Question

how does sin(-13pi/4) = -5pi/4

we were asked to find the exact value of sin(-13pi/4)
how does sin(-13pi/4) = -5pi/4

we were asked to find the exact value of sin(-13pi/4)
@Mathematics

Answer 1

sin(-13pi/4) = sin(-5pi/4) = sin (3pi/4) because sin(x + 2pi) = sin(x)

Answer 2

could you demonstrate that pls im having a hard time understanding

Answer 3

-13pi/4 is over one revolution so add 2pi to it.

Answer 4

Think about the unit circle. Go around one more complete revolution then you're still at the same point on the unit circle, but the angle has changed by 2pi

Answer 5

\[-\frac{13 \pi}{4} + \frac{8\pi}{4} = -\frac{5\pi}{4}\]

Answer 6

oh tyler...now i see

Answer 7

and also could you demonstrate how cos(5pi/3)= -sqrt2/2 and cos(-270) = 0

Answer 8

\[\sin(\frac{-13 \pi}{4})=\sin(\frac{-4}{4}\pi+\frac{-9}{4}\pi)=\sin(-\pi-\frac{4}{4}\pi-\frac{5}{4} \pi)=\sin(-\pi-\pi-\frac{5}{4} \pi)\]
\[\sin(-2\pi-\frac{5}{4}\pi)=\sin(-(2\pi+\frac{5}{4}\pi))\]

Answer 9

cos(5pi/3) should be 1/2 not what you said

Answer 10

and change -270 into positive degrees so you do 360 + -270 = 90

and cos(90) = 0

Answer 11

oh thanks for the help again Tyler. how have you been?

Answer 12

I've been good.. how about you?

Answer 13

i have been good as well...preparing for the test. its on wednesday. I almost have everything down. I may have one or two questions that I need help, will you be around this evening?

Answer 14

yeah i can help if you want

Answer 15

ok i need help with inverses.

first I have tan(x/2) + (pi/6) and i need to find the inverse

Answer 16

Note by the way that even if we have shown sin(-13pi/4) = sin (3pi/4), that's not the final answer. You need to evaluate sin(3pi/4)

Answer 17

oh ok thanks for letting me know

Answer 18

Ok inverses.. just a sec

Answer 19

ok

Answer 20

to be honest, ive never had to do that before so im not sure on how to do it.

Answer 21

oh ok thats fine.

Answer 22

cos x = 1 - 2 sin^2 (x/2) hence
\[ \sin(x/2) = \pm \sqrt{1 - \cos x} \]

Answer 23

remember last time we did amp, domain range,period,etc...? there was one thing else i need to ask...how do you find horizontal shift

Answer 24

horizontal shift is the phase shift

Answer 25

if you're given y=3csc(x + pi/4)+ 2

Answer 26

3pi/4 is in the first quadrant, so we will want sin(3pi/4) to be positive. Thus:
\[ \sin(3\pi/4) = \sqrt{1 - cos(3\pi/2)} \]

Answer 27

James do you know how to do this??

first I have tan(x/2) + (pi/6) and i need to find the inverse

Answer 28

pi/4 is the horizontal shift aka phase shift

Answer 29

thanks my teacher got the horizontal shift as -pi/2 for y=-2cos (2x + pi) - 1...im guessing she made a mistake right?

Answer 30

I would think.. because horizontal means going left and right.. so the horizontal shift would be -pi

Answer 31

yeah that makes sense.

Answer 32

Horizontal Shifts

To shift a graph horizontally, a constant must be added to the function within parentheses--that is, the constant must be added to the angle, not the whole function.

Answer 33

so james do you know how to find that inverse?

Answer 34

so yeah the teacher made a mistake

Answer 35

well hold on.. i think your teacher factored out a 2

Answer 36

\[y = -2\cos2(x + \frac{\pi}{2})\]

Answer 37

hmm thats what it now looks like

Answer 38

so how did she get 2 as the denominator in the phase shift

Answer 39

factoring something out is like dividing by that number so if you have \[(2x + \pi)\] and you factor out a 2 you do \[(\frac{2x}{2} + \frac{\pi}{2})\] and simplify \[(x + \frac{\pi}{2})\]

Answer 40

oh now i see

Answer 41

sorry, what are you asking now?

Answer 42

if you've got it I'll bugger off

Answer 43

first I have tan(x/2) + (pi/6) and i need to find the inverse

Answer 44

james this is what we needed help with

first I have tan(x/2) + (pi/6) and i need to find the inverse

Answer 45

So if y = tan(x/2) + pi/6, what is x?

Answer 46

no, what is the inverse of y=tan(x/2) + pi/6

Answer 47

We just say the same thing.

Assuming that \( -\pi/2 < x < \pi/2 \), then

\[ x = 2 \arctan(y - \pi/6) \]

Answer 48

Ugh, I can't type today: "We just said the same thing."

Answer 49

And actually we need x/2 to be in that range -pi/2 to pi/2, hence -pi < x < pi

Answer 50

ok james i have two more

what about f(x) arcos(x/2)-7

f(x)= sin(3x)

Answer 51

thanks for the help btw

Answer 52

Let y = arccos(x/2) - 7. Now solve for x. That gives you the inverse function. The only wrinkle is to think carefully about the domain and range.

Answer 53

2cos(x+7)??

Answer 54

oh ok thanks for the help to both of your...can i get a help with the last one

f(x)= sin(3x)

Answer 55

\[\frac{1}{3}\arcsin(x)\]

Answer 56

thanks tyler

Answer 57

yeah lol.. im getting the hang of these lol

Answer 58

lol...now i just have basic questions. if intervals = period/4 and my period is 2pi/3...how do i end up with pi/6

Answer 59

Im confused :/ lol

Answer 60

oh wait no im not.

Answer 61

you do \[\frac{\frac{2\pi}{3}}{4}\]

Answer 62

oh i see

Answer 63

\[\frac{2\pi}{3} * \frac{1}{4}\] which equals \[\frac{2\pi}{12}\] which equals \[\frac{\pi}{6}\]

Answer 64

and how do you get 1/2 of 2pi/3 to equal pi/3

Answer 65

\[\frac{\frac{2\pi}{3}}{2}\]

Answer 66

\[\frac{2\pi}{3} * \frac{1}{2}\]\[\frac{2\pi}{6}\] \[\frac{\pi}{3}\]

Answer 67

woow man thanks for the help i greatly appreciate it.

Answer 68

no problem. Do you understand??

Answer 69

yes very well.

Answer 70

cool :D

Answer 71

Anything else??

Answer 72

thats it. maybe if i come across something later on that i dont understand as im studying. but those were the only ones i couldnt comprehend

Answer 73

Cool, good luck on your test :)

Answer 74

thanks for the help once again

Answer 75

No problem. See you around sometime!! :D

Answer 76

you too