i. Prove that sin 3A=3sin A−4sin^3 A. ii.Show that θ=54° satisfies the equation sin 3θ=−cos 2θ. Hence or otherwise, find the value of sin 54°.


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Question Number 122007 by ZiYangLee last updated on 13/Nov/20

$i . Prove that \sin 3 A = 3 \sin A - {4 \sin}^{3} A .$ $ii . Show that θ = 54 ° satisfies the equation$ $\sin 3 θ = - \cos 2 θ . Hence or otherwise,$ $find the value of \sin 54 ° .$
	Answered by TANMAY PANACEA last updated on 13/Nov/20

	$s i n (3 A) = s i n 2 A c o s A + c o s 2 A s i n A$ $= 2 s i n A (1 - {s i n}^{2} A) + (1 - 2 {s i n}^{2} A) s i n A$ $= 2 s - 2 s^{3} + s - 2 s^{3} = 3 s - 4 s^{3} [s = s i n A]$
	Answered by TANMAY PANACEA last updated on 13/Nov/20

	$L H S s i n (3 \times 54) = s i n (162) = s i n (180 - 18) = s i n 18$ $R H S - c o s (108) = - c o s (90 + 18) = - (- s i n 18) = s i n 18$ $s o s i n 3 θ = - c o s 2 θ p r o v e d$ $s i n 54 = 3 s i n 18 - 4 {s i n}^{3} 18$ $s i n 54 = 3 (\frac{\sqrt{5} - 1}{4}) - 4 {(\frac{\sqrt{5} - 1}{4})}^{3}$
	Answered by $@y@m last updated on 13/Nov/20
	$(i) sin (x+y)=sin xcos y+cos xsin y Put x=2A, y=A sin 3A=sin 2Acos A+cos 2Asin A =(2sinAcos A)cos A+(1−2sin^2 A)sin A =2sinAcos^2 A+sin A−2sin^3 A =2sinA(1−sin^2 A)+sin A−2sin^3 A =2sinA−2sin^3 A+sin A−2sin^3 A =3sinA−4sin^3 A ■■■ (ii) sin (3×54)=sin 162=sin (180−18)=sin 18 −cos (2×54)=−cos 108=−cos (90+18)=sin 18 ■■■ sin 54=sin (3×18) =3sin 18−4sin^3 18 =sin 18(3−4sin^2 18) = (((√5)−1)/4){3−4×((5+1−2(√5))/(16))} = (((√5)−1)/4){3−((6−2(√5))/4)} = (((√5)−1)/4)×((6+2(√5))/4) = ((6(√5)−6+10−2(√5))/(16)) = ((4+4(√5))/(16)) =(((√5)+1)/4)$
	$(i) \sin (x + y) = \sin x \cos y + \cos x \sin y$ $P u t x = 2 A, y = A$ $\sin 3 A = \sin 2 A \cos A + \cos 2 A \sin A$ $= (2 \sin A \cos A) \cos A + (1 - 2 \sin^{2} A) \sin A$ $= 2 \sin A \cos^{2} A + \sin A - {2 \sin}^{3} A$ $= 2 \sin A (1 - \sin^{2} A) + \sin A - {2 \sin}^{3} A$ $= 2 \sin A - {2 \sin}^{3} A + \sin A - {2 \sin}^{3} A$ $= 3 \sin A - {4 \sin}^{3} A$ $◼ ◼ ◼$ $(i i) \sin (3 \times 54) = \sin 162 = \sin (180 - 18) = \sin 18$ $- \cos (2 \times 54) = - \cos 108 = - \cos (90 + 18) = \sin 18$ $◼ ◼ ◼$ $\sin 54 = \sin (3 \times 18)$ $= 3 \sin 18 - 4 \sin^{3} 18$ $= \sin 18 (3 - {4 \sin}^{2} 18)$ $= \frac{\sqrt{5} - 1}{4} {3 - 4 \times \frac{5 + 1 - 2 \sqrt{5}}{16}}$ $= \frac{\sqrt{5} - 1}{4} {3 - \frac{6 - 2 \sqrt{5}}{4}}$ $= \frac{\sqrt{5} - 1}{4} \times \frac{6 + 2 \sqrt{5}}{4}$ $= \frac{6 \sqrt{5} - 6 + 10 - 2 \sqrt{5}}{16}$ $= \frac{4 + 4 \sqrt{5}}{16}$ $= \frac{\sqrt{5} + 1}{4}$
	Commented by ZiYangLee last updated on 16/Nov/20

	$thanks!$
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