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Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab?
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Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab?
Problem:
Given that sin inverse (x) = a b and sin inverse (y) = a - b, we need to find the value of ab.
Solution:
To find the value of ab, we need to analyze the given information and use the properties of inverse trigonometric functions. Let's break down the solution into smaller steps:
Step 1: Understand the meaning of sin inverse (x) = a b and sin inverse (y) = a - b.
- The notation sin inverse (x) represents the inverse sine function, also denoted as arcsin(x). It gives the angle whose sine is x.
- Similarly, sin inverse (y) represents the inverse sine function of y, i.e., arcsin(y). It gives the angle whose sine is y.
- In this context, a and b are the angles (in radians or degrees) whose sine values are x and y, respectively.
Step 2: Relate the given information to the properties of inverse trigonometric functions.
- The sine function is periodic with a period of 2π radians or 360 degrees. This means that for any real number k, sin(a + 2πk) = sin(a).
- The inverse sine function, arcsin(x), gives the principal value of the angle whose sine is x in the range [-π/2, π/2] or [-90°, 90°].
- Using the above properties, we can see that sin(a + 2πk) = sin(a - 2πk) = sin(a) for any integer k.
Step 3: Determine the relationship between a, b, x, and y.
- From sin inverse (x) = a b, we know that the angle whose sine is x is represented by a b.
- From sin inverse (y) = a - b, we know that the angle whose sine is y is represented by a - b.
- Since the sine function is periodic, we can write sin(a b) = sin(a - b) by adding or subtracting multiples of 2π from a b and a - b.
Step 4: Use trigonometric identities to simplify the equation.
- Applying the sum-to-product identity for sine, we have sin(a)cos(b) + cos(a)sin(b) = sin(a)cos(b) - cos(a)sin(b).
- Simplifying further, sin(a)cos(b) - sin(a)cos(b) = 0.
- This implies that sin(a)cos(b) = 0.
Step 5: Determine the possible values of ab.
- The equation sin(a)cos(b) = 0 holds true for two cases:
1. sin(a) = 0 and cos(b) ≠ 0
2. sin(a) ≠ 0 and cos(b) = 0
- In the first case, sin(a) = 0 implies a = 0 or a = π. Since the range of arcsin is [-π/2, π/2], a = π is not in the valid range.
- In the second case, cos(b) = 0 implies b = π/2 or b = 3π/2. Both values fall within the valid range of arcs
Given that sin inverse (x) = a b and sin inverse (y) = a - b, we need to find the value of ab.
Solution:
To find the value of ab, we need to analyze the given information and use the properties of inverse trigonometric functions. Let's break down the solution into smaller steps:
Step 1: Understand the meaning of sin inverse (x) = a b and sin inverse (y) = a - b.
- The notation sin inverse (x) represents the inverse sine function, also denoted as arcsin(x). It gives the angle whose sine is x.
- Similarly, sin inverse (y) represents the inverse sine function of y, i.e., arcsin(y). It gives the angle whose sine is y.
- In this context, a and b are the angles (in radians or degrees) whose sine values are x and y, respectively.
Step 2: Relate the given information to the properties of inverse trigonometric functions.
- The sine function is periodic with a period of 2π radians or 360 degrees. This means that for any real number k, sin(a + 2πk) = sin(a).
- The inverse sine function, arcsin(x), gives the principal value of the angle whose sine is x in the range [-π/2, π/2] or [-90°, 90°].
- Using the above properties, we can see that sin(a + 2πk) = sin(a - 2πk) = sin(a) for any integer k.
Step 3: Determine the relationship between a, b, x, and y.
- From sin inverse (x) = a b, we know that the angle whose sine is x is represented by a b.
- From sin inverse (y) = a - b, we know that the angle whose sine is y is represented by a - b.
- Since the sine function is periodic, we can write sin(a b) = sin(a - b) by adding or subtracting multiples of 2π from a b and a - b.
Step 4: Use trigonometric identities to simplify the equation.
- Applying the sum-to-product identity for sine, we have sin(a)cos(b) + cos(a)sin(b) = sin(a)cos(b) - cos(a)sin(b).
- Simplifying further, sin(a)cos(b) - sin(a)cos(b) = 0.
- This implies that sin(a)cos(b) = 0.
Step 5: Determine the possible values of ab.
- The equation sin(a)cos(b) = 0 holds true for two cases:
1. sin(a) = 0 and cos(b) ≠ 0
2. sin(a) ≠ 0 and cos(b) = 0
- In the first case, sin(a) = 0 implies a = 0 or a = π. Since the range of arcsin is [-π/2, π/2], a = π is not in the valid range.
- In the second case, cos(b) = 0 implies b = π/2 or b = 3π/2. Both values fall within the valid range of arcs
Community Answer
Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab?
sin−1(x)=a+b and sin−1(y)=a-b
--> x=sin(
a+b
),y=sin(
a+b
)
1+xy =1+sin(a+b)sin(a−b)
We know that
sin(A
−
B)=sinAcosB
−
sinBcosA
sin(A+B)=sinAcosB+sinBcosA
-->1+xy=1+[(sin
a
cos
b
+cos
a
sin
b
)(sin
a
cos
b−
cos
a
sin
b
)]
=1+[(sinacosb)2−(cosasinb)2] using a2−b2=(a−b)(a+b)
=sin2(a)+cos2(a)+(sinacosb)2−(cosasinb)2 [sin�x+cos�x=1]
=sin2(a)[1+cos2b]+cos2(a)[1−sin2b]
=sin2(a)+sin2(a)cos2(b)+cos2(a)(cos2b)
=sin2(a)+cos2(b)[sin2(a)+cos2(a)= 1]
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Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab?
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Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab?.
Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Sin inverse (x)=a+b and sin inverse(y)= a-b, then find 1+ab?.
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