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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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sin (tan⁻¹x), where |x| < 1, is equal to
We are given to determine sin(tan⁻¹x), where |x| < 1 . Step 1: Let us take θ = tan⁻¹x . This gives us the following equations: tan(θ) = x Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below: -Opposite side= x Adjacent siRead more
We are given to determine sin(tan⁻¹x), where |x| < 1 .
Step 1: Let us take θ = tan⁻¹x .
This gives us the following equations:
tan(θ) = x
Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below:
-Opposite side= x
Adjacent side = 1
Step2: Applying Pythagorean theorem
To find the hypotenuse, use the Pythagorean theorem:
Hypotenuse = √(1² + x²) = √(1 + x²)
Step 3: Calculate sin(θ)
We know that:
sin(θ) = Opposite / Hypotenuse = x / √(1 + x²)
Final Answer:
Thus, the value of sin(tan⁻¹x) is:
x / √(1 + x²)
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If sin⁻¹ x > cos⁻¹ x, then x, should lie in the interval
We are given to find the interval where sin⁻¹x > cos⁻¹x. Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π]. For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy: sin⁻¹x > cos⁻¹x Step 2:Read more
We are given to find the interval where sin⁻¹x > cos⁻¹x.
Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π].
For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy:
sin⁻¹x > cos⁻¹x
Step 2: Use the identity sin⁻¹x + cos⁻¹x = π/2 From the identity:
sin⁻¹x + cos⁻¹x = π/2 We can subtract cos⁻¹x from both sides to get:
sin⁻¹x = π/2 – cos⁻¹x Thus, for sin⁻¹x > cos⁻¹x, we need:
π/2 – cos⁻¹x > cos⁻¹x That is:
π/2 > 2cos⁻¹x which gives:
cos⁻¹x 1/√2
Hence, the required condition sin⁻¹x > cos⁻¹x holds when x belongs to:
(1/√2, 1)
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