Sin (60°+theta)-cos (30°-theta)=?

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Sin (60°+theta)-cos (30°-theta)=?

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Answer 1 · 2018-03-10T05:59:21

$0$ $0$

Explanation:

Given: $sin (60^{o} + θ) - cos (30^{o} - θ)$ $sin(60^o + theta) - cos(30^o-theta)$

Use the sum and difference formulas:
$sin (u + v) = sin u cos v + cos u sin v$ $sin(u+v) = sin u cos v + cos u sin v$
$cos (u - v) = cos u cos v + sin u sin v$ $cos(u-v) = cos u cos v + sin u sin v$

${sin 30}^{o} = \frac{1}{2}; {cos 30}^{o} = \frac{\sqrt{3}}{2}$ $sin 30^o = 1/2; " "cos 30^o = (sqrt(3))/2$

${sin 60}^{o} = \frac{\sqrt{3}}{2}; {cos 60}^{o} = \frac{1}{2}$ $sin 60^o = (sqrt(3))/2; " "cos 60^o = 1/2$

$sin (60^{o} + θ) = {sin 60}^{o} cos θ + {cos 60}^{o} sin θ$ $sin(60^o + theta) = sin60^o cos theta + cos 60^o sin theta$
$= \frac{\sqrt{3}}{2} cos θ + \frac{1}{2} sin θ$ $= (sqrt(3))/2 cos theta + 1/2 sin theta$

$cos (30^{o} - θ) = {cos 30}^{o} cos θ + {sin 30}^{o} sin θ$ $cos(30^o-theta) = cos 30^o cos theta + sin 30^o sin theta$
$= \frac{\sqrt{3}}{2} cos θ + \frac{1}{2} sin θ$ $= (sqrt(3))/2 cos theta + 1/2 sin theta$

$sin (60^{o} + θ) - cos (30^{o} - θ) =$ $sin(60^o + theta) - cos(30^o-theta) =$

$\frac{\sqrt{3}}{2} cos θ + \frac{1}{2} sin θ - \frac{\sqrt{3}}{2} cos θ - \frac{1}{2} sin θ = 0$ $(sqrt(3))/2 cos theta + 1/2 sin theta - (sqrt(3))/2 cos theta - 1/2 sin theta = 0$

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Sin (60°+theta)-cos (30°-theta)=?

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