Prove arcsin(a) + arcsin(b)= arcsin (a(square root (1 - b squared))+b(square root (1 - a squared)))

neela | Student

To prove arcsinA+arc sinb = arcxina*

Let arcsi(a) = x  and arc sin(b) = y so that

a=sinx and b sin y. Yherefore, cosx = sqrt(1-a^2) and cosy = sqrt(1-b^2)

Therfore, sin(x+y) = sinx*cosy+cosxsiny = a*sqrt(1-b^2) +b*sqrt(1-a^2) or by taking inverse,

x+y  =

arc sin(a)+arcsin(b)  = arc sine{a*sqrt(1-b^2)+b*sqrt(1-a^2)}

giorgiana1976 | Student

As you know, arcsin(a) and arcsin(b) are angles.

For example:

arcsin (1/2)= 30 degrees

arcsin (1/2)= pi/6

Let's make the substitution:



Also we know that:

sin (arcsin a)=a (e.g. sin (arcsin 1/2)=sin 30=1/2)

cos (arcsin a)= cos (u)

From the fundamental relation:

(sin a)^2+(cos a)^2=1

cos (u)= sqrt [1- (sin u)^2]

But u=arcsin a

cos (u)= sqrt [1- (sin arcsin a)^2]=sqrt (1-a^2)

cos (v)= sqrt [1- (sin arcsin b)^2]=sqrt (1-b^2)

Let's apply the function sine to the (arcsin(a) + arcsin(b))


sin (arcsin(a) + arcsin(b))=sin (u+v)=

=sin u*cos v + sin v* cosu=

=sin (arcsin(a))*sqrt (1-b^2)+sin(arcsin(b))*sqrt (1-b^2)=

=a*sqrt (1-b^2)+b*sqrt (1-b^2)

But applying the function sine in the left side of the equal, we have to apply it on the right side too, for keeping up the balance of the identity.

So, after applying the function sine:

sin{arcsin[ a(sqrt (1 - b^2))+b(sqrt (1 - a^2)]}=

=a(sqrt (1 - b^2))+b(sqrt (1 - a^2)

We've noticed that after applying the function sine, both side of the identity, the result is the same!

sin (arcsin(a) + arcsin(b))=sin{arcsin[ a(sqrt (1 - b^2))+b(sqrt (1 - a^2)]}

a*sqrt (1-b^2)+b*sqrt (1-b^2)=a*sqrt (1-b^2)+b*sqrt (1-b^2)

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