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

Apr 28, 2018

$\sqrt{3} - \frac{3}{4}$

#### Explanation:

$\sin \left({420}^{o}\right) + \cos {390}^{o} - \cos \left(- {300}^{o}\right) \sin \left(- {330}^{o}\right)$

=$\sin \left({420}^{o}\right) + \cos {390}^{o} - \cos \left({300}^{o}\right) \left(- \sin \left(330\right)\right)$

=$\sin \left({420}^{o}\right) + \cos \left({390}^{o}\right) + \cos \left({300}^{o}\right) \sin \left({330}^{o}\right)$

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

$= \cos {30}^{0} + \sin {60}^{0} + \left(\sin {60}^{o}\right) \left(- \cos {30}^{o}\right)$

$= \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2.} \frac{\sqrt{3}}{2}$

$= \sqrt{3} - \frac{3}{4}$

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 $\frac{1}{2}$.

#### 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 ${180}^{\circ} .$

$\sin {420}^{\circ} \cos {390}^{\circ} - \cos \left(- {300}^{\circ}\right) \sin \left(- {330}^{\circ}\right)$

$= \sin \left({420}^{\circ} - {360}^{\circ}\right) \cos \left({390}^{\circ} - {360}^{\circ}\right) - \cos \left(- {300}^{\circ} + {360}^{\circ}\right) \sin \left(- {330}^{\circ} + {360}^{\circ}\right)$

$= \sin \left({60}^{\circ}\right) \cos \left({30}^{\circ}\right) - \cos \left({60}^{\circ}\right) \sin \left({30}^{\circ}\right)$

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

$\sin \left(a - b\right) = \sin a \cos b - \cos a \sin b$

$= \sin \left({60}^{\circ}\right) \cos \left({30}^{\circ}\right) - \cos \left({60}^{\circ}\right) \sin \left({30}^{\circ}\right)$

$= \sin \left({60}^{\circ} - {30}^{\circ}\right) = \sin {30}^{\circ} = \frac{1}{2}$

Apr 28, 2018

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

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

=>Sin420°Cos390°+Cos300°Sin330°

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Now split the angle measurement in terms of ${360}^{\circ}$

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

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=>Sin60°Cos30°- Cos60°Sin30°

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

$\implies \sin {30}^{\circ} = \frac{1}{2}$