If 𝐴 = #sqrt(1+cos 320°)/2# and 𝐵 = #sqrt( 1−cos 320°)/ 2# , then 𝐴 + 𝐵 =?

If 𝐴 =
#sqrt(1+cos 320°)/2#

and 𝐵 = #sqrt( 1−cos 320°)/ 2#
, then 𝐴 + 𝐵 =

a) 𝑐𝑜𝑠160° − 𝑠𝑖𝑛160°
b) −𝑐𝑜𝑠160° + 𝑠𝑖𝑛160°
𝑐) 𝑐𝑜𝑠160° + 𝑠𝑖𝑛160°
d) −𝑐𝑜𝑠160° − 𝑠𝑖𝑛160°
e) 0

1 Answer
Apr 4, 2018

The correct answer is

#-\cos(160°)+\sin(160°)#.

I cannot see the choices while answering so please find the corresponding letter.

Explanation:

Where we run into trouble is with signs. To get out of trouble we use the symmetry relations of trigonometric functions to get the argument between 0° and 90°.

We start with

#\cos(360°-x)=\cos(x)#

#\cos(320°)=\cos(40°)#

Then

#\sqrt{{1+\cos(320°)}/2}+\sqrt{{1-\cos(320°)}/2}=\sqrt{{1+\cos(40°)}/2}+\sqrt{{1-\cos(40°)}/2}#

Now apply the half angle formulas:

#sin(x/2)=\pm\sqrt{{1-cos(x)}/2}#

#cos(x/2)=\pm\sqrt{{1+cos(x)}/2}#

By making the argument between 0° and 90°, we assure tgat both signs are positive and thus we beat the sign problems:

#\sqrt{{1+cos(40°)}/2}=\cos(20°)#

#\sqrt{{1-cos(40°)}/2}=\sin(20°)#

So

#\sqrt{{1+\cos(320°)}/2}+\sqrt{{1-\cos(320°)}/2}=\sqrt{{1+\cos(40°)}/2}+\sqrt{{1-\cos(40°)}/2}#

#=\cos(20°)+\sin(20°)#

Now we can get an angle of 160° using these symmetry relations:

#\sin(180°-y)=\sin(y)#

#\cos(180°-y)=-\cos(y)#

Then

#\cos(20°)+\sin(20°)=-\cos(160°)+\sin(160°)#.

I cannot see the multiple choices whilevwriting the answer so please pick the right letter accordingly.