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

Question

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

3 Answers

Impact of this question

18239 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-28T10:20:50

$\sqrt{3} - \frac{3}{4}$ $sqrt3-3/4$

Explanation:

$sin (420^{o}) + {cos 390}^{o} - cos (- 300^{o}) sin (- 330^{o})$ $sin(420^o)+cos390^o-cos(-300^o)sin(-330^o)$

= $sin (420^{o}) + {cos 390}^{o} - cos (300^{o}) (- sin (330))$ $sin(420^o)+cos390^o-cos(300^o)(-sin(330))$

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

$= sin (5 \times 90^{o} - 30^{o}) + cos (5 \times 90^{o} - 60^{o}) + cos (4 \times 90^{o} - 60^{o}) sin (4 \times 90^{o} - 30^{o})$ $=sin(5xx90^o-30^o)+cos(5xx90^o-60^o)+cos(4xx90^o- 60^o)sin(4xx90^o-30^o)$

$= {cos 30}^{0} + {sin 60}^{0} + ({sin 60}^{o}) (- {cos 30}^{o})$ $=cos30^0+sin60^0+(sin60^o)(-cos30^o)$

$= \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} . \frac{\sqrt{3}}{2}$ $=sqrt3/2+sqrt3/2-sqrt3/2.sqrt3/2$

$= \sqrt{3} - \frac{3}{4}$ $=sqrt3-3/4$

Answer 2 · 2018-04-28T13:20:12

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

Explanation:

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

Let's get all of these in the range $- 180^{\circ}$ $-180^circ$ to $180^{\circ} .$ $180^circ.$

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

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

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

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

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

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

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

Answer 3 · 2018-04-28T13:43:49

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

$color(white)(ww$

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

$=>Sin420°Cos390°+Cos300°Sin330°$

$color(white)(ww$

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, $color(red)(SinACosB- CosASinB = sin(A-B)$

$=>sin30^@ =1/2$

Science

Math

Humanities

... and beyond

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

3 Answers

Explanation:

Explanation:

Related questions

Impact of this question