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The domain of the function sin⁻¹ (3x -1) is
Let us now solve this function step by step to find out the domain of sin⁻¹(3x - 1). Step 1: Domain of sin⁻¹(y) We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1. In the case of sin⁻¹(3x - 1), we must have: -1 ≤ 3x - 1 ≤ 1 Step 2: Solve the inequality 1. Add 1 to all sides: 0 ≤ 3x ≤ 2 2. DivideRead more
Let us now solve this function step by step to find out the domain of sin⁻¹(3x – 1).
Step 1: Domain of sin⁻¹(y)
We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1.
In the case of sin⁻¹(3x – 1), we must have:
-1 ≤ 3x – 1 ≤ 1
Step 2: Solve the inequality
1. Add 1 to all sides:
0 ≤ 3x ≤ 2
2. Divide through by 3:
0 ≤ x ≤ 2/3
Step 3: Domain
The domain of sin⁻¹(3x – 1) is:
[0, 2/3]
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sin[π/3 + sin⁻¹(1/2)] is equal to
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps: Step 1: Simplify sin⁻¹(1/2) The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is: sin⁻¹(1/2) = π/6 Step 2: Substitute into the given expression Now substitute sin⁻¹(1/2) = π/6 into the expression: sin[π/3 + sin⁻¹(1/2)] =Read more
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps:
Step 1: Simplify sin⁻¹(1/2)
The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is:
sin⁻¹(1/2) = π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(1/2) = π/6 into the expression:
sin[π/3 + sin⁻¹(1/2)] = sin(π/3 + π/6)
Step 3: Simplify π/3 + π/6
Find a common denominator:
π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2
Step 4: Simplify sin(π/2)
From the unit circle, sin(π/2) = 1.
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Which of the following is the principal value branch of cosec⁻¹x?
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties: Step 1: Definition of the principal value branch The principal value branch of cosec⁻¹x is defined such that: 1. It includes all possible values of the inverse cosecant function. 2. It avoids discontinuities or undefiRead more
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties:
Step 1: Definition of the principal value branch
The principal value branch of cosec⁻¹x is defined such that:
1. It includes all possible values of the inverse cosecant function.
2. It avoids discontinuities or undefined values (like when cosec x = 0).
Step 2: Range of cosec⁻¹x
For cosec⁻¹x, the principal value is taken from the range:
[-π/2, π/2] – {0}
This excludes 0 because cosec x is undefined at sin x = 0.
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sin [π/3 – sin⁻¹ (-1/2)] is equal to
We are asked to find sin[π/3 – sin⁻¹(-1/2)]. Step 1: Simplify sin⁻¹(-1/2) The value sin⁻¹(-1/2) is the angle whose sine is -1/2, and the range of sin⁻¹ is [-π/2, π/2]. The angle that satisfies this condition is: sin⁻¹(-1/2) = -π/6 Step 2: Substitute into the given expression Now substitute sin⁻¹(-1/Read more
We are asked to find sin[π/3 – sin⁻¹(-1/2)].
Step 1: Simplify sin⁻¹(-1/2)
The value sin⁻¹(-1/2) is the angle whose sine is -1/2, and the range of sin⁻¹ is [-π/2, π/2].
The angle that satisfies this condition is:
sin⁻¹(-1/2) = -π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(-1/2) = -π/6 into the expression:
sin[π/3 – (-π/6)] = sin[π/3 + π/6]
Step 3: Simplify π/3 + π/6
Find a common denominator:
π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2
Step 4: Calculate sin(π/2)
From the unit circle, sin(π/2) = 1.
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If tan⁻¹ x = y, then:
To solve the problem given, let's go through the properties of the inverse tangent function: Step 1: Define tan⁻¹x = y The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x Step 2: Range of tan⁻¹x The principalRead more
To solve the problem given, let’s go through the properties of the inverse tangent function:
Step 1: Define tan⁻¹x = y
The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x
Step 2: Range of tan⁻¹x
The principal range of the inverse tangent function tan⁻¹x (or arctan) is:
-π/2 ≤ y ≤ π/2
This signifies that y is in the interval (-π/2, π/2).
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