Sigma Percentile
JEE Main 2022 (26 June Shift 2)
LEVELBoard

Animated Solution for Mathematics - Trigonometry: is equal to :

Select Answer:

Visualized Solution

Analyze the Expression

  • Given expression:
  • Objective: Simplify the product of these three sine functions.

The Triple Angle Identity

  • Core Identity:
  • This is a standard trigonometric shortcut for JEE.

Map the Base Angle

  • Let the smallest angle be our base:

The Reference Angle

  • The identity revolves around a reference axis.

Verify

  • Check:
  • This matches our second term perfectly!

Verify

  • Check:
  • This matches our third term perfectly!

Substitute into the Identity

  • Expression:
  • Substitute:

Simplify the Multipliers

  • Multiply the constants:
  • Expression becomes:

Calculate the Final Angle

  • Multiply the angle:
  • Expression becomes:

Value of

  • Recall standard value:
  • Substitute:

Final Result

  • Simplify:
  • The correct option is (2).

The Sigma Insight: Trigonometric Functions of Compound Angles

Solution Diagram

Analyzing the Setup

Welcome, future engineers! Today, we are going to unravel a problem that, at first glance, looks like a nightmare of irrational numbers. We are asked to evaluate .
If you try to punch these into a calculator, you will get a mess of decimals. But in the world of JEE Advanced, we do not use calculators; we use insight and the beauty of mathematical structure.
Look closely at the angles: , , and . In mathematics, whenever you see angles that seem to cluster around a specific value, stop and look for symmetry.
Notice that is and is . We have a base angle of , and the other two angles are perfectly positioned around a reference axis. This is not a coincidence; it is a setup.

The Triple Angle Identity

Our toolkit for this problem is a powerful trigonometric identity:
This identity is a secret weapon. It allows us to compress a product of three sine terms into a single sine term of triple the angle. It is elegant, efficient, and exactly what we need.

The Execution

Imagine you are standing on a circle, marking these angles. You have your angle, with the angle sitting just below the reference line and the angle sitting just above it. The symmetry is perfect.
Now, we substitute into our identity. The expression transforms into:
Look at how the complexity vanishes. We are left with . The constant divided by gives us .
So, we are simply calculating . We know that .
Substituting this in, we get , which simplifies beautifully to . The problem that seemed impossible is now solved with grace.

Similar Questions

JEE Advanced 1988
LEVELBoard

The value of the expression is equal to

(A)
(B)
(C)
(D)
JEE Main 2019 (8 April Shift 1)
LEVELJEE Main

If and , then is equal to :

(A)
(B)
(C)
(D)
JEE Main 2024 (01 Feb Shift 1)
LEVELJEE Main

If , and , , then is equal to :

(A)
(B)
(C)
(D)
JEE Main 2010
LEVELJEE Main

Let and , where . Then

(A)
(B)
(C)
(D)
JEE Advanced 1978
LEVELBoard

If and , find the possible values of .

JEE Main 2006
LEVELJEE Main

If the roots of the quadratic equation are and , respectively, then the value of is

(A)
2
(B)
3
(C)
0
(D)
1
JEE Main 2005
LEVELJEE Main

In a triangle . If and are the roots of then

(A)
(B)
(C)
(D)
JEE Advanced 1983
LEVELJEE Main

If , then has the value

(A)
(B)
(C)
(D)
none of these
JEE Advanced 1993
LEVELJEE Main

Determine the smallest positive value of (in degrees) for which .

JEE Advanced 1979
LEVELJEE Main

(a) Draw the graph of from to . (b) If , and lies between 0 and , find .