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How do you find an equivalent expression free of trig funtions for

**g(b)=cos(arctan(b))?**- Thread starter princiebebe57
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How do you find an equivalent expression free of trig funtions for **g(b)=cos(arctan(b))?**

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AKG

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HallsofIvy

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Since "[itex]tan(\theta)[/itex]" is "opposite side over near side" you can think of b as b/1. Draw a right triangle have "opposite side" b and "near side" 1. Now [itex]cos(\theta)[/itex] is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenuse and find [itex]cos(\theta)[/itex].

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arildno

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Alternatively, remember the identity: [itex]\sec^{2}(\theta)=\tan^{2}(\theta)+1[/itex]

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