How do you evaluate the expression $sin (u + v)$ $sin(u+v)$ given $sin u = \frac{3}{5}$ $sinu=3/5$ with $\frac{π}{2} < u < p$ $pi/2<u<p$ and $cos v = - \frac{5}{6}$ $cosv=-5/6$ with $π < v < \frac{3 π}{2}$ $pi<v<(3pi)/2$ ?

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How do you evaluate the expression $sin (u + v)$ $sin(u+v)$ given $sin u = \frac{3}{5}$ $sinu=3/5$ with $\frac{π}{2} < u < p$ $pi/2<u<p$ and $cos v = - \frac{5}{6}$ $cosv=-5/6$ with $π < v < \frac{3 π}{2}$ $pi<v<(3pi)/2$ ?

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Answer 1 · 2016-12-03T14:35:28

$sin (u + v) = \frac{- 15 + 4 \sqrt{11}}{36} = - 0.0481$ $sin(u+v)=(-15+4sqrt11)/36=-0.0481$

Explanation:

As $sin u = \frac{3}{5}$ $sinu=3/5$ and domain of $u$ $u$ is given by $\frac{π}{2} < u < π$ $pi/2 < u < pi$ i.e. $u$ $u$ is in second quadrant and $cos u$ $cosu$ is negative and

$cos u = - \sqrt{1 - {(\frac{3}{5})}^{2}} = - \sqrt{1 - \frac{9}{25}} = - \sqrt{\frac{16}{25}} = - \frac{4}{5}$ $cosu=-sqrt(1-(3/5)^2)=-sqrt(1-9/25)=-sqrt(16/25)=-4/5$

Further as $cos v = - \frac{5}{6}$ $cosv=-5/6$ with $π < v < \frac{3 π}{2}$ $pi < v < (3pi)/2$ , hence $v$ $v$ is in third quadrant and $sin v$ $sinv$ is negative and

$sin v = - \sqrt{1 - {(- \frac{5}{6})}^{2}} = - \sqrt{1 - \frac{25}{36}} = - \sqrt{\frac{11}{36}} = - \frac{\sqrt{11}}{6}$ $sinv=-sqrt(1-(-5/6)^2)=-sqrt(1-25/36)=-sqrt(11/36)=-sqrt11/6$

Now $sin (u + v) = sin u cos v + cos u sin v$ $sin(u+v)=sinucosv+cosusinv$

= $\frac{3}{5} \times (- \frac{5}{6}) + (- \frac{4}{5}) \times (- \frac{\sqrt{11}}{6})$ $3/5xx(-5/6)+(-4/5)xx(-sqrt11/6)$

= $\frac{- 15}{30} + \frac{4 \sqrt{11}}{30}$ $(-15)/30+(4sqrt11)/30$

= $\frac{- 15 + 4 \sqrt{11}}{36}$ $(-15+4sqrt11)/36$

= $\frac{- 15 + 4 \times 3.317}{36} = - \frac{1.732}{36} = - 0.0481$ $(-15+4xx3.317)/36=-1.732/36=-0.0481$

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How do you evaluate the expression $sin (u + v)$ $sin(u+v)$ given $sin u = \frac{3}{5}$ $sinu=3/5$ with $\frac{π}{2} < u < p$ $pi/2<u<p$ and $cos v = - \frac{5}{6}$ $cosv=-5/6$ with $π < v < \frac{3 π}{2}$ $pi<v<(3pi)/2$ ?

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How do you evaluate the expression $sin (u + v)$ $sin(u+v)$ given $sin u = \frac{3}{5}$ $sinu=3/5$ with $\frac{π}{2} < u < p$ $pi/2<u<p$ and $cos v = - \frac{5}{6}$ $cosv=-5/6$ with $π < v < \frac{3 π}{2}$ $pi<v<(3pi)/2$ ?