Given: sine (a) = four-fifths, startfraction pi over 2 endfraction < a < pi and sine (b) = startfraction negative 2 startroot 5 endroot over 5 endfraction, pi, pi < b < startfraction 3 pi over 2 endfraction what is the value of cos(a – b)?



Answer :

The value of Cos ( a – b ) is ( [tex]-\sqrt{5} / 5[/tex] ) .

Given :

sine (a) = four-fifths, start fraction pi over 2 end fraction < a < pi and sine (b) = start fraction negative 2 start root 5 end root over 5 end fraction, pi, pi < b < start fraction 3 pi over 2 end fraction .

Sin ( a ) = 4/5

Sin ( b ) = [tex]-2\sqrt{5} / 5[/tex]

Identity :

Cos( a - b ) = cos a * cos b + sin a * sin b

substitute

cos b = -1 / [tex]\sqrt{5}[/tex]

cos a = -3/5

cos ( a - b ) = -3/5 * -1 / [tex]\sqrt{5}[/tex] + 4/5 * [tex]-2\sqrt{5} / 5[/tex]

on solving

cos ( a - b ) = [tex]-\sqrt{5} / 5[/tex]

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