please help ^.^

- anonymous

please help ^.^

- schrodinger

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- anonymous

##### 1 Attachment

- geekfromthefutur

That's devil math xD

- anonymous

thanks anyway XD

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## More answers

- Nnesha

can u use unit circle ?

- anonymous

all that I know have to do is draw the angle given, and then find cosine and sine, I don't think it matters how I get it done I just have to show my work

- Nnesha

i like degrees so if you want you can convert radians to degree 7pi/4 times 180/pi

- anonymous

ok so 315 degrees?

- Nnesha

yes right which is |dw:1433361087764:dw|in 4th quadrant

- anonymous

alright

- Nnesha

|dw:1433361160024:dw|
draw a right traingle near to the axis
we need that reference angle
which is 45
360-315 = 45
now you have to use 45-45-90 triangle

- anonymous

ok I think I understand...

- Nnesha

http://prntscr.com/7crrp5 right here |dw:1433361396623:dw|
sin = opposite over hypotenuse
cos = adjacent over hypotenuse

- Nnesha

|dw:1433361539015:dw|
what is opposite side of 45 angle ?

- Nnesha

do you now how to find 45-45-90 triangle from the unit circle ?

- anonymous

I'm a little lost.. I know that 1 is opposite the 45

- Nnesha

yes right and hypotenuse is the longest side of right triangle which is ?

- anonymous

the squroot 2

- anonymous

square root*

- anonymous

oh whatever you know what I mean

- Nnesha

yes right so \[\huge\rm sin =\frac{ 1 }{ \sqrt{2} }\]
no radicals should be at the denominator so that's why we have to multiply both top and bottom by sqrt{2}
\[\huge\rm sin =\frac{ 1 }{ \sqrt{2} } \times \frac{\sqrt{2}}{\sqrt{2}}\]
sovle that

- Nnesha

solve*

- anonymous

umm... \[\frac{ \sqrt{2} }{ 2 }\] ?

- Nnesha

yes right

- Nnesha

-.- guess? well that's right
now cos = adjacent over hyp
so cos= ?

- anonymous

oh yay!

- anonymous

so \[\frac{ 45 }{ \sqrt{2} }\]

- anonymous

(and it wasn't a guess, i was just trying to remember when I had learned that, hoping I did it right from my memory)

- Nnesha

nope adjacent (side that touchz|dw:1433362268317:dw| the angle )

- Nnesha

adjacent = 1

- anonymous

oh so \[\frac{1 }{ \sqrt{2} }\]

- anonymous

...right?

- Nnesha

yes right solve that

- anonymous

wouldn't it be the same as the other one?

- Nnesha

yes right so (x ,y) x represent cosine y coordinate represent sin
for signs remember CAST rule |dw:1433362536321:dw|

- anonymous

whoa just got a little lost...

- Nnesha

alright it's okay
so what part u didn't understand ?? :-)

- anonymous

why do we need (x, y) what part of the problem does that fit into?

- Nnesha

yes that's what you need
cosine and sin value |dw:1433362805558:dw|

- Nnesha

x coordinate represent cosine
y represent sin
and that order pair we just found
which is
cosine = sqrt{2}/2
sin =sqrt{2}/2

- anonymous

|dw:1433362859876:dw| so you're saying this is x, and this is y?

- Nnesha

nope all angles ha' one x and one y value |dw:1433362920755:dw|
unit circle is almost like normal x y graph
CAST rule is easy to remember negative and plus signs
like we are in 4th quadrant where only cos values are positive
sin is negative and tan is negative so
angle 315
cosine and sin value are \[(\frac{ \sqrt{2} }{ 2 } , -\frac{ \sqrt{2} }{ 2 })\]

- Nnesha

http://prntscr.com/7cs6pe here is a unit cricle

- Nnesha

circle*

- anonymous

\[\left( \frac{ \sqrt{2} }{ 2 }, \frac{ \sqrt{2} }{ 2 } \right)\]
so then wouldn't it be like this?

- anonymous

I don't get why the second is negative....

- Nnesha

yes right but it's in 4th quadrant where y values are negative right so sin suppose to be negative

- anonymous

OH because sine is negative... sorry

- Nnesha

|dw:1433363153305:dw|

- Nnesha

it's like normal x y graph |dw:1433363292356:dw|

- anonymous

OMG IT JUST CLICKED!!! that took my brain way too long >.<

- Nnesha

:-)

- anonymous

@geekfromthefutur perhaps you'd like a math lesson? XD

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