PreCal Honors - Unit Test #7

PreCalculus Honors

Unit Test #7

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Use the fundamental identities to simplify $({cos}^{2} x) ({tan}^{2} x + 1)$ as much as possible.

(4)

$({cos}^{2} x) ({tan}^{2} x + 1) = ({cos}^{2} x) (\frac{{sin}^{2} x}{{cos}^{2}} + \frac{{cos}^{2} x}{{cos}^{2} x}) = {sin}^{2} x + {cos}^{2} x = 1$

Verify: $\frac{2 - {sin}^{2} (x)}{cos (x)} = sec (x) + cos (x)$

(6)

Working the right-hand side…

$sec (x) + cos (x) = \frac{1}{cos (x)} + \frac{{cos}^{2} (x)}{cos (x)} = \frac{1 + {cos}^{2} (x)}{cos (x)} = \frac{1 + (1 - {sin}^{2} (x))}{cos (x)} = \frac{2 - {sin}^{2} (x)}{cos (x)}$

QED.

Verify: $\frac{{cos}^{2} (x)}{1 - sin (x)} = 1 + sin (x)$

(6)

Working the left-hand side…

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

QED.

Find the exact value of $cos (\frac{2 π}{9}) cos (\frac{π}{36}) - sin (\frac{2 π}{9}) sin (\frac{π}{36})$ .

(4)

This smells like the sum formula for cosine…

$cos (\frac{2 π}{9} + \frac{π}{36}) = cos (\frac{8 π}{36} + \frac{π}{36}) = cos (\frac{9 π}{36}) = cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$

Find the exact value of $sin (\frac{5 π}{8})$ .

(5)

This will require the half-angle formula for sine. Note that $\frac{5 π}{8}$ is in the second quadrant, so the value must be positive.

$\sqrt{\frac{1 - cos (\frac{5 π}{4})}{2}} = \sqrt{\frac{1 - (- \frac{1}{\sqrt{2}})}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}}$

Solve $sin (2 x) = - \frac{1}{2}$ for $x \in [0, 2 π)$ .

(6)

Note that $x \in [0, 2 π) \Rightarrow 2 x \in [0, 4 π)$ . $sin (2 x) = - \frac{1}{2} \Rightarrow 2 x \in \{\frac{7 π}{6}, \frac{11 π}{6}, \frac{19 π}{6}, \frac{23 π}{6}\} \Rightarrow x \in \{\frac{7 π}{12}, \frac{11 π}{12}, \frac{19 π}{12}, \frac{23 π}{12}\}$

Solve ${sin}^{2} (x) + 2 sin (x) + 1 = 0$ for $x \in [0, 2 π)$ .

(5)

${sin}^{2} (x) + 2 sin (x) + 1 = {(sin (x) + 1)}^{2}$ ; thus, ${(sin (x) + 1)}^{2} = 0 \Rightarrow (sin (x) + 1) = 0 \Rightarrow sin (x) = - 1 \Rightarrow x = \frac{3 π}{2}$ .

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