3, 8 Find dy/dx in, sin2 x + cos2 y = 1 Class 12
Sin2 X Cos2 X. = x 8 − 1 8 × sin4x 4 +c. X ∈ { π 4, 3π 4, 5π 4, 7π 4 } answer link
Ex 5.3, 8 Find dy/dx in, sin2 x + cos2 y = 1 Class 12
From the last step, you can just arctan both sides twice, once with 3 and another with − 3 and you will have both of your answers. Web 3(sin2x +cos2x) = 3 explanation: Web sin 2x cos 2x is one such trigonometric identity that is important to solve a variety of trigonometry questions. ∫sin2xcos2xdx = 1 4 ∫(4sin2xcos2x)dx. Sine function of an angle (theta) is the ratio of the opposite side to the hypotenuse. = x 8 − 1 8 ∫cos4xdx. Sin 2 ( x) = 3 cos 2 ( x) sin 2 ( x) cos 2 ( x) = 3 tan 2 x = 3 tan x = ± 3. (image will be uploaded soon) sine (sin): Web sin2(x) − cos2(x) = − cos(2x) in general, cos(u) = 0 ⇔ u = nπ 2 for some n ∈ z thus we have sin2(x) − cos2(x) = 0 ⇒ −cos(2x) = 0 ⇒ 2x = nπ 2 for n ∈ z ⇒ x = nπ 4 for n ∈ z restricting our values to the interval [0,2π] gives our final result: Apply the pythagorean identity sin2θ + cos2θ = 1.
Sine function of an angle (theta) is the ratio of the opposite side to the hypotenuse. We know, (sin2x +cos2x = 1) (−(sin2x +cos2x = −(1))). How do you simplify the expression −(sin2x +cos2x) ? Web sin2(x) − cos2(x) = − cos(2x) in general, cos(u) = 0 ⇔ u = nπ 2 for some n ∈ z thus we have sin2(x) − cos2(x) = 0 ⇒ −cos(2x) = 0 ⇒ 2x = nπ 2 for n ∈ z ⇒ x = nπ 4 for n ∈ z restricting our values to the interval [0,2π] gives our final result: In other words, sinθ is the opposite side divided by the hypotenuse. Sin 2 ( x) = 3 cos 2 ( x) sin 2 ( x) cos 2 ( x) = 3 tan 2 x = 3 tan x = ± 3. Apply the pythagorean identity sin2θ + cos2θ = 1. ∫sin2xcos2xdx = 1 4 ∫(4sin2xcos2x)dx. = x 8 − sin4x 32 +c. = x 8 − 1 8 × sin4x 4 +c. X ∈ { π 4, 3π 4, 5π 4, 7π 4 } answer link
