please help... calc will be the death of me lol

- anonymous

please help... calc will be the death of me lol

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- anonymous

##### 1 Attachment

- anonymous

@ikram002p heyy any idea here lol

- idku

this is calculus, are you sure?

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## More answers

- idku

|dw:1440615503189:dw|

- idku

sin(30)=1/2
mutliply both sides times -1 and use the rule
-sin(k)=sin(-k)

- anonymous

wait so how did you decide to do sin 30

- idku

As for your second problem
|dw:1440615688714:dw|

- idku

how you decide to do for sin(30)? what do you want to know?

- anonymous

lets just work on the first one im wayy too lost lol so you drew a reference triangle?

- welshfella

when you have negative values of a trig ratio you might find it easier to remember their signs in the 4 quadrants 0 - 360 degrees

- welshfella

|dw:1440615941625:dw|

- dumbcow

google unit circle

- welshfella

so they are ALL positive in first quadrant (0 to 90 degrees)
Only the sine is positive in 2nd quadrant, only the tangent in the third and only the cosine in the 4th quadrant - otherwise they are negative

- anonymous

okay i get that

- welshfella

so if you have sine x = -0.5
Sine 30 is 0.5 ( thats first quadant)
its negative in 2nd and 3rd
so 2 values for the angle would be 180-30 = 150 and 180+ 30 = 210
these are the results if the question asks for values of the angle in range 0 to 360 degrees.

- welshfella

oh i messed up there sorry!
sine is negative in 3rd and 4th quadrants so thats 210 and 330 dgrees!!

- anonymous

phew i got nervous for a sec i was like wuttt... but i get that so far

- welshfella

|dw:1440616444926:dw|

- welshfella

|dw:1440616515975:dw|

- welshfella

personally i find it easier to use pictures with these problems.

- welshfella

theres 2 triangles worth remembering
the one that idku posted and also the 45-45-90 triangle

- welshfella

|dw:1440616730872:dw|

- anonymous

alrigh but for this [prob im dealing with the 30 60 90

- welshfella

2 cos x = sqrt3
cos x = sqrt3/2
so what is the value of angle x ( from the 30 60 90 triangle)?

- welshfella

remember cosine = adjacent / hypotenuse

- welshfella

maybe the presence of the S is confusing you ?

- anonymous

so would i tak inverse cos (sqrt3/2) to find the angle

- welshfella

yes - or you can see it on the diagram
|dw:1440617639361:dw|

- anonymous

when i do it the way i said i get a domain error

- anonymous

wait can we stick with #1 for now im trying to understand it

- anonymous

so i get how you found 210 and 330

- anonymous

|dw:1440617866896:dw|

- phi

when you have time, you might want to review
https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/Trig-unit-circle/v/unit-circle-definition-of-trig-functions-1

- anonymous

ignore that drawing. i just watched a tutorial. so the question is asking me for the angle measure correct?

- phi

personally I would solve this
\[ \sin x = - \frac{1}{2} , \text{ for } 0 \le x \lt 2 \pi\]
but graphing (sketching actually) sin x from 0 to 2pi
|dw:1440619769364:dw|

- phi

looking at the curve, it should be clear there are two angles where sin x = -1/2
we should have memorized some "standard angles"
in particular sin 30º = 1/2
thus the solutions will be 30º *in the correct quadrants*
in particular 180+30 =210 (but put in radians)
and 360-30 = 330º, but put in radians.

- phi

for 2 cos x = sqr(3)
\[ \cos x = \frac{\sqrt{3}}{2} \]
|dw:1440620036903:dw|
again, two solutions
we should know cos 30º = sqr(3)/2
thus 30 and 360-30 = 330º

- anonymous

okay i understand much bette. the way i learned from the tutorial is to isolate the x and then look for it on the unit circle to find possible soulutions and then determine the correct quadrants

- phi

yes, that will work also.

- freckles

|dw:1440621102930:dw|

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