Question

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

Answer 1

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.