5.2

4. Determine the exact value of each trigonometric expression.

• a) $\sin 30^{\circ }\times \tan 60^{\circ }-\cos 30^{\circ }$$$
  \sin 30^\circ \times \tan 60^\circ - \cos 30^\circ = \frac{1}{2} \times \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0

• b) $2\cos 45^{\circ }\times \sin 45^{\circ }$

  $$2\cos 45^{\circ }\times \sin 45^{\circ }=2\times \frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{2}=2\times \frac{2}{4}=1$$

• c) ${\tan }^{2}30^{\circ }-{\cos }^{2}45^{\circ }$

  $${\tan }^{2}30^{\circ }-{\cos }^{2}45^{\circ }={\left(\frac{1}{\sqrt{3}}\right)}^2-{\left(\frac{\sqrt{2}}{2}\right)}^2=\frac{1}{3}-\frac{2}{4}=\frac{1}{3}-\frac{1}{2}=\frac{2}{6}-\frac{3}{6}=-\frac{1}{6}$$

• d) $1-\frac{\sin 45^{\circ }}{\cos 45^{\circ }}$

1 - \frac{\sin 45^\circ}{\cos 45^\circ} = 1 - \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 - 1 = 0 $$ **5. Using exact values, show that $\sin^2 \theta + \cos^2 \theta = 1$ for each angle.** * **a) $\theta = 30^\circ$** $$ \sin^2 30^\circ + \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1 $$ * **b) $\theta = 45^\circ$** $$ \sin^2 45^\circ + \cos^2 45^\circ = \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} + \frac{2}{4} = \frac{4}{4} = 1 $$ * **c) $\theta = 60^\circ$** $$ \sin^2 60^\circ + \cos^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 $$ **6. Using exact values, show that $\frac{\sin \theta}{\cos \theta} = \tan \theta$ for each angle.** * **a) $\theta = 30^\circ$** $$ \frac{\sin 30^\circ}{\cos 30^\circ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{2} \times \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \tan 30^\circ $$ * **b) $\theta = 45^\circ$** $$ \frac{\sin 45^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 = \tan 45^\circ $$ * **c) $\theta = 60^\circ$** $$ \frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3} = \tan 60^\circ $$ **7. Using the appropriate special triangle, determine $\theta$ if $0^\circ \le \theta \le 90^\circ$.** * **a) $\sin \theta = \frac{\sqrt{3}}{2}$** We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Therefore, $\theta = 60^\circ$. * **b) $\sqrt{3} \tan \theta = 1$** Dividing both sides by $\sqrt{3}$, we get $\tan \theta = \frac{1}{\sqrt{3}}$. We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Therefore, $\theta = 30^\circ$. * **c) $2\sqrt{2} \cos \theta = 2$** Dividing both sides by $2\sqrt{2}$, we get $\cos \theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. We know that $\cos 45^\circ = \frac{\sqrt{2}}{2}$. Therefore, $\theta = 45^\circ$. * **d) $2 \cos \theta = \sqrt{3}$** Dividing both sides by 2, we get $\cos \theta = \frac{\sqrt{3}}{2}$. We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$. Therefore, $\theta = 30^\circ$.