Question

help with 11 -20 please? :(

Answer 1

Hey :)
So uhhhhhh...
oooo trig! my fav!

Answer 2

Where we stuck? Numbrero 11?

Answer 3

Soooo, where you stuck at?
\[\large\rm \cos \theta=0.5563\]Not sure how to find theta?

Answer 4

Yes 11 :( I don't know how to work them out

Answer 5

Hold on hold on.
Lemme explain this really quick,
then we'll just use the shortcut every time after that.

Answer 6

Ok thank you :)

Answer 7

\(\large\rm x+4=7\)
How would we solve for x here?
We would perform the `inverse of addition`.
We would apply subtraction of 4 to each side, yes?

How bout this problem?
\(\large\rm 3x=9\)
We would apply the `inverse of multiplication`,
So applying division of 3 to each side, yes? :)

Notice that in both cases, we're trying to "undo" to thing happening to x.

So that's what we're doing in our problem to isolate this angle theta.
Inverse sine is the `inverse of sine`.
So it will "undo" the sine being applied to the angle.

Answer 8

Oh oh, sorry inverse cosine :)

Answer 9

I'm following :)

Answer 10

So taking inverse cosine of each side gives us,\[\large\rm \arccos(\cos \theta)=\arccos(0.5563)\]On the left side,
things simplify nicely, sine and inverse sine "undo" one another.
You can sort of think of it like "cancelling out", but that's not totally accurate.\[\large\rm \theta=\arccos(0.5563)\]

Answer 11

If you look back at what we had written originally,
then maybe you can see the shortcut,
all we do is...
`rewrite sine and inverse sine`
`switch the positions of the two other items`\[\large\rm \text{if }\sin(a)=b\qquad\text{then }\arcsin(b)=a\]

Answer 12

Is it ok if I use this notation?
arcsine? Have you seen this before?
Or do you prefer \(\large\rm \sin^{-1}\) ?
They mean the same thing

Answer 13

sorry typo in my shortcut
~ rewrite sine as* inverse sine
~ switch the position of the other two items

Answer 14

okok i think i get it let me try it :)

Answer 15

so on the calculator I just input what? \[\sin^{-1} (.5563)\]

Answer 16

woops, \(\large\rm \cos^{-1}(.5563)\) you mean? +_+
and yes, make sure you're in `degree mode` first.

This will get us ONE SOLUTION.
Two solutions exist within one full rotation of the unit circle (within 360 degrees).
We'll figure that out in a moment.

What do you get for this first angle though?

Answer 17

I gotta stop using sine in my examples -_-
I think that was confusing you.

Answer 18

oh right my bad, I should've know as well. No big deal, anyways i got 56.200 if I round correct?

Answer 19

Ok great so one solution is \(\large\rm \theta=56.2^o\)

Answer 20

For the cosine function,
since it is an even function,
we get mirrored outputs in quadrants 1 and 4.

Since this angle is in quadrant 1,
our other angle must be in quadrant 4.

Our other solution will be given by \(\large\rm 360^o-\theta\)

Answer 21

So what do you get for your other solution?
\(\large\rm 360^o-56.2^o=?\)

Answer 22

303.8 degrees :) ?

Answer 23

Mmmm looks good.
And a nice easy way to verify that you've done everything correct,
is to plug your angles into your cosine function:
\(\large\rm \cos(303.8^o)\approx0.5562956\)
\(\large\rm \cos(56.2^o)\approx0.5562956\)

Answer 24

Ok :) and I do the same for the rest correct?

Answer 25

Well are 15-18 done a different way?

Answer 26

Unfortunately, each of these is going to be a different trick :(

Answer 27

It's the same idea with the inverse,
you rewrite your function as inverse function,
and switch the pieces.

That part is the same.

But after that, finding the second angle will be different in each case.

Answer 28

Lemme summarize this way:

After you find your first angle \(\large\rm \theta\),
here is what you'll do to find the other angle:

For cosine:
\(\large\rm 360^o-\theta\)

For tangent:
\(\large\rm 180^o+\theta\)

For sine:
\(\large\rm 180^o-\theta\)

Answer 29

oh my gosh I hate this so much :( i'll never be done :(

Answer 30

yah it's a lot of stuff :'c

Answer 31

So for problem 12:
We rewrite using inverse function, and switch the parts,\[\large\rm \tan\theta=-4.2568\]\[\large\rm \tan^{-1}(-4.2568)=\theta=-76.78^o\]That's the angle the calculator gives us.

Answer 32

But notice, this angle is negative!
It's not between 0 and 360 like we need.
So it doesn't even count as our first angle.

We'll add 180 to it, to get our first angle.
\(\large\rm 180+-76.78=103.22\)

Answer 33

We'll add 180 again to find our next angle,
\(\large\rm 180+103.22=283.22\)

So those are the two solutions we were looking for between 0 and 360.
103.22 and 283.22

Answer 34

It seems so easy when you do it :(

Answer 35

How do you input CSC , SEC COT and their inverses on a calculator? :(

Answer 36

Ok so you've probably noticed that there is no \(\large\rm \csc^{-1}\) button on the calculator :) So this one requires some special work.

Answer 37

We'll need to make use of this identity\[\large\rm \csc\theta=\frac{1}{\sin\theta}\]

Answer 38

So in problem 13,\[\large\rm \csc \theta=3.2745\]will become\[\large\rm \frac{1}{\sin \theta}=3.2745\]and we can try to get sin theta out of the denominator from that point.

Answer 39

cross multiply, or whatever other algebra tricks you can remember.

Answer 40

Multiplying both sides by sin(theta),
dividing both sides by 3.2745,
gives us,\[\large\rm \frac{1}{3.2745}=\sin \theta\]Let's input this division into our calculator to get a different decimal value.

Answer 41

\[\large\rm 0.30539=\sin \theta\]And from here,
we can use the usual tricks that we've been talking about.

Answer 42

So yes, when you're given csc, cot, sec,
it's going to be a bit of extra work :d

Answer 43

because you always need to relate them back to sine, cosine and tangent.

Answer 44

Oh no :( but I appreciate your help SO much! You are every helpful and kind, but mostly patient!

Answer 45

Ahhh I need a math break :)
You wanna try 13 and 14 maybe?

They shouldn't be too bad.

Answer 46

Try to get some answers for 13 and 14,
then @zepdrix to page me,
or @ someone else if I'm not online so someone can check your work :D

Answer 47

I will try my best with them, thank you so much :)

Answer 48

it is easiest to remember i think, if you can imagine the unit circle, and how it has symmetries for sin and cos, where sin is the y coord and cos is the x coord on the circle..

this may help practice https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/trig-identities/v/trigonometry-unit-circle-symmetry?_escaped_fragment_=.

Answer 49

Thank you @DanJS :)