Question

Heya!! ❤️

Prove that :-

cosα + cos( 120° + α ) + cos ( 120° - α ) = 0​

Answer 1

Solution :

L.H.S. = cosα + cos(120° + α) + cos(120° - α)

= cosα + cos120° cosα - sin120° sinα + cos120° cosα + sin120° sinα

= cosα + 2 cos120° cosα

= cosα + 2 cosα cos(90° + 30°)

= cosα + 2 cosα (- sin30°)

= cosα + 2 cosα (- ½)

= cosα - cosα

= 0

= R.H.S.

Hence, proved.

More about Trigonometry :

 It is the study of angles and the relationships between the angles and ratios sine, cosine, tangent, cosectant, sectant, cotangent.

Some formulas are -

 sin²A + cos²A = 1

 sec²A - tan²A = 1

 cosec²A - cot²A = 1

 sin(90° - A) = cosA

 cos(90° - A) = sinA

 sin(90° + A) = cosA

 cos(90° + A) = - sinA

 sinA * cosecA = 1

 cosA * secA = 1

 tanA * cotA = 1

 cos(A + B) = cosA cosB - sinA sinB

 cos(A - B) = cosA cosB + sinA sinB

 sin(A + B) = sinA cosB + cosA sinB

 sin(A - B) = sinA cosB - cosA sinB

Answer 2

QUESTION :

cosα + cos( 120° + α ) + cos ( 120° - α ) = 0

Proof:

Taking L. H. S

=cosα + cos (120-α) + cos (120+α)

=cosα + 2cos [(120-α+120+α) /2] cos [(120-α-120-α) /2]

=cosα + 2cos120 cos(-α)

=cosα+ 2cos(180-60) cosα

=cosα + 2 (-cos 60) cosα

= cosα - 2 × 1/2 cosα

= cosα - cosα

=0 = R. H. S