Find principal value of sin -1 -1/2
WebThe principal values of trigonometric functions can be computed through a sequence of steps. First, we find the quadrant in which this principle value exists. Here θ is the acute angle and the principal values in the four quadrants can be obtained as follows. In the first quadrant, the angle is either θ, or π/2 - θ. Websin -1 (1) = Π/2 (in radian) Since the inverse sin -1 (1) is 90° or Π/2. ‘1’ denotes the maximum value of the sine function. Hence, for every 90 degrees it will happen, such as at Π/2, 3Π/2, and so on. By this we can conclude that; sin -1 (1) = Π/2+2Πk (for any integer k) Value Of Sin 15 Value Of Sin 180 Sin 90 Degrees Law of Sines Value of sin 0
Find principal value of sin -1 -1/2
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WebMay 24, 2024 · May 24, 2024 The inverse sine is multivalued, so we need to include 2π 3, its supplement which shares a sine, and all coterminal angles: arcsinsin( 2π 3) = 2π 3 +2πk or π 3 +2πk integer k The principal value is π 3 Answer link WebJul 20, 2024 · Find the principal value of `sin^ (-1)` ` (1/ (sqrt (2)))`... Doubtnut 2.69M subscribers Subscribe 2.4K views 4 years ago Question From - NCERT Maths Class 12 Chapter 2 SOLVED …
WebJun 23, 2011 · or. Note: We could also find the sine of 15 degrees using sine (45° − 30°). sin 75°: Now using the formula for the sine of the sum of 2 angles, sin ( A + B) = sin A … WebJul 25, 2016 · arcsin(sin(2rad × 180∘ πrad)) = arcsin[sin(( 360 π)∘)] = arcsin(sin(114.59∘)) The sin(114.59∘) evaluates to about 0.9093, and the arcsin of that would then be …
WebFind, without the use of a calculator, the principal value of sin⁻¹ (-0.5) in radians. Principal values: − π 2 ≤ sin − 1 x ≤ π 2 sin π 6 = 1 2 sin π 6 = 0.5 sin ( − π 6) = − sin π 6 = − 0.5 ∴ sin − 1 ( − 0.5) = − π 6 WebFind the principal value of sin - 1 - 1 2 Solution Solve using trigonometric rules: The given inverse trigonometric function is sin - 1 - 1 2. Let, sin - 1 - 1 2 = y ⇒ - 1 2 = sin y [ ∵ sin - 1 x = y ⇒ x = sin y] ⇒ sin y = - 1 2 As we know, the range of principal value of sin - 1 x lies in π π π π - π 2, π 2 ⇒ π π sin y = sin - π 6 ⇒ π π y = - π 6
WebMar 30, 2024 · Ex 2.1, 10 (Method 1) Find the principal value of cosec–1 (–√2) Let y = cosec–1 (– √2) y = −cosec–1 (√2) y = − 𝝅/𝟒 Since Range of cosec−1 is [−π/2,π/2] − {0} Hence, Principal Value is (−𝝅)/𝟒 We know that cosec−1 (−x) = − cosec −1 x Since cosec 𝜋/4 = √2 𝜋/4 = cosec−1 (√2) Ex 2.1, 10 (Method 2) Find the principal value of cosec–1 (–√2) Let y = …
WebMar 22, 2024 · Transcript Example 1 Find the principal value of sin–1 (1/√2). Let y = sin–1 (1/√2) sin y = 1/√2 sin y = sin (𝝅/𝟒) ∴ y = 𝝅/𝟒 Since range of principal value of sin −1 is [ (−𝝅)/𝟐, ( 𝝅)/𝟐] Hence, Principal Value is 𝝅/𝟒 (Since sin 𝜋/4 = 1/√2) Next: Example 2 → Ask a doubt Chapter 2 Class 12 Inverse Trigonometric Functions Serial order wise Examples swpl live scoresWebThe first step is to fill in the sin (x) value in the blank text field. Select the type of angle which can either be in degrees Celsius (°) or radians (rad). Afterwards, click the equal sign (=) or the ‘Calculate’ button to perform the calculation. The results will be displayed at the bottom platform of the calculator below the two controls. swpl facebookWebMay 6, 2024 · To Find : Value of Sin⁻¹ ( Sin 4π/5 ) Solution: Sin⁻¹ ( Sin 4π/5 ) as we know that Sinα = Sin (π - α) α = 4π/5 => Sin (4π/5) = Sin ( π - 4π/5) => Sin (4π/5) = Sin ( (5π - 4π)/5) => Sin (4π/5) = Sin ( π/5) Sin⁻¹ ( Sin 4π/5 ) = Sin⁻¹ ( Sin π/5 ) = π/5 Sin⁻¹ ( Sin 4π/5 ) = π/5 = 36° Learn more: swp live tickerWebThe computation is straightforward. The values of $z^w$ are all the possible values of $e^{w\log z}$. So here, you want the possible values of $e^{\frac{2}{\pi}\log i}$. text from unknown number saying helloWebThis calculator will find the inverse trigonometric values for principal values in the ranges listed in the table. You can view the ranges in the Inverse Trigonometric Function Graphs. swp leather welding maksWebApr 18, 2024 · Explanation: Recall the following Definition of sin−1 function : sin−1x = θ, x ≤ 1 ⇔ sinθ = x,θ ∈ [ − π 2, π 2]. Replacing x by sinθ, we get, sin−1(sinθ) = θ,θ ∈ [ − π 2, π 2]. Now, sin−1(sin(3 π 4)) = sin−1{sin(π− π 4)} = sin−1(sin( π 4)), = π 4, ∵, π 4 ∈ [ − π 2, π 2]. ⇒ sin−1(sin(3 π 4)) = π 4 ≠ 3 π 4. Answer link text from unknown number scamWebSep 9, 2016 · The range of arcsin is defined as between − π 2 radians and π 2 radians. arcsin (x) is the one angle within this range whose sine is x. The one angle between − π … text from unknown number that says hello