Sin420°+Cos390°-Cos(-300°)Sin(-330°) Solve And Answer Value?

3 Answers
Apr 28, 2018

#sqrt3-3/4#

Explanation:

#sin(420^o)+cos390^o-cos(-300^o)sin(-330^o)#

=#sin(420^o)+cos390^o-cos(300^o)(-sin(330))#

=#sin(420^o)+cos(390^o)+cos(300^o)sin(330^o)#

#=sin(5xx90^o-30^o)+cos(5xx90^o-60^o)+cos(4xx90^o- 60^o)sin(4xx90^o-30^o)#

#=cos30^0+sin60^0+(sin60^o)(-cos30^o)#

#=sqrt3/2+sqrt3/2-sqrt3/2.sqrt3/2#

#=sqrt3-3/4#

Apr 28, 2018

Adding or subtracting #360^circ# each angle brings them in range, where the form is recognized to be the sine difference angle formula, and works out to #1/2#.

Explanation:

Angles that differ by a multiple of #360^circ# are called coterminal , and have the same value for their trig functions.

Let's get all of these in the range #-180^circ# to #180^circ. #

# sin 420^circ cos 390^circ - cos(-300^circ) sin (-330^circ) #

# = sin (420^circ -360^circ) cos (390^circ-360^circ) - cos(-300^circ+360^circ) sin (-330^circ+360^circ) #

# = sin (60^circ) cos (30^circ) - cos(60^circ) sin (30^circ) #

We know what all of those are so we could work this out, or recognize it as the sine difference angle formula:

# sin(a-b) = sina cos b - cos a sin b#

# = sin (60^circ) cos (30^circ) - cos(60^circ) sin (30^circ)#

# = sin (60^circ - 30^circ) = sin 30^circ = 1/2#

Apr 28, 2018

#[Sin420°Cos390°]-[Cos(-300°)Sin(-330°)] #

#color(white)(ww#

#"As "[ color(magenta)(cos(-x) = cosx ; sin(-x)= -sinx)]#

#=>Sin420°Cos390°+Cos300°Sin330° #

#color(white)(ww#

Now split the angle measurement in terms of #360^@#

#=>Sin(360+60)°Cos(360+30)°+Cos(360-60)°Sin(360-30)° #

#color(white)(ww#

#=>Sin60°Cos30°- Cos60°Sin30° #

#color(white)(ww#

This is of the form, #color(red)(SinACosB- CosASinB = sin(A-B) #

#=>sin30^@ =1/2#