Considering the principal values of the inverse trigonometric functions, sin−1(23x+211−x2),−21<x<21 is equal to:
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Step-by-Step Solution
Step 1: Substitute x with a trigonometric function
To simplify the expression, we can use a trigonometric substitution. Let x=sinθ. This substitution is valid because the domain of x is given as −21<x<21, which corresponds to a valid range for sinθ.
Step 2: Rewrite the expression using the substitution
Substitute x=sinθ into the given expression. We also know that 1−sin2θ=cos2θ=∣cosθ∣. Since −21<x<21, we have −21<sinθ<21. This implies −6π<θ<4π. In this interval, cosθ is positive, so 1−sin2θ=cosθ.
Step 3: Apply trigonometric identity
We recognize that 23=sin3π and 21=cos3π. Substituting these values, the expression becomes sin−1(sin3πsinθ+cos3πcosθ). This is in the form of the cosine addition formula.
Step 4: Simplify using cosine addition formula
Using the cosine addition formula, cos(A−B)=cosAcosB+sinAsinB, we can simplify the expression inside the inverse sine function to cos(θ−3π).
Step 5: Convert cosine to sine
To further simplify, we use the identity cosx=sin(2π−x). Applying this, we get sin−1(sin(2π−(θ−3π))). Simplifying the argument gives sin−1(sin(65π−θ)).
Step 6: Final simplification
Since −6π<θ<4π, we have −4π<−θ<6π. Adding 65π to the inequality, we get 65π−4π<65π−θ<65π+6π, which simplifies to 127π<65π−θ<π. This range is within the principal value branch of sin−1y, which is [−2π,2π]. However, the range 127π<65π−θ<π is not within the principal value branch. Let's re-evaluate the identity. Instead of cosx=sin(2π−x), we can use cosx=sin(2π+x) if the argument is in the correct range. Let's check the range of θ−3π. Since −6π<θ<4π, then −6π−3π<θ−3π<4π−3π, which means −2π<θ−3π<−12π. In this range, cos(θ−3π) is positive.
We need to ensure that the argument of sin−1(sinA) is in [−2π,2π].
Let A=θ−3π. Then cosA=sin(2π−A).
So, sin−1(cos(θ−3π))=sin−1(sin(2π−(θ−3π))).
2π−(θ−3π)=2π−θ+3π=65π−θ.
The range of 65π−θ:
Since −6π<θ<4π, then −4π<−θ<6π.
Adding 65π: 65π−4π<65π−θ<65π+6π.
1210π−3π<65π−θ<66π.
127π<65π−θ<π.
This range is not in [−2π,2π].
We know that sin−1(sinx)=x if x∈[−2π,2π].
If x∈[2π,23π], then sin−1(sinx)=π−x.
Since 127π<65π−θ<π, the argument is in the interval (2π,π).
Therefore, sin−1(sin(65π−θ))=π−(65π−θ)=π−65π+θ=6π+θ.
Substitute back θ=sin−1x.