How do you simplify $cos(arcsinx - arctan2x)$ ?

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Answer 1 · 2016-07-14T07:17:17

Reqd. Value $={sqrt(1-x^2)+2x^2}/sqrt(1+4x^2)$

Let $arcsinx=A, &, arctan2x=B$ , so that, $sinA=x, &, tanB=2x, A in [-pi/2,pi/2], B in (-pi/2,pi/2)$

We will consider the case $A, B in [0,pi/2]$ , so that all trigonometric ratios will be $+ve$ . The other case can be dealt with similarly.

Now reqd. value $=cos(A-B)=cosAcosB+sinAsinB.......(I)$

$sinA=x rArr cosA =sqrt(1-x^2)$

$tanB=2x rArr sec^2B=1+tan^2B=1+4x^2rArr secB=sqrt(1+4x^2) rArr cosB=1/sqrt(1+4x^2)$

$tanB=2xrArrcotB=1/(2x)rArrcsc^2B=1+cot^2B=1+1/(4x^2)=(4x^2+1)/(4x^2)rArrcscB=sqrt(1+4x^2)/(2x)rArrsinB=(2x)/sqrt(1+4x^2)$

Sub.ing all these in $(I)$ , we have,

The Reqd. Value $=sqrt(1-x^2)/sqrt(1+4x^2)+(2x^2)/sqrt(1+4x^2)$

$={sqrt(1-x^2)+2x^2}/sqrt(1+4x^2)$

Hope, this is Helpful! Enjoy Maths. , &, yes, don't forget to consider

the case $A,B in [-pi/2,0] !$

How do you simplify cos(arcsinx - arctan2x)cos(arcsinx - arctan2x)?