$\cos^2 (\cos^{-1} x)$ $=x^2$ How?

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$$\cos^2 (\cos^{-1} x)$$$$=x^2$$

Which trigonometric rule was used here?

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2 Answers

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By definition, $\cos^{-1} x$ is a real number so that $\cos(\cos^{-1} x) = x.$ Thus, $\cos^2(\cos^{-1} x) = (\cos(\cos^{-1} x))^2 = x^2.$

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In general, for a function $f:D_f\to R_f$, where $f$ is a bijective function for $x\in D_f$, and its inverse $g:R_f\to D_f$, we have the identities:$$f(g(x))=x, x\in R_f$$ and $$g(f(x))=x, x\in D_f$$It is worth mentioning this, because this implies that the statement in your question, $\cos^2(\cos^{-1}(x))=x^2$, is only true for the domain of $\cos^{-1} x$, which of course is $x\in [-1,1]$.

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