# inverses......

• May 21st 2008, 09:33 PM
sillylittletimes
inverses......
okay the ^-1 means the inverse of it.
and when i write 'pi' its supposed to be the pi symbol i just couldnt find it.
please help i need to understand how to do these i learn very well by example and i need answers to study these for finals

1) sin^-1(-1/2)
2)csc^-1(-√2)
3)cot^-1 √3
4)cos^-1(-1)
5)sec^-1(-√2)
6)sin^-1(cos pi/6)
7)cos^-1(tan pi/4)
8)sin[tan^-1 (-1)]
9)tan^-1(sec pi)
10)cos(sec^-1 √2)
11)sec^-1(cot 3pi/4)
12)csc^-1(cos 5pi/6)

• May 21st 2008, 09:48 PM
angel.white
Quote:

Originally Posted by sillylittletimes
1) sin^-1(-1/2)

$y = arcsin(-1/2)$

take the sine of both sides

$sin(y) = -1/2$

what y value will make this true? (look at your unit circle)

Quote:

Originally Posted by sillylittletimes
2)csc^-1(-√2)

$y=arccsc(-\sqrt{2})$

take cosecant of both sides

$csc(y)=-\sqrt{2}$

convert to sine notation

$\frac 1{sin(y)}=-\sqrt{2}$

$-\frac 1{\sqrt{2}}=sin(y)$

What value of y makes this true? look at your unit circle, note that 1/sqrt(2) is the same as sqrt(2)/2

Quote:

Originally Posted by sillylittletimes
8)sin[tan^-1 (-1)]

remember that arctangent is an angle, it is the angle that when you take the tangent of it, you will get negative one.

So draw a right triangle, and mark the opposite side as -1, and the adjacent side as 1. Then use the Pythagorean theorem to mark the hypotenuse sqrt(2)

This triangle is the triangle that if you took the tangent of it, you would get -1, so it is the triangle represented by tan^-1(-1)

Now we want the sine of this triangle, so that will be opposite over hypotenuse, which gives us -1/sqrt(2)
• May 24th 2008, 04:12 AM
nandu11
inverse just doesnt means 1 upon something it means inverse of a function which is an entirely diff thing form reciprocals.....like we have sin30 degree or sin(pi/6)[pi/6 is the radian form of 30 degree] sin30=1/2 but if we'll have inverse of it then sin(inverse)1/2 wud be 30 degree..likewise for sin(inverse)-1/2 we'll have the answers as -pi/2..and this wud depend upon principle value of that particular function...for sin it is -pi/3 to pi/2..that means answer of any sin inverse function can only be got within this interval...like for other trigo ratios u can find principal values....