Prove that $sin to the power of negative 1 end exponent x plus sin to the power of negative 1 end exponent x y equals sin to the power of negative 1 end exponent open square brackets x square root of 1 minus y squared end root plus square root of 1 minus x squared end root close square brackets$

Solution

Put sin^–1.x = θ, sin^–1 y = ϕ
∴x = sin θ, y = sin ϕ
Now sin (θ + ϕ) = sin θ cos ϕ + cos θ sinϕ
$therefore space space sin left parenthesis straight theta plus straight ϕ right parenthesis sinθ square root of 1 minus sin squared straight ϕ end root plus square root of 1 minus sin squared straight theta. sinϕ end root rightwards double arrow space space space straight theta plus straight ϕ equals sin to the power of negative 1 end exponent open square brackets sinθ square root of 1 minus sin squared straight ϕ end root plus sinϕ square root of 1 minus sin squared straight theta end root close square brackets rightwards double arrow space space space sin to the power of negative 1 end exponent straight x plus sin to the power of negative 1 end exponent straight y equals sin to the power of negative 1 end exponent open square brackets straight x square root of 1 minus straight y squared end root plus straight y square root of 1 minus straight x squared end root close square brackets$

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Inverse Trigonometric Functions

Prove that $sin to the power of negative 1 end exponent x plus sin to the power of negative 1 end exponent x y equals sin to the power of negative 1 end exponent open square brackets x square root of 1 minus y squared end root plus square root of 1 minus x squared end root close square brackets$

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