Question

NO LINKS!!!

Use the fact that the trigonometric functions are periodic to find the exact value of each expression. Don't use a calculator. This is NOT MULTIPLE CHOICE!!

a. sin (405°)

b. cot (390°)

c. cos (33π/4)

d. sec (17π/4)

c. ​

Answer 1

$\\ \rm\rightarrowtail sin(405)$

$\\ \rm\rightarrowtail sin(360+45)$

$\\ \rm\rightarrowtail sin45$

$\\ \rm\rightarrowtail \dfrac{1}{\sqrt{3}}$

#2

$\\ \rm\rightarrowtail cot(390)$

$\\ \rm\rightarrowtail cot(360+30)$

$\\ \rm\rightarrowtail cot30$

$\\ \rm\rightarrowtail \sqrt{3}$

#3

$\\ \rm\rightarrowtail cos(33\pi/4)$

$\\ \rm\rightarrowtail cos(8\pi+\pi/4)$

$\\ \rm\rightarrowtail cos\pi/4$

$\\ \rm\rightarrowtail \dfrac{1}{\sqrt{2}}$

#4

$\\ \rm\rightarrowtail sec(17\pi/4)$

$\\ \rm\rightarrowtail sec(4\pi +\dfrac{\pi}{4})$

$\\ \rm\rightarrowtail sec\dfrac{\pi}{4}$

$\\ \rm\rightarrowtail \sqrt{2}$

Answer 2

Answer:

$\sin(405)=\dfrac{\sqrt{2}}{2}$

$\cot(390)=\sqrt{3}$

$\cos\left(\dfrac{33}{4}\pi\right)=\dfrac{\sqrt{2}}{2}$

$\sec\left(\dfrac{17}{4}\pi\right)=\sqrt{2}$

Step-by-step explanation:

Trig identities used:

$\sin(A\pm B)=\sin A \cos B \pm \cos A \sin B$

$\cos(A \pm B)=\cosA \cos B \mp \sin A \sin B$

$\tan(A \pm B)=\dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

$\cot A=\dfrac{1}{\tan A}$

$\sec A=\dfrac{1}{\cos A}$

Standard trig angles

$\sin(0 \pm 360n)=\sin(0 \pm 2 \pi n)=0\\\\\cos(0 \pm 360n)=\cos(0 \pm 2 \pi n)=1\\\\\tan(0 \pm 180n)=0\\\\\sin(45 \pm 360n)=\sin(\frac14 \pi \pm 2\pi n)=\dfrac{\sqrt{2}}{2}\\\\\cos(45 \pm 360n)=\cos(\frac14 \pi \pm 2\pi n)=\dfrac{\sqrt{2}}{2}\\\\\tan(30 \pm 180n)=\dfrac{\sqrt{3}}{3}$

Question a

$\sin (405) = \sin (360 + 45)$

             $= \sin(360)\cos(45) + \cos(360)\sin(45)$

             $=0 \cdot \dfrac{\sqrt{2}}{2}+1 \cdot \dfrac{\sqrt{2}}{2}$

             $=\dfrac{\sqrt{2}}{2}$

Question b

$\cot(390)=\dfrac{1}{\tan(390)}$

              $=\dfrac{1}{\tan(360+30)}$

              $=\dfrac{1-\tan(360) \tan(30)}{\tan(360)+\tan(30)}$

              $=\dfrac{1-0 \cdot \frac{\sqrt{3}}{3}}{0+\frac{\sqrt{3}}{3}}$

              $=\dfrac{1}{\frac{\sqrt{3}}{3}}$

              $=\sqrt{3}$

Question c

$\cos\left(\dfrac{33}{4}\pi\right)=\cos(8\pi+\frac14\pi)$

                  $=\cos(8\pi)\cos(\frac14\pi)}-\sin(8\pi)\sin(\frac14\pi)}$

                  $=1 \cdot \dfrac{\sqrt{2}}{2}-0 \cdot \dfrac{\sqrt{2}}{2}$

                  $=\dfrac{\sqrt{2}}{2}$

Question d

$\sec\left(\dfrac{17}{4}\pi\right)=\dfrac{1}{\cos(4\pi+\frac14\pi)}$

                 $=\dfrac{1}{\cos(4\pi)\cos(\frac14\pi)-\sin(4\pi)\sin(\frac14\pi)}}$

                 $=\dfrac{1}{1 \cdot \dfrac{\sqrt{2}}{2} -0 \cdot \dfrac{\sqrt{2}}{2}}$

                 $=\dfrac{1}{\dfrac{\sqrt{2}}{2} }$

                 $=\sqrt{2}$