How do you find the exact value of #sin(alpha-beta)# if #sinalpha=-5/13# and #sinbeta=1/6# if the terminal side of #alpha# lies in QIII and the terminal side of #beta# lies in QI?

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Answer 1 · 2017-06-20T08:13:42

# (12 - 5 sqrt 35)/78 = 0.225#

#sin alpha = -5/13# since #alpha# lie in QIII, therefore #cos alpha = -12/13 and tan alpha = 5/12#

#sin beta = 1/6# since #beta# lie in QI, therefore #cos beta = sqrt 35/6 and tan alpha = 1/sqrt 35#

#sin (alpha - beta) = sin alpha cos beta - sin beta cos alpha# #->i#

plug in the above values in #->i#

#sin (alpha - beta) = (-5/13)( sqrt 35/6) - (1/6)( -12/13)#

#sin (alpha - beta) = (-5sqrt 35)/78 + 12/78 = (12 - 5 sqrt 35)/78 = 0.225#