prove that? sin 3θ sin^3 θ + cos 3θ cos^3 θ = cos^3 (2θ)


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Question Number 83430 by jagoll last updated on 02/Mar/20

$prove that ?$ $\sin 3 θ \sin^{3} θ + \cos 3 θ \cos^{3} θ =$ $\cos^{3} (2 θ)$
	Answered by mind is power last updated on 02/Mar/20

	$s i n (3 θ) {s i n}^{3} (θ) + c o s (3 θ) {c o s}^{3} (θ)$ $= s i n (θ) s i n (3 θ) {s i n}^{2} (θ) + c o s (θ) c o s (3 θ) ({c o s}^{2} (θ))$ $= s i n (θ) s i n (3 θ) (1 - {c o s}^{2} (θ)) + c o s (θ) c o s (3 θ) {c o s}^{2} (θ)$ $= s i n (θ) s i n (3 θ) + {c o s}^{2} (θ) (c o s (θ) c o s (3 θ) - s i n (θ) s i n (3 θ))$ $= \frac{c o s (2 θ) - c o s (4 θ)}{2} + {c o s}^{2} (θ) (c o s (4 θ))$ $= {c o s}^{2} (θ) - {c o s}^{2} (2 θ) + {c o s}^{2} (θ) (2 {c o s}^{2} (2 θ) - 1)$ $= \frac{1 + c o s (2 θ)}{2} - {c o s}^{2} (2 θ) + (\frac{1 + c o s (2 θ)}{2}) (2 {c o s}^{2} (2 θ) - 1)$ $= \frac{1 + c o s (2 θ)}{2} - {c o s}^{2} (2 θ) - \frac{1}{2} - \frac{c o s (2 θ)}{2} + {c o s}^{3} (2 θ) + {c o s}^{2} (2 θ)$ $= {c o s}^{3} (2 θ)$
	Commented by mind is power last updated on 02/Mar/20

	$w i t h e p l e a s u r$
	Commented by jagoll last updated on 02/Mar/20

	$thank you sir$
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