Question

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

Answer 1

Answer:

1/2

Step-by-step explanation:

i hope it will help you

Answer 2

$\large\underline{\sf{Solution-}}$

Given Trigonometric function is

$\rm :\longmapsto\:cos24\degree + cos55\degree + cos125\degree + cos204$

can be rewritten as

$\rm \: = \: \: (cos24\degree + cos204\degree ) + (cos55\degree + cos125\degree )$

Consider,

$\red{\rm :\longmapsto\:cos204\degree }$

$\rm \: = \: \: cos(90\degree \times 2 + 24\degree )$

Now, it implies cos204° lies in third quadrant,

So, cos is negative and as its even multiple of 90°, so there is no change in Trigonometric ratio.

Thus,

$\rm \: = \: \: - \: cos24\degree$

$\red{\rm :\longmapsto\:cos204\degree = - \: cos24\degree }$

Again, Consider

$\red{\rm :\longmapsto\:cos125\degree }$

$\rm \: = \: \: cos(180\degree - 55\degree )$

$\rm \: = \: \: cos(90\degree \times 2 - 55\degree )$

It implies, cos155° lies in second quadrant and we know in second quadrant cos is negative and as its have even multiple of 90°, so there is no change in Trigonometric ratio.

Thus,

$\rm \: = \: \: - \: cos55\degree$

$\red{\rm :\longmapsto\:cos125\degree = - \: cos55\degree }$

Hence,

$\rm \: = \: \: (cos24\degree + cos204\degree ) + (cos55\degree + cos125\degree )$

can be rewritten as

$\rm \: = \: \: (cos24\degree - cos24\degree ) + (cos55\degree - cos55\degree )$

$\rm \: = \: \: 0 + 0$

$\rm \: = \: \: 0$

Hence,

$\blue{\bf :\longmapsto\:cos24\degree + cos55\degree + cos125\degree + cos204 = 0}$

Additional Information :-

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ