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demonsking52801:
can u check your question is it correctly written?
then on rhs it will become sin 3A
but in lhs its given sin³A
how is it
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Dekho agar samjha ma nahi aaya to puche lo
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please yaar
bhaiya ji
number sharing here is not a good idea
why bro
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Proof: sin 3A
= sin (2A + A)
= sin 2A cos A + cos 2A sin A
= 2 sin A cos A ∙ cos A + (1 - 2 sin^2 A) sin A
= 2 sin A (1 - sin^2 A) + sin A - 2 sin^3 A
= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A
= 3 sin A - 4 sin^3 A
Therefore, sin 3A = 3 sin A - 4 sin^3 A
= sin (2A + A)
= sin 2A cos A + cos 2A sin A
= 2 sin A cos A ∙ cos A + (1 - 2 sin^2 A) sin A
= 2 sin A (1 - sin^2 A) + sin A - 2 sin^3 A
= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A
= 3 sin A - 4 sin^3 A
Therefore, sin 3A = 3 sin A - 4 sin^3 A
ha samjha gay
thanks for explanation
OK
aur kuch boliye
sleep bye gn subah school hai
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