This is a standard trigonometric shortcut for JEE.
Map the Base Angle
Let the smallest angle be our base: θ=20∘
The Reference Angle
The identity revolves around a 60∘ reference axis.
Verify 60∘−θ
Check: 60∘−θ=60∘−20∘=40∘
This matches our second term perfectly!
Verify 60∘+θ
Check: 60∘+θ=60∘+20∘=80∘
This matches our third term perfectly!
Substitute into the Identity
Expression: 16[sin(20∘)sin(40∘)sin(80∘)]
Substitute: 16×[41sin(3×20∘)]
Simplify the Multipliers
Multiply the constants: 16×41=4
Expression becomes: 4sin(3×20∘)
Calculate the Final Angle
Multiply the angle: 3×20∘=60∘
Expression becomes: 4sin(60∘)
Value of sin(60∘)
Recall standard value: sin(60∘)=23
Substitute: 4×23
Final Result
Simplify: 4×23=23
The correct option is (2).
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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 16sin(20∘)sin(40∘)sin(80∘).
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: 20∘, 40∘, and 80∘. In mathematics, whenever you see angles that seem to cluster around a specific value, stop and look for symmetry.
Notice that 40∘ is 60∘−20∘ and 80∘ is 60∘+20∘. We have a base angle of θ=20∘, and the other two angles are perfectly positioned around a 60∘ 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:
sin(θ)sin(60∘−θ)sin(60∘+θ)=41sin(3θ)
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 20∘ angle, with the 40∘ angle sitting just below the 60∘ reference line and the 80∘ angle sitting just above it. The symmetry is perfect.
Now, we substitute θ=20∘ into our identity. The expression 16sin(20∘)sin(40∘)sin(80∘) transforms into:
16×[41sin(3×20∘)]
Look at how the complexity vanishes. We are left with 16×41sin(60∘). The constant 16 divided by 4 gives us 4.
So, we are simply calculating 4sin(60∘). We know that sin(60∘)=23.
Substituting this in, we get 4×23, which simplifies beautifully to 23. The problem that seemed impossible is now solved with grace.