Question

Question #58bec

Answer 1

`sin(pi/3+x)-cos(pi/6+x)=sin x`

Explanation:

`"let " sin(pi/3+x)-cos(pi/6+x) " be equal to 'k' "`

`sin(a+b)=sin a*cos b +cos a * sin b`

`sin(pi/3+x)=sin( pi/3)*cos x+cos(pi/3)*sin x`

`sin(pi/3+x)=sqrt 3/2*cos x+1/2*sin x`

`cos(a+b)=cos a*cos b-sin a*sin b`

`cos(pi/6+x)=cos(pi/6)*cos x-sin(pi/6)*sin x`

`cos(pi/6+x)=sqrt 3/2*cos x-1/2* sin x`

`sqrt 3/2*cos x+1/2*sin x-(sqrt 3/2*cos x-1/2* sin x)`

`cancel(sqrt 3/2*cos x)+1/2*sin x-cancel(sqrt 3/2*cos x)+1/2 *sin x=k`

`1/2*sin x+1/2*sin x=k`

`1/2 sin x(1+1)=k`

`1/cancel(2) *cancel(2) sin x=k`

`sin x=k`

`sin(pi/3+x)-cos(pi/6+x)=k`

`sin(pi/3+x)-cos(pi/6+x)=sin x`

Answer 2

Refer to the Explanation Section below.

Explanation:

As a Second Method, let us solve this as under :

Let `(pi/3+x)=A, and, (pi/6+x)=B rArr A+B=pi/2+2x.`

`:. A=pi/2+(2x-B).`

Hence, the R.H.S. `=sinA-cosB,`

`=sin{pi/2+(2x-B)}-cosB,`

`=cos(2x-B)-cosB,....[because, sin(pi/2+theta)=costheta].`

But, `cosC-cosD=-2sin((C+D)/2)sin((C-D)/2).`

`:."The R.H.S.="-2sin{((2x-B)+B)/2}sin{((2x-B)-B)/2}`

`=-2sinxsin(x-B)`

Here, `B=pi/6+x rArr x-B=-pi/6.`

`:."The R.H.S.="-2sinxsin(-pi/6),`

`=(-2sinx)(-sin(pi/6)),`

`=(-2sinx)(-1/2),`

`=sinx`

`="The L.H.S."`

Enjoy Maths.!