How do you calculate #sin^-1(sin2)#?

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Answer 1 · 2016-07-25T18:53:18

Inverses cancel each other out. #sin^(-1)(x)# is just another way of writing an inverse, or #arcsin(x)#.

Note that #arcsin# returns an angle, and if the angle is in degrees, then

#color(blue)(arcsin(sin(2^@)) = 2^@)#

If the #2# is in radians, then in terms of degrees:

#arcsin(sin(2 cancel"rad" xx 180^@/(pi cancel"rad"))) = arcsin[sin((360/pi)^@)]#

#= arcsin(sin(114.59^@))#

The #sin(114.59^@)# evaluates to about #0.9093#, and the #arcsin# of that would then be #1.14159cdots#, i.e.

#color(blue)(arcsin(sin("2 rad")) = pi - 2 " rad")#.

Note that this is NOT:

#1/(sin(sin2))#

which is not the same thing. If you did have #1/(sin(sin(2))#, it would be equal to #(sin(sin2))^(-1)#.

However, even though #sin^2(x) = (sinx)^2#, it does not mean that #sin^(-1)(x) = (sinx)^(-1)#.

Answer 2 · 2017-08-03T06:33:13

Refer to the Explanation Section.

Recall the following Defn. of #sin^-1# fun.,

#sin^-1x=theta, |x| <=1 iff sintheta=x, theta in [-pi/2,pi/2].#

Substituting the value #x=sintheta,# recd. from the R.H.S., into

the L.H.S., we get,

# sin^-1(sintheta)=theta, theta in [-pi/2,pi/2]..........(star)#

Now, regarding the Soln. of the Problem, we note that, there is

no mention about the Measure of the Angle #2,# i.e., it is

not clear, it is #2^@,# or #2" radian."#

If it is #2^@,#then, it follows from #(star)# that,

#sin^-1(sin2^@)=2^@.#

In case, it is #2" radian,"# we note that,

#sin2=sin(pi-(pi-2))=sin(pi-2),#

where, since #(pi-2) in [-pi/2,pi/2],# we have, by #(star),#

#sin^-1(sin2)=pi-2.#