Question

prove that cos(A-B) = cosA. cosB + sinA. sinB​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

By Euler's form

$\rm :\longmapsto\: {e}^{ix} = cosx + isinx$

So, if we replace x by - x, we get

$\rm :\longmapsto\: {e}^{ - ix} = cos( - x) + isin( - x)$

$\rm :\longmapsto\: {e}^{ - ix} = cosx - isinx$

Now,

Consider,

$\rm :\longmapsto\: {e}^{ix} = cosx + isinx$

Replace x by A - B, we get

$\rm :\longmapsto\: {e}^{i(A - B)}$

$\rm \:  =  \:  \: cos(A - B) + isin(A - B) - - (1)$

Now,

Again consider,

$\rm :\longmapsto\: {e}^{i(A - B)}$

can be rewritten as

$\rm \:  =  \:  \: {e}^{iA} \times {e}^{ - iB}$

$\rm \:  =  \:  \: (cosA + isinA)(cosB - isinB)$

$\rm \:  =cosAcosB - icosAsinB + isinAcosB - {i}^{2}sinAsincos$

$\rm \:  =cosAcosB - icosAsinB + isinAcosB + sinAsincos$

$\rm \:  =  \:(cosAcosB + sinAsinB) + i(sinAcosB - sinBcosA)$

Thus, from the above calculations, we get

$\rm :\longmapsto\: {e}^{i(A - B)} \: = \: cos(A - B) + isin(A - B)$

and

$\rm \:  {e}^{i(A - B)} =  \:(cosAcosB + sinAsinB) + i(sinAcosB - sinBcosA)$

So, on comparing above two equations we get

$\rm \: cos(A - B) + isin(A - B) = (cosAcosB + sinAsinB) + i(sinAcosB - sinBcosA)$

On comparing real parts, we get

$\rm :\longmapsto\:cos(A - B) = cosAcosB + sinAsinB$

Hence, Proved

Additional Information ;-

If we compare imaginary parts, we get

$\rm :\longmapsto\:sin(A - B) = sinAcosB - sinBcosA$

If we replace, B by - B, we get

$\rm :\longmapsto\:sin(A + B) = sinAcos( - B) - sin( - B)cosA$

$\rm :\longmapsto\:sin(A + B) = sinAcosB + sinBcosA$