
What is the value of $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$?
A . $\cos A$
B . 0
C . $\sqrt{3}\sin A$
D . $\sqrt{3}\cos A$
Answer
233.1k+ views
Hint: In this question, we are given an identity $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$. First, we use the formula of $\cos (A+B)+\cos (A-B)$ to solve the question and then we use the complimentary angles of trigonometric ratios and simplify them to get the desirable answer.
Formula Used:
In this question, we use the identity:-
$\cos (A+B)+\cos (A-B)=2\cos A\cos B$
And the complimentary angle
$\cos ({{270}^{\circ }}-A)=-\sin A$ to solve the question.
Complete step- by- step Solution:
We have given an identity $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$………………………… (1)
Now we know the identity
$\cos (A+B)+\cos (A-B)=2\cos A\cos B$
$\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)=2\cos {{240}^{\circ }}\cos A$
Now by putting the value in equation (1), we get
$\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$ = $\cos A+2\cos {{240}^{\circ }}\cos A$
$\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1+2\cos ({{270}^{\circ }}-{{30}^{\circ }}))$
We know the complimentary angles of trigonometric ratios the value of
$\cos ({{270}^{\circ }}-A)=-\sin A$
So $\cos ({{270}^{\circ }}-{{30}^{\circ }})=-\sin {{30}^{\circ }}$
Then $\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1+2(-\sin {{30}^{\circ }}))$
And value of $\sin {{30}^{\circ }}=\dfrac{1}{2}$
Then the value of $\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1-2\times \dfrac{1}{2}))$
Which is equal to $\cos A(1-1)$ = 0
Hence, $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$ = 0
Thus, Option ( B ) is the correct answer.
Note: In trigonometry, different type of questions can be solved with the help of different trigonometry formulas. These questions may include trigonometric ratios, Pythagoras theorem, product identities etc. There are many trigonometry formulas. We choose the formula according to our question demands. Learning and remembering these formulas help the student to solve the questions accurately and able to choose the correct option.
Formula Used:
In this question, we use the identity:-
$\cos (A+B)+\cos (A-B)=2\cos A\cos B$
And the complimentary angle
$\cos ({{270}^{\circ }}-A)=-\sin A$ to solve the question.
Complete step- by- step Solution:
We have given an identity $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$………………………… (1)
Now we know the identity
$\cos (A+B)+\cos (A-B)=2\cos A\cos B$
$\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)=2\cos {{240}^{\circ }}\cos A$
Now by putting the value in equation (1), we get
$\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$ = $\cos A+2\cos {{240}^{\circ }}\cos A$
$\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1+2\cos ({{270}^{\circ }}-{{30}^{\circ }}))$
We know the complimentary angles of trigonometric ratios the value of
$\cos ({{270}^{\circ }}-A)=-\sin A$
So $\cos ({{270}^{\circ }}-{{30}^{\circ }})=-\sin {{30}^{\circ }}$
Then $\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1+2(-\sin {{30}^{\circ }}))$
And value of $\sin {{30}^{\circ }}=\dfrac{1}{2}$
Then the value of $\cos A+2\cos {{240}^{\circ }}\cos A$ = $\cos A(1-2\times \dfrac{1}{2}))$
Which is equal to $\cos A(1-1)$ = 0
Hence, $\cos A+\cos ({{240}^{\circ }}+A)+\cos ({{240}^{\circ }}-A)$ = 0
Thus, Option ( B ) is the correct answer.
Note: In trigonometry, different type of questions can be solved with the help of different trigonometry formulas. These questions may include trigonometric ratios, Pythagoras theorem, product identities etc. There are many trigonometry formulas. We choose the formula according to our question demands. Learning and remembering these formulas help the student to solve the questions accurately and able to choose the correct option.
Recently Updated Pages
JEE Main 2026 Session 2 Registration Open, Exam Dates, Syllabus & Eligibility

JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

Trending doubts
JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding Average and RMS Value in Electrical Circuits

Understanding Collisions: Types and Examples for Students

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding Atomic Structure for Beginners

JEE Main Syllabus 2026: Download Detailed Subject-wise PDF

Other Pages
JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions For Class 11 Maths Chapter 6 Permutations and Combinations (2025-26)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines (2025-26)

Statistics Class 11 Maths Chapter 13 CBSE Notes - 2025-26

Inductive Effect and Its Role in Acidic Strength

Degree of Dissociation: Meaning, Formula, Calculation & Uses

