Question

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

Answer 1

`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`

Answer 2

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`

Answer 3

`[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`