Advertisements
Advertisements
प्रश्न
If θ = 30°, verify that: 1 - sin 2θ = (sinθ - cosθ)2
Advertisements
उत्तर
Given: θ = 30°
1 - sin2θ
= 1 - sin2 x 30°
= 1 - sin60°
= `1 - sqrt(3)/(2)`
= `(2 - sqrt(3))/(2)`
(sinθ - cosθ)2
= sin2θ + cos2θ - 2sinθ cosθ
= 1 - 2 x sin30° x cos30
= `1 - 2 xx (1)/(2) xx sqrt(3)/(2)`
= `1 - sqrt(3)/(2)`
= `(2 - sqrt(3))/(2)`
⇒ 1 - sin2θ = (sinθ - cosθ)2.
APPEARS IN
संबंधित प्रश्न
If sin x + cos y = 1 and x = 30°, find the value of y
Solve the following equation for A, if sec 2A = 2
Calculate the value of A, if (cosec 2A - 2) (cot 3A - 1) = 0
If θ = 30°, verify that: sin2θ = `(2tanθ)/(1 ++ tan^2θ)`
Evaluate the following: `((1 - cosθ)(1 + cosθ))/((1 - sinθ)(1 + sinθ)` if θ = 30°
If θ < 90°, find the value of: `tan^2θ - (1)/cos^2θ`
If ΔABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC =7units, find ∠B, AB and AC.
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cosec64° + sec70°
Evaluate the following: cos39° cos48° cos60° cosec42° cosec51°
If cos3θ = sin(θ - 34°), find the value of θ if 3θ is an acute angle.
