Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : Dec 2013

Let, $ u=sinx \cdot sin y \cdot sin(x+y) \\ \; \\ \;=\; \dfrac{1}{2} \Big[ (2sinx \cdot siny) \cdot sin(x+y) \Big] \\ \; \\ \; \\ \;=\; \dfrac{1}{2} \Big[ (cos(x-y)-cos(x+y)) \cdot sin(x+y) \Big] \; \; \\ \ldots \{ cos(A-B)-cos(A+B) \;=\; 2sinAsinB \} \\ \; \\ \; \\ \;=\; \dfrac{1}{2} \Big[ cos(x-y)sin(x+y)-cos(x+y)sin(x+y) \Big] \\ \; \\ \; \\ \;=\; \dfrac{1}{4} \Big[ 2cos(x-y)sin(x+y)-2cos(x+y)sin(x+y) \Big] \\ \; \\ \; \\ \;=\; \dfrac{1}{4} \Big[ sin2x + sin2y -sin(2(x+y)) \Big] \\ \{ \because sin(A+B)+sin(A-B) \;=\; 2sinA \;cosB\} \\ \; \\ \; \\ $

Diff. u partially w.r.t.x,

$ \therefore \dfrac{\partial u}{\partial x} \;=\; \dfrac{1}{4} \bigg\{ 2cos2x+0-cos[2(x+y)] \dfrac{\partial }{\partial x} (2x+2y) \bigg\} \\ \; \\ $

Equating it to zero,

$ \therefore \dfrac{1}{4} \bigg\{ 2cos2x-2cos[2(x+y)] \bigg\} \;=\; 0 \\ \; \\ 2cos2x-2cos[2(x+y)] \;=\; 0 \\ \; \\ \; \\ \therefore 2sin(2x+y) \; siny \;=\; 0 \;\;\;\;\; \ldots (ii) \\ \bigg\{ \because cosC-cosD \;=\; 2sin\Big( \dfrac{C+D}{2} ) sin\Big( \dfrac{C-D}{2} ) \bigg\} \\ \; \\ \; \\ \; \\ $

Diff. y partially w.r.t.x,

$ \therefore \dfrac{\partial u}{\partial y} \;=\; \dfrac{1}{4} \bigg\{ 2cos2y-cos[2(x+y)] \dfrac{\partial }{\partial y} (2x+2y) \bigg\} \\ \; \\ $

Equating it to zero,

$ \therefore \dfrac{1}{4} \bigg\{ 2cos2y-2cos[2(x+y)] \bigg\} \;=\; 0 \\ \; \\ 2cos2y-2cos[2(x+y)] \;=\; 0 \; \; \; \ldots(iii) \\ \; \\ \; \\ \therefore 2cos2y-2cos[2(x+y)] \;=\; 0 \; \; \; \ldots From \; (i) \\ \; \\ \; \\ \therefore 2 sin\Big( \dfrac{2y+2x}{2} ) sin\Big( \dfrac{2x-2y}{2} ) \;=\; 0 \\ \; \\ 2 sin\Big( x+y ) sin\Big( x-y ) \;=\; 0 \\ \; \\ \; \\ $

From (i) and (iii), we can conclude that $ cos2x \;=\; cos2y$

$ \\ or \; x=y \\ \; \\ \therefore x \;= \; \pi, 2\pi, 3\pi, \ldots \{ \because cos2\pi \;=\; cos4\pi\ldots =1 \} \\ \; \\ \therefore x \;=\; y \;= \; n\pi $