Inverse Sine and Cosine

[giraffie]

Ok, so I'm having some trouble with trig. =( Here's an example problem:

Cos^-1(cos5?/3)

I first simplified the cos5?/3 to 1/2. From reading my textbook, I saw that you're supposed to replace the 'outside term' with a variable, so:
v=Cos^-1(1/2)
and
cos(v)=1/2

However, I don't know where to go from here. I looked on Hotmath and it said the answer was ?/3, because '–?<y<?'. I don't understand where they got ?/3 from. Can anyone help?

 

[arthur ohlsten]

2 suggestions



Sketch a curve of cos v 0<v<2pi
From the curve you can determine the following

If you had cosv=0
then v=pi/2 or 3pi/2

if you had cos v = 1
then v= 0 or 2pi

if you had cos v = -1
then v= pi

==================================================================================================
sketch a right triangle
mark the hypoteneuse 1
mark a interior angle v
then the adjacent side becomes 1/2 , because the cos v = adjacent side / hypotenuese
cos v= 1/2 / 1
cos v=1/2 given

then the other interior angle is 30 degrees WHY? because the sin u = opposite / hypotenuse
sin u= 1/2 / 1
sin u =1/2
u=30 degrees
then v=60 degrees

if 360 degrees = 2pi
then 360/60 = 2pi/x
x= [[2 pi]60]/360
x=pi/3 answer

Arthur
x=

 

[Deleted member 4993]

Ok, so I'm having some trouble with trig. =( Here's an example problem:

Cos^-1(cos5?/3)

I first simplified the cos5?/3 to 1/2. From reading my textbook, I saw that you're supposed to replace the 'outside term' with a variable, so:
v=Cos^-1(1/2)
and
cos(v)=1/2 = cos (60°) = cos(?/3)

However, I don't know where to go from here. I looked on Hotmath and it said the answer was ?/3, because '–?<y<?'. I don't understand where they got ?/3 from. Can anyone help?

You know that, if

cos[sup:1nonao1p]-1[/sup:1nonao1p](x) = ? .... where 0 ? ? ? ?

then

cos(2? ± ?) = x

Now set up

\(\displaystyle cos^{-1}[cos(\frac{5\pi}{3})] \, = \, \theta\)

and continue...

 

[giraffie]

Ooh, I get it. Thanks