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Example 1
If $f(\theta)=tan(2\theta+\dfrac{\pi}{4})$ and $h(\theta)=\cos(\theta + \dfrac{\pi}{4})$, then $f(h(\dfrac{\pi}{4}))=$
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$-2$
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$-1$
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$0$
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$1$
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$2$
Solution
- First examine the 5 possible answer; we are asked to evaluate $f(h(\dfrac{\pi}{4}))$ given $f$ and $h$. In order to evaluate $f(h(\dfrac{\pi}{4}))$, we first evaluate $h(\dfrac{\pi}{4})$
$h(\dfrac{\pi}{4}) =\cos(\dfrac{\pi}{4} + \dfrac{\pi}{4}) = \cos(\dfrac{\pi}{2})= 0 $
- We now substitute $h(\dfrac{\pi}{4})$ by $0$ in $f(h(\dfrac{\pi}{4}))$ and calculate $f(h(\dfrac{\pi}{4}))$ as follows
$f(h(\dfrac{\pi}{4})) = f(0) = \tan (2(0)+\dfrac{\pi}{4})= \tan (\dfrac{\pi}{4})=1$
Answer D
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