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If sin⁻1x+sin⁻1y=(2π/3), then cos⁻1x+cos⁻1y=
- a)2π/3
- b)π/3
- c)π/6
- d)π
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
If sin1x+sin1y=(2π/3), then cos1x+cos1y=a)2π/3b)π/3c)π/6d)...
Explanation:
Sin1x + Sin1y = 2π/3 implies that both Sin1x and Sin1y can take values in the range [-1,1]. Since the sum of two sin values is less than 2, it indicates that both Sin1x and Sin1y are positive.
Using Trigonometric Identities:
We know that Sin^2θ + Cos^2θ = 1. Let's square the given equation to get rid of the square roots.
Expanding the Squared Equation:
(Sin1x)^2 + 2(Sin1x)(Sin1y) + (Sin1y)^2 = (2π/3)^2
1 + 2(Sin1x)(Sin1y) + 1 = 4π^2/9
2(Sin1x)(Sin1y) = 4π^2/9 - 2
(Sin1x)(Sin1y) = 2π^2/9 - 1
Using the Pythagorean Identity:
Cos^2θ = 1 - Sin^2θ
Since both Sin1x and Sin1y are positive, Cos1x and Cos1y are both negative.
Calculating Cos1x + Cos1y:
Cos1x + Cos1y = -√(1 - (Sin1x)^2) - √(1 - (Sin1y)^2)
= -√(1 - (2π^2/9 - 1)) - √(1 - (2π^2/9 - 1))
= -√(2 - 2π^2/9) - √(2 - 2π^2/9)
= -√(18/9 - 2π^2/9) - √(18/9 - 2π^2/9)
= -√(16/9) - √(16/9)
= -4/3 - 4/3
= -8/3
= -π/3
Therefore, Cos1x + Cos1y = -π/3, which is equivalent to option b) π/3.
Sin1x + Sin1y = 2π/3 implies that both Sin1x and Sin1y can take values in the range [-1,1]. Since the sum of two sin values is less than 2, it indicates that both Sin1x and Sin1y are positive.
Using Trigonometric Identities:
We know that Sin^2θ + Cos^2θ = 1. Let's square the given equation to get rid of the square roots.
Expanding the Squared Equation:
(Sin1x)^2 + 2(Sin1x)(Sin1y) + (Sin1y)^2 = (2π/3)^2
1 + 2(Sin1x)(Sin1y) + 1 = 4π^2/9
2(Sin1x)(Sin1y) = 4π^2/9 - 2
(Sin1x)(Sin1y) = 2π^2/9 - 1
Using the Pythagorean Identity:
Cos^2θ = 1 - Sin^2θ
Since both Sin1x and Sin1y are positive, Cos1x and Cos1y are both negative.
Calculating Cos1x + Cos1y:
Cos1x + Cos1y = -√(1 - (Sin1x)^2) - √(1 - (Sin1y)^2)
= -√(1 - (2π^2/9 - 1)) - √(1 - (2π^2/9 - 1))
= -√(2 - 2π^2/9) - √(2 - 2π^2/9)
= -√(18/9 - 2π^2/9) - √(18/9 - 2π^2/9)
= -√(16/9) - √(16/9)
= -4/3 - 4/3
= -8/3
= -π/3
Therefore, Cos1x + Cos1y = -π/3, which is equivalent to option b) π/3.
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If sin1x+sin1y=(2π/3), then cos1x+cos1y=a)2π/3b)π/3c)π/6d)πCorrect answer is option 'B'. Can you explain this answer?
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