PreCalculus Honors Unit Test #7

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No Calculator Allowed

 1 Use the fundamental identities to simplify $\left({cos}^{2}x\right)\left({tan}^{2}x+1\right)$ as much as possible. (4)

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

 2 Verify: $\frac{2-{sin}^{2}\left(x\right)}{cos\left(x\right)}=sec\left(x\right)+cos\left(x\right)$ (6)

Working the right-hand side…

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

QED.

 3 Verify: $\frac{{cos}^{2}\left(x\right)}{1-sin\left(x\right)}=1+sin\left(x\right)$ (6)

Working the left-hand side…

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

QED.

 4 Find the exact value of $cos\left(\frac{2\pi }{9}\right)cos\left(\frac{\pi }{36}\right)-sin\left(\frac{2\pi }{9}\right)sin\left(\frac{\pi }{36}\right)$. (4)

This smells like the sum formula for cosine…

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

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

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

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

 6 Solve $sin\left(2x\right)=-\frac{1}{2}$ for $x\in \left[0,2\pi \right)$. (6)

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

 7 Solve ${sin}^{2}\left(x\right)+2sin\left(x\right)+1=0$ for $x\in \left[0,2\pi \right)$. (5)

${sin}^{2}\left(x\right)+2sin\left(x\right)+1={\left(sin\left(x\right)+1\right)}^{2}$; thus, ${\left(sin\left(x\right)+1\right)}^{2}=0⇒\left(sin\left(x\right)+1\right)=0⇒sin\left(x\right)=-1⇒x=\frac{3\pi }{2}$.

Page last updated 11:10 2021-06-03