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Maths Question 11 – JEE-MAIN 2026

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If sin⁡(π18)⋅sin⁡(5π18)⋅sin⁡(7π18)=K, then the value of sin⁡(10Kπ3) is:

The angles π18, 5π18, and 7π18 can be expressed in the form θ, 60∘−θ, and 60∘+θ.

● Video Walkthrough
Step 1: Evaluate K using trigonometric identity✦ Active

The given product is sin⁡(π18)sin⁡(5π18)sin⁡(7π18). Let θ=π18=10∘. Then the angles are 10∘, 50∘, 70∘. These can be written as θ, 60∘−θ, 60∘+θ. We use the identity sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡(3θ).

K=sin⁡(10∘)sin⁡(50∘)sin⁡(70∘)=14sin⁡(3×10∘)=14sin⁡(30∘) K=14×12=18
Step 2: Substitute K into the expression to be found○ Expand

Now we need to find the value of sin⁡(10Kπ3). Substitute K=18 into the expression.

sin⁡(10Kπ3)=sin⁡(10×18×π3)=sin⁡(10π24)=sin⁡(5π12)
Step 3: Calculate the final sine value○ Expand

Convert 5π12 to degrees and evaluate the sine value. 5π12=5×180∘12=5×15∘=75∘. We can use the sum formula for sine.

sin⁡(75∘)=sin⁡(45∘+30∘)=sin⁡(45∘)cos⁡(30∘)+cos⁡(45∘)sin⁡(30∘) =12×32+12×12 =322+122=3+122

This matches option 1.

💡 Teacher's Secret Hint

Remember common trigonometric values for angles like 30∘,45∘,60∘ and how to combine them for angles like 75∘.

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