Write the trigonometric expression as an algebraic expression
Write the trigonometric expression as an algebraic expression.
$6 \cos(2 \cos^{-1} x)$
Can someone explain to me how to do this? I tried it on my own after watching a youtube video from patrickjmt but it's confusing.
My work:
3 Answers
$\begingroup$We have: $6\cos (2\cos^{-1}x) = 6(2\cos^2(\cos^{-1}x)-1)=6(2x^2-1)$
$\endgroup$ 2 $\begingroup$Hint:
use: $$ \cos 2 \alpha= \cos^2 \alpha-\sin^2 \alpha= 2\cos^2 \alpha-1 $$ for $\alpha=\cos^{-1} x$. Than note that , by definition, $\cos (\cos^{-1} x)=x$
So you find: $$ \cos (2 \cos^{-1} x)=2\cos^2(\cos^{-1} x)-1=2x^2-1 $$
$\endgroup$ $\begingroup$This is $\cos\Big(2\times\text{something}\Big)$. There's a standard identity for that: \begin{align} \cos\Big(2\times\text{something}\Big) & = \cos^2\Big(\text{something}\Big) - \sin^2\Big(\text{something}\Big) \\[10pt] & = \Big(\cos(\text{something})\Big)^2 - \Big(\sin^2(\text{something})\Big)^2 \end{align} Then you need to deal with $\cos(\arccos(x))$ (that part is easy) and $\sin(\arccos(x))$. The latter can be done via a diagram like the one you've drawn.
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