How do I solve this?

Question

find sinα if cosα = 1/5 and α is in quadrant IV

317 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-19T09:17:22

#cosα = 1/5#

by the identity, we know, #cos^2x +sin^2x=1#

therefore, #sin^2α= 1- cos^2α#

#sin^2α= 1- (1/5)^2#

#sin^2α= 1- (1/25)#

#sin^2α= 24/25#

#sinα= +-sqrt(24/25)#

since, α is in quadrant IV, #sinalpha# must be negative,

#sinα= -sqrt24/5= - (2sqrt6)/5#