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watermelon14 · Answer

$$\cos (arccsc(-2))$$

watermelon14 · Answer

the answer is $$\sqrt{3}/2  nott  -\sqrt{3}/2$$

watermelon14 · Answer

not*

watermelon14 · Answer

ik the restrictions of csc is -pi/2 to pi/2 but i just don't see how to apply that to finding the answer

watermelon14 · Answer

@jim_thompson5910 ?

hyuna301 · Answer

convert arcsec into arccos
suppose arcsec(x) = @        , here @ = theta
                     x   = sec@
now , cos@ = 1/x
,

watermelon14 · Answer

what about for arcsin(pi/2) the answer is -1 not 1. i don't understand the restriction for sin is -pi/2 to pi/2. 1 is less than pi/2 so why can't it work

daniel.ohearn1 · Answer

You said the answer is sqrt(3)/2.  You can draw a corresponding triangle. |dw:1462766332282:dw|

daniel.ohearn1 · Answer

Now what must theta be given your answer?

watermelon14 · Answer

yeah but i know that that is the answer after i learned that i am wrong. i am not given it at first

peachpi · Answer

The restriction places the angle in the 4th quadrant, since csc is negative.
cos(arccsc(-2))
The angle between -π/2 and π/2 with a csc of -2 is -π/6.
Then cos (-π/6) = (√3)/2

|dw:1462766662067:dw|

watermelon14 · Answer

couldn't also be in quad 3?

daniel.ohearn1 · Answer

There are many ways to represent the inverse csc function but the question is do you need to or can you just plug it in to your calculator to figure that it's sqrt(3)/2?

watermelon14 · Answer

what about for arcsin(pi/2) the answer is -1 not 1. i don't understand the restriction for sin is -pi/2 to pi/2. 1 is less than pi/2 so why can't it work 
can we do this first, it is easier when i get this i'll probably get the rest

peachpi · Answer

no. angles in the 3rd quadrant are outside the restriction

peachpi · Answer

I'm not sure how you'd find arcsin(pi/2). 
Usually you'd say arcsin 1 = pi/2

daniel.ohearn1 · Answer

1/sin(-2)  does not equal csc^-1(-2)    even though   sin(-2) = 1/csc(-2)

daniel.ohearn1 · Answer

Reciprocal identities do not translate over inverse trig functions.

daniel.ohearn1 · Answer

That tells you that the restrictions for sin and inverse sine are different in this case inverse sin has a wavelength from -1 to 1 while sine as you know 0 to pi

daniel.ohearn1 · Answer

That's sin inverse of -1 to sin inverse of 1 which is -pi/2 to pi/2.

daniel.ohearn1 · Answer

The quadrant where you draw the triangle won't matter if you maintain the magnitude of sqrt(3)/2, for purposes of pinpointing theta because theta will always be positive.

daniel.ohearn1 · Answer

assuming theta is the interior angle of the triangle.