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 (1)/(√(3))`

#2

`\\ \rm\rightarrowtail cot(390)`

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

`\\ \rm\rightarrowtail cot30`

`\\ \rm\rightarrowtail √(3)`

#3

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

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

`\\ \rm\rightarrowtail cos\pi/4`

`\\ \rm\rightarrowtail (1)/(√(2))`

#4

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

`\\ \rm\rightarrowtail sec(4\pi +(\pi)/(4))`

`\\ \rm\rightarrowtail sec(\pi)/(4)`

`\\ \rm\rightarrowtail √(2)`

Answer 2

`\sin(405)=(√(2))/(2)`

`\cot(390)=√(3)`

`\cos\left((33)/(4)\pi\right)=(√(2))/(2)`

`\sec\left((17)/(4)\pi\right)=√(2)`

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)=(\tan A \pm \tan B)/(1 \mp \tan A \tan B)`

`\cot A=(1)/(\tan A)`

`\sec A=(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)=(√(2))/(2)\\\\\cos(45 \pm 360n)=\cos(\frac14 \pi \pm 2\pi n)=(√(2))/(2)\\\\\tan(30 \pm 180n)=(√(3))/(3)`

Question a

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

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

`=0 \cdot (√(2))/(2)+1 \cdot (√(2))/(2)`

`=(√(2))/(2)`

Question b

`\cot(390)=(1)/(\tan(390))`

`=(1)/(\tan(360+30))\\\\`

`=(1-\tan(360) \tan(30))/(\tan(360)+\tan(30))`

`=(1-0 \cdot (√(3))/(3))/(0+(√(3))/(3))`

`=(1)/((√(3))/(3))`

`=√(3)`

Question c

`\cos\left((33)/(4)\pi\right)=\cos(8\pi+\frac14\pi)`

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

`=1 \cdot (√(2))/(2)-0 \cdot (√(2))/(2)`

`=(√(2))/(2)`

Question d

`\sec\left((17)/(4)\pi\right)=(1)/(\cos(4\pi+\frac14\pi))`

`=(1)/(\cos(4\pi)\cos(\frac14\pi)-\sin(4\pi)\sin(\frac14\pi))}`

`=(1)/(1 \cdot (√(2))/(2) -0 \cdot (√(2))/(2))`

`=(1)/((√(2))/(2) )`

`=√(2)`