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

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Answer 1 · 2018-03-06T08:54:22

$sin(arcsin(3/5)+arctan(-2))=-1/sqrt5$

Let us note that the range of $arcsinx$ and $arctanx$ is $[-pi/2,pi/2]$

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

then $sinA=3/5$ and $tanB=-2$

therefore $cosA=sqrt(1-(3/5)^2)=sqrt(16/25)=4/5$

and $secB=sqrt(1+(-2)^2)=sqrt5$ i,e. $cosB=1/sqrt5$

and as $sinA=tanBcosB=-2/sqrt5$

we have $sin(arcsin(3/5)+arctan(-2))$

= $sin(A+B)$

= $sinAcosB+cosAsinB$

= $3/5xx1/sqrt5+4/5xx(-2/sqrt5)$

= $(3-8)/(5sqrt5)$

= $-5/(5sqrt5)$

= $-1/sqrt5$

How do you simplify the expression sin(arcsin(3/5)+arctan(-2))sin(arcsin(3/5)+arctan(-2))?