log(x-y%2)=0.5*(logx+logy)then prove that xy-1+yx-1=6
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But it is clear that if we restrict the domain to [-π/2,π/2] and range to
[–1, 1], then. Y = sin x is one-one onto and hence it is invertible.
So, y = sin x x ∈ [-π/2,π/2], y ∈ [–1, 1]
⇒ x = sin–1 y y ∈ [–1, 1]. x ∈ [-π/2,π/2]
This value of x is called the principle value, i.e. belonging to [-π/2,π/2] and [-π/2,π/2] range and it is called principal value range.
Note: The smallest numerical angle is called principal value.
In general the inverse circular functions with their domain and range can be tabulated as:
[–1, 1], then. Y = sin x is one-one onto and hence it is invertible.
So, y = sin x x ∈ [-π/2,π/2], y ∈ [–1, 1]
⇒ x = sin–1 y y ∈ [–1, 1]. x ∈ [-π/2,π/2]
This value of x is called the principle value, i.e. belonging to [-π/2,π/2] and [-π/2,π/2] range and it is called principal value range.
Note: The smallest numerical angle is called principal value.
In general the inverse circular functions with their domain and range can be tabulated as:
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