If A + B + C = 180°, Prove That sin 2A + sin 2B − sin 2C = 4 cos A cos B sin C
Prove that sin2A+sin2B−sin2C=4cosA cosB sinC when A+B+C=180°, using the sum-to-product formula twice along with the supplementary-angle identities.
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The Value of cos(π/2²)·cos(π/2³)⋯cos(π/2¹⁰)·sin(π/2¹⁰) Is
Find the value of the product cos(π/4)cos(π/8)···cos(π/1024)sin(π/1024) using the repeated identity sinθcosθ=½sin2θ to telescope it down.
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The Value of (sin 70°)(cot 10° cot 70° − 1) Is Equal To
A short JEE Main 2025 trigonometry problem solved by converting cotangents to cosines and sines, then applying the cos(A+B) identity.
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If sin²10°sin20°sin40°sin50°sin70°=α−1/16sin10°, Find 16+1/α
A five-term trigonometric product from JEE Main 2022, reduced step by step using complementary angles and the sin a cos a identity.
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Value of 36(4cos²9°-1)(4cos²27°-1)(4cos²81°-1)(4cos²243°-1)
A JEE Main 2023 trigonometric product that telescopes to a single value once each factor is rewritten using the sin 3θ triple-angle identity.