Question

sin(a) cos(a) = (1/2)sin (2a)
prove this identity​

Answer 1

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

We know,

By Euler's Form, we have

$\rm :\longmapsto\: {e}^{ia} = cosa + i \: sina$

So,

$\rm :\longmapsto\: {e}^{ib} = cosb + i \: sinb$

Also,

$\rm :\longmapsto\: {e}^{i(a + b)} = cos(a + b) + i \: sin(a + b) - - - (1)$

Now,

$\rm :\longmapsto\: {e}^{i(a + b)}$

can be rewritten as

$\rm \:  =  \:  \: {e}^{ia} \times {e}^{ib}$

$\rm \:  =  \:  \: (cosa + i \: sina) \times (cosb + i \: sinb)$

$\rm \:  =  \:  \: cosacosb + icosasinb + isinacosb + {i}^{2}sinasinb$

$\rm \:  =  \:  \: cosacosb + i(cosasinb + sinacosb) - sinasinb$

$\rm \:  =  \:  \: (cosacosb - sinasinb) + i(cosasinb + sinacosb)$

From equation (1) and from above we concluded that

$\rm :\longmapsto\:cos(a + b) + i \: sin(a + b)$

$\rm \:  =  \:  \: (cosacosb - sinasinb) + i(cosasinb + sinacosb)$

So, on comparing Imaginary parts on both sides, we get

$\red{\rm :\longmapsto\:sin(a + b) = sina \: cosb + sinb \: cosa}$

On substituting b = a, we get

$\red{\rm :\longmapsto\:sin(a + a) = sina \: cosa + sina \: cosa}$

$\red{\rm :\longmapsto\:sin(2a) = 2 \: sina \: cosa }$

$\red{\bf\implies \:sina \: cosa \: = \: \dfrac{1}{2} \: sin2a}$

Hence, Proved