Suppose I want to find $\sin^6x+\cos^6x$. What stops me from saying that $\sin^2t=\sin^6x$, and $\cos^2t=\cos^6x$? Of course this is wrong because $\sin^2t+\cos^2t=1$ and $\sin^6x+\cos^6x$ does not equal 1. So what stops me from making this substitution? The domain $and$ range of the sixth-degree functions are completely within the quadratics.
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7You are making the incorrect assumption that the $\textbf{same}$ $t$ will satisfy both $\sin^2 t = \sin^6 x$ and $\cos^2 t=\cos^6 x$, though it is true that any one of them can be satisfied separately without a problem. – Milind Hegde Jun 01 '13 at 17:49
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Oh ok thanks you – Ovi Jun 01 '13 at 17:55
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If you put $\sin^2 t = \sin^6 x$ you are not free to choose a value for $\cos^2 t$ - you have already constrained this by your initial choice.
Mark Bennet
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You are making the incorrect assumption that the $\textbf{same }t$ will satisfy both $\sin^2 t=\sin^6 x$ and $\cos^2 t=\cos^6 x$, though it is true that any one of them can be satisfied separately without a problem.
Milind Hegde
- 3,914