How do you simplify the expression $sin (arcsin (\frac{3}{5}) + arctan (- 2))$ $sin(arcsin(3/5)+arctan(-2))$ ?

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Answer 1 · 2018-03-06T08:54:22

$sin (arcsin (\frac{3}{5}) + arctan (- 2)) = - \frac{1}{\sqrt{5}}$ $sin(arcsin(3/5)+arctan(-2))=-1/sqrt5$

Let us note that the range of $arcsin x$ $arcsinx$ and $arctan x$ $arctanx$ is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$

Now let $arcsin (\frac{3}{5}) = A$ $arcsin(3/5)=A$ and $arctan (- 2) = B$ $arctan(-2)=B$ ,

then $sin A = \frac{3}{5}$ $sinA=3/5$ and $tan B = - 2$ $tanB=-2$

therefore $cos A = \sqrt{1 - {(\frac{3}{5})}^{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ $cosA=sqrt(1-(3/5)^2)=sqrt(16/25)=4/5$

and $sec B = \sqrt{1 + {(- 2)}^{2}} = \sqrt{5}$ $secB=sqrt(1+(-2)^2)=sqrt5$ i,e. $cos B = \frac{1}{\sqrt{5}}$ $cosB=1/sqrt5$

and as $sin A = tan B cos B = - \frac{2}{\sqrt{5}}$ $sinA=tanBcosB=-2/sqrt5$

we have $sin (arcsin (\frac{3}{5}) + arctan (- 2))$ $sin(arcsin(3/5)+arctan(-2))$

= $sin (A + B)$ $sin(A+B)$

= $sin A cos B + cos A sin B$ $sinAcosB+cosAsinB$

= $\frac{3}{5} \times \frac{1}{\sqrt{5}} + \frac{4}{5} \times (- \frac{2}{\sqrt{5}})$ $3/5xx1/sqrt5+4/5xx(-2/sqrt5)$

= $\frac{3 - 8}{5 \sqrt{5}}$ $(3-8)/(5sqrt5)$

= $- \frac{5}{5 \sqrt{5}}$ $-5/(5sqrt5)$

= $- \frac{1}{\sqrt{5}}$ $-1/sqrt5$

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