Calculate the expressions E(x)= 1+ sinx + cosx E(x)= sinx - cosx E(x)= 1 - cosx

For the third expression, we'll transform the
difference of 2 trigonometric function, with the same name, into a
product.

We'll write the expression E(x)= 1 - cosx as
a difference of 2 function with the same name, in this case,
"cosine".

We'll substitute the value 1 = cos 0, so
the 

E(x)=1-cos x <=> E(x)=cos 0-cos
x.

We'll transform, the difference into a
product:

E(x)=2sin[(0+x)/2]*sin[(0-x)/2]=2sin(x/2)*sin(-x/2),

But
sin(-a)= -sin a, so E(x)=
-2sin(x/2)*sin(x/2),

E(x) =
-2sin(x/2)]^2

 

In
order to solve the expression E(x)=sin x-cos x, we'll try to  write the value cos x=sin
[(pi/2)-x]. So,

E(x)=sin x-cos
x

E(x)=sin x-sin
[(pi/2)-x]

E(x)=2cos{[x+(pi/2)-x]/2}sin{[x-(pi/2)+x]/2},

We'll
eliminate like
terms

E(x)=2cos(pi/4)sin{[(2x/2)-(pi/4)]}

E(x)=2*(sqrt2)/2*sin{[(2x/2)-(pi/4)]}

E(x)=(sqrt2)*sin{[(2x/2)-(pi/4)]}

 

Now,
we'll solve the first expression:

E(x)= 1+
sinx + cosx

We'll write the expression  1 +
cosx as a sum of 2 function with the same name, in this case,
"cosine".

We'll substitute the value 1 = cos 0, so
the 

1+cos x <=> cos 0+cos
x.

We'll transform, the difference into a
product:

2cos[(0+x)/2]*cos[(0-x)/2]=2cos(x/2)*cos(-x/2),

But
cos(-a)= cos a, so 
2cos(x/2)*cos(-x/2)=2cos(x/2)*cos(x/2),

2cos(x/2)*cos(-x/2)=2cos(x/2)]^2

So,
the expression will become:

E(x)= 2cos(x/2)]^2+
sinx

We'll write sin x = sin 2(x/2) =
2sin(x/2)*cos(x/2)

E(x)= 2[cos(x/2)]^2+
2sin(x/2)*cos(x/2)

We'll factorize by
2cos(x/2):

E(x)= 2[cos(x/2)]*[cos(x/2) + sin
(x/2)]

We'll write cos (x/2) = sin
[(pi/2)-(x/2)]

E(x)= 2[cos(x/2)]*{sin [(pi/2)-(x/2)] + sin
(x/2)}

E(x)= 2[cos(x/2)]*{2sin [(pi/2 - x/2 +
x/2)/2]*cos[(pi/2 - x/2 - x/2)/2]

E(x)=
4[cos(x/2)]*[sin(pi/4)]*[cos(pi/4 -
x/2)]

E(x)=2sqrt2*cos(x/2)*cos(pi/4 -
x/2)