How do you evaluate #sin(-175) * tan(185°) * cos(355)+sin(-85°) * cos(365°)#?

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Answer 1 · 2017-09-12T14:49:59

#sin(-175^@)*tan(185^@)*cos(355^@)+sin(-85^@)*cos(365^@)=-1#

#sin(-175^@)*tan(185^@)*cos(355^@)+sin(-85^@)*cos(365^@)#

and as #sin(-A)=-sinA#, this is equal to

#-sin175^@*tan(185^@)*cos(355^@)-sin85^@*cos(365^@)#

= #-sin(180^@-5^@)*tan(180^@+5^@)*cos(360^@-5^@)-sin(90^@-5^@)cos(360^@+5^@)#

= #-sin5^@*tan5^@*cos5^@-cos5^@*cos5^@#

= #-sin5^@*(sin5^@)/(cos5^@)*cos5^@-cos^2 5^@#

= #-sin^2 5^@-cos^2 5^@#

= #-(sin^2 5^@+cos^2 5^@)#

=#-1#