Question

The system of equations sin x cos y = a’ and
siny.cosx = a
has a solution if lalsi
1-15
1+15
<a<
has a solution only if
2
2.
1-15
has a solution only if
2
as +1+v5
<
2
can hold if a = -1​

Answer 1

Step-by-step explanation:

Given that:

The system of equations 'sin x cos y = a’ and siny.cosx = a'

To find: Solution of equation

Solution:

Let

$sinx \: cos \: y = a \: \: \: \: ...eq1 \\ \\ siny \: cos \: x = a \: \: \: \: ...eq2$

Add both eq1 and eq2

$sinx \: cos \: y + siny \: cos \: x = a + a \\ \\ sin(x + y) = 2a \\ \\ or \\ \\ x + y = {sin}^{ - 1}2a \: \: \: ...eq3 \\ \\ \because \\ \\ \boxed{sinA \: cos \: B+ cos A \: sin \: B = sin(A+ B)}$

Subtract both equations

$sinx \: cos \: y - siny \: cos \: x = a - a \\ \\ sin(x - y) = 0 \\ \\ or \\ \\ x - y = {sin}^{ - 1}0 = 0 \: \: \: ...eq4 \\ \\ \because \\ \\ \boxed{sinA \: cos \: B- cos A \: sin \: B = sin(A-B)}$

From eq3 and eq4

add both eq3 and eq4

$x + y = {sin}^{ - 1} 2a \\ x - y = 0 \\ - - - - - - \\ 2x = {sin}^{ - 1} 2a \\ \\ \bold{x = \frac{1}{2} {sin}^{ - 1} 2a }$

Put thus value of x in eq4

$x - y = 0 \\ \\ \frac{1}{2} {sin}^{ - 1} 2a - y = 0 \\ \\ \bold{ y = \frac{1}{2} {sin}^{ - 1} 2a}$

Hope it helps you.