コレクション cosθ=√3/2 804006-Sinθ+cosθ/sinθ-cosθ=3+2√2
Precalculus Solve for x cos (x)= ( square root of 3)/2 cos (x) = √3 2 cos ( x) = 3 2 Take the inverse cosine of both sides of the equation to extract x x from inside the cosine x = arccos( √3Take the inverse cosine of both sides of the equation to extract θ θ from inside the cosine θ = arccos(− √3 2) θ = arccos ( 3 2) Simplify the right side Tap for more steps θ = 5π 6 θ = 5 πCosθ =√5/3 Stepbystep explanation The formula for cscθ =hypotenuse/opposite cscθ =1/sinθ If cscθ =3/2 this means the hypotenuse is 3 and the opposite side value is 2Find the adjacent
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Sinθ+cosθ/sinθ-cosθ=3+2√2
Sinθ+cosθ/sinθ-cosθ=3+2√2-Solution Given , √3 sin θ= cos θ ⇒ sinθ cosθ= 1 √3 or, tan θ= 1 √3 ⇒ tan θ= tan 30∘ ∵tan30∘ = 1 √3 ⇒ θ=30∘ On simplifying the given equation, we get 3 cos2θ2 cos θ 3 cos θ2 = cos sin A = √ ( 3^2 (√2)^2 ) / 3 = √ 9 2 / 3 = √7 / 3 tan B = 4/3 cos B = 3 / sqrt (3^2 4^2) = 3 / sqrt 25 = 3/5 sin B = 4/sqrt (3^2 4^2) = 4 /sqrt 25 = 4/5 So we have sin
Given Costheta 3 2 Which Of The Following Are The Possible Values Of Sin 2 Theta
Calculation sinθ = ± √ (1 – cos 2 θ) ⇒ √ 1 – (√3/2) 2 } = √ (4 – 3)/4 = ± (1/2) Here θ lies in the third quadrant, value of sinθ is negative So, sinθ = (1/2) Sinθ = 1/cosecθ ⇒ cosecθ = – 2If α and β are the zeros of the quadratic polynomial f(x)=3x25x2 The length of a rectangle exceeds its breadth by 5m If the width is increased by 1 m and the length is decreased by 2m1 2 sinθ cosθ = (3/4) (1/4) (√3/2) 2 sinθ cosθ = √3/2 sin2θ = √3/2 (2 sinθ cosθ = sin2θ) ⇒ 2θ = 60° ⇒ θ = 30° Now, tanθ = tan30° ⇒ 1/√3 ∴ the value of tanθ will be 1/√3 Shortcut Trick
1 Answer 1 vote answered by Anaswara (315k points) selected by Dheeya Best answer Given cos θ = −√3 2 − 3 2 Since, θ is in IIIrd Quadrant P(sinθ, cosθ) (0 ≤ θ ≤2π) be a point in triangle with vertices (0, 0), ( √3/2,0) and (0, √3/ 2) if If sin θ cos θ = √2, then tan θ cot θ = (a) 1 (b) 2 (c) 3 (d) 4 This question is Similar to Question 8 CBSE Class 10 Sample Paper for Boards Maths Standard Support
Answer (1 of 7) If cosθ=√3/2, Then,squaring on both sides,we get =>cos²θ=(√3/2)² =>cos²θ=3/4 =>1/cos²θ =4/3 =>sec²θ =4/3, because 1/cosθ=secθ =>1tan²θ =4/3, becauseSinθ=√3/2 =>θ is 60° (Q1) or1° (Q2) cosθ = 1/2 =>θ is 1° (Q2) or 360–1 = 240° (Q3) By comparing the two conditions, θ lies in 2nd quadrant (Q2) SCALER Helping Software & Data √3sinθ cosθ = √2 Dividing both sides by 2 , We get ( √3sinθ cosθ ) / 2 = √2/2 (√3/2) sinθ (1/2) cosθ = 1/√2 cos30° sinθ sin30° cosθ = sin45° sin( θ 30° ) = sin45 ° using
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