Question

Find the exact value of:

Answer 1

\[ \sin(\sin^{-1} (-1/2)+\cos^{-1} (-1/2))\]

Answer 2

1 is the answer I got

Answer 3

you mght have to use the sin formula for the addition of angles

sin(a+b) = sina cosb + sinb cosa

Answer 4

and a triangle labeled helps too

Answer 5

first what we will do is that sin(-x) = -sinx and cos(-x) = cosx and also that cos theta =a and cos^-1 (a) = theta alright these are some rules

Answer 6

@amistre64 , It is arcsin and not sin :D

Answer 7

arctrigs are angles in disguise

Answer 8

@amistre64 , my reply was in reference to this statement
"you mght have to use the sin formula for the addition of angles sin(a+b) = sina cosb + sinb cosa"

Answer 9

a = arcsin(-1/2) ; b = arccos(-1/2)

they are still angles to work out

Answer 10

sin(a+b) = sina cosb + sinb cosa
= sin(arcsin(-1/2)) cos(arccos(-1/2))
+ sin(arccos(-1/2)) cos(arcsin(-1/2))

Answer 11

Why complicate things ? @amistre64

We should keep two things in mind
1) arcsin(x) always lies between (-pi/2 , pi/2)
2) arccos(x) always lies between (0 , pi)

and just apply them

Answer 12

I mean that they are the principal solutions ofcourse :D

Answer 13

so in this case sin(sin−1(−1/2)+cos−1(−1/2)) so let us do this in steps and solve sin(sin−1(−1/2) first ok and arcsin(-1/2) = theta and sin theta = -1/2 and this is true when theta = 7pi/6 and now sin(7pi/6) = -1/2

and now we we will do cos−1(−1/2) = theta and cos theta = -1/2 and this is only true when theta = 2pi/3

and now let us add these 2 terms to get -1/2 +2pi/3 = (4pi -3)/6 :)

Answer 14

for sin(sin−1(−1/2) we could have used a defintion where sin(arcsin(x)) this is just equal to x

Answer 15

hamza; arc sin doesnt have 7pi/6 in its range ....

Answer 16

why nott

Answer 17

sin 7pi/6 = -1/2 ?

Answer 18

This is probably the easiest, \(\arcsin x + \arccos x = \frac \pi 2 \) then the required exact value is 1.

Answer 19

OMG lol i read the question totally wrong i thought it was sin(sin−1(−1/2) + cos-1(-1/2)

Answer 20

i am so sorry

Answer 21

NOTE: \(\large \arcsin( x) + \arccos( x) = \frac \pi 2 \forall x \in [-1,1] \)

Answer 22

@Foolformath, I wanted to confirm my answer. Thanx to everyone for their input

Answer 23

Glad to help :)