How do you find #sin^-1(sin ((5pi)/6))#?

Question

Delilah Aguilar · Accepted Answer

#sin^-1(sin(5pi)/6)=pi/6#

#sin((5pi)/6)=1/2#, so we're really being asked for #sin^-1(1/2).#

Let

#y=sin^-1(1/2)#

Then, from the definition of arcsine,

#1/2=sin(y)#

( #y=sin^-1(a) hArr a=sin(y)# )

In other words, we want to know where the sine function returns #1/2.#

Sine is equal to #1/2# for #pi/6+2npi, (5pi)/6+2npi,# where #n# is any integer. But really, since the domain of arcsine is #[-pi/2, pi/2], pi/6# is the only valid answer.

Quinn Anderson · Answer

#pi/6#

Hint: We know that , #color(red)(sin^-1(sintheta)=theta,# where, #color(red)(theta in [-pi/2,pi/2]# Even though, #color(blue)(sin^-1(sin((5pi)/6))!=((5pi)/6),# because, #color(blue)((5pi)/6 !in [-pi/2,pi/2]# So, we reduce #theta=(5pi)/6.# But how ? By using Addition and Subtraction Formulas for sine and cosine .Now enjoy the answer.

Answer:

Here,

#(5pi)/6=pi-pi/6#

#(sin^-1(sin((5pi)/6))=sin^-1(sin(pi-pi/6))#

#=sin^-1(sin(pi/6))....to#, Apply #color(red)([sin(pi-theta)=sintheta ]#

#=pi/6......to#, where, #pi/6in[-pi/2,pi/2]#

Delilah Aguilar · Answer

undefined To find $ \sin^{-1}(\sin(\frac{5\pi}{6})) $, first, calculate $ \sin(\frac{5\pi}{6}) $, which equals $ \frac{\sqrt{3}}{2} $. Then, since $ \sin^{-1}(\sin(\theta)) = \theta $ only if $ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $, and $ \frac{5\pi}{6} $ lies in the second quadrant, the answer is $ \frac{\pi}{6} $.