Question

cos 24° + cos 55° + cos 125° + cos 204° =
1)-1 2) 0
3) 1 4) 2​

Answer 1

Answer:

0 (Option 2)

Step-by-step explanation:

As per the information provided in the question, We have :

• cos 24° + cos 55° + cos 125° + cos 204°.

We are asked to add these. In order to do that, We will simplify the given equation using the identity mentioned below.

${\longmapsto \bf \cos(A) + \cos(B) = 2 \cos\bigg(\dfrac{A + B}{2}\bigg) \cos\bigg(\dfrac{A - B}{2}\bigg)}$

Applying this to cos 55° + cos 125°,

${\longmapsto \rm \cos(24 ^ \circ) + 2 \cos\bigg(\dfrac{55^ \circ + 125^ \circ}{2}\bigg) \cos\bigg(\dfrac{55^ \circ - 125^ \circ}{2}\bigg)+ \cos(204^ \circ)}$

On simplifying,

${\longmapsto \rm \cos(24 ^ \circ) + 2 \cos\bigg(\dfrac{180^ \circ}{2}\bigg) \cos\bigg(\dfrac{ - 70^ \circ}{2}\bigg)+ \cos(204^ \circ)}$

${\longmapsto \rm \cos(24 ^ \circ) + 2 \cos(90 ^ \circ)\cos ( - 35)+ \cos(204^ \circ)}$

Cos(-b) = cos(b). Thus,

${\longmapsto \rm \cos(24 ^ \circ) + 2 \cos(90 ^ \circ)\cos ( 35)+ \cos(204^ \circ)}$

The value of Cos(90°) is 0, Thus,

${\longmapsto \rm \cos(24 ^ \circ) + 2 \times 0 \times \cos ( 35)+ \cos(204^ \circ)}$

${\longmapsto \rm \cos(24 ^ \circ) + 0 \times \cos ( 35)+ \cos(204^ \circ)}$

${\longmapsto \rm \cos(24 ^ \circ) + 0 + \cos(204^ \circ)}$

${\longmapsto \rm \cos(24 ^ \circ) + \cos(204^ \circ)}$

Applying the same identity again,

${\longmapsto \bf \cos(A) + \cos(B) = 2 \cos\bigg(\dfrac{A + B}{2}\bigg) \cos\bigg(\dfrac{A - B}{2}\bigg)}$

${\longmapsto \rm 2 \cos\bigg(\dfrac{24^ \circ + 204^ \circ}{2}\bigg) \cos\bigg(\dfrac{24^ \circ - 204^ \circ}{2}\bigg)}$

On simplifying,

${\longmapsto \rm 2 \cos\bigg(\dfrac{228}{2}\bigg) \cos\bigg(\dfrac{ - 180}{2}\bigg)}$

${\longmapsto \rm 2 \cos(114 ^{ \circ} ) \cos(90^{ \circ})}$

${\longmapsto \rm 2 \cos(114 ^{ \circ} ) \times 0}$

${\longmapsto \rm 2 \times 0}$

${\longmapsto \rm 0}$

∴ cos 24° + cos 55° + cos 125° + cos 204° is 0.

Answer 2

Question :

cos 24° + cos 55° + cos 125° + cos 204°

Solution :

We have to arrange them first

→ cos 24° + cos 204° + cos 55° + cos 125°

→ cos 24° + cos(90° × 2 + 24°) + cos (90° - 35°) + cos(90° + 35°)

→ cos 24° - cos 24° + sin 35° - sin 35°

→ 0

So the correct answer is : option 2 i.e. 0