This identity can be proven easily by applying sin to both sides of the original equation: 1.) 2.) 3.) We continue by using implicit differentiation, keeping in mind to use the chain rule on sin y: 4.) Solve for : 5.) Then, substitution with our original equation yields in terms of x: 6.) This may not appear to be a good idea at first, but it can be simplified if the identity is remembered. . 7.) This is a good definition to memorize, along with and , since they appear quite frequently in differentiation problems.
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