Question

dy
The for the function cos
${}^{2}$
x + cos² y = cos(2x +2y)
da
where y is a function of x, is
OPTIONS
a.
cos 2x + 2 cos(2x + 2y)
- 2 cos(2x + 2y) - cos 2y
b.
sin 2x - 2 sin(2x + 2y)
2 sin(2x + 2y) - sin 2y
C.
cos 2x - 2 cos(2. + 2y)
2 cos(2x + 2y) - cos 2y
d.
sin 2x + 2 sin(2x + 2y)
-2 sin(2x + 2y) - sin 2y​

Answer 1

Answer:

z=x+y

cos

2

x+cos

2

y+cos

2

z−2cosx.cosy.cosz

=cos

2

x+cos

2

y+cos

2

(x+y)−2cosx.cosy.cos(x+y)

=cos

2

x+cos

2

y+(cosxcosy−sinxsiny)

2

−2cosxcosy[cosxcosy−sinxsiny]

=cos

2

x+cos

2

y+cos

2

xcos

2

y+sin

2

xsin

2

y−2cosxcosysinxsiny−2cos

2

xcos

2

y+2cosxcosysinxsiny

=cos

2

x+cos

2

y−cos

2

xcos

2

y+sin

2

xsin

2

y

=cos

2

x+cos

2

y−cos

2

xcos

2

y+sin

2

x(1−cos

2

y)

=cos

2

x+cos

2

y−cos

2

xcos

2

y+sin

2

x−sin

2

xcos

2

y

=cos

2

x+sin

2

x+cos

2

y−cos

2

y(cos

2

x+sin

2

x)

=1