sin^(10) x + cos^(10) x = ((61)/(256))


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Question Number 140382 by john_santu last updated on 07/May/21

$\sin^{10} x + \cos^{10} x = \frac{61}{256}$
	Answered by MJS_new last updated on 07/May/21
	$sin x =s∧cos x =(√(1−s^2 )) 5s^8 −10s^6 +20s^4 −5s^2 +1=((61)/(256)) s^8 −2s^6 +2s^4 −s^2 +((39)/(256))=0 ⇒ sin x =±(1/2) ∨ sin x =±((√3)/2) within 0≤x<2π we get x∈{(π/6), (π/3), ((2π)/3), ((5π)/6), ((7π)/6), ((4π)/3), ((5π)/3), ((11π)/6)}$
	$\sin x = s \land \cos x = \sqrt{1 - s^{2}}$ $5 s^{8} - 10 s^{6} + 20 s^{4} - 5 s^{2} + 1 = \frac{61}{256}$ $s^{8} - 2 s^{6} + 2 s^{4} - s^{2} + \frac{39}{256} = 0$ $\Rightarrow$ $\sin x = \pm \frac{1}{2} \lor \sin x = \pm \frac{\sqrt{3}}{2}$ $within 0 ⩽ x < 2 π we get$ $x \in {\frac{π}{6}, \frac{π}{3}, \frac{2 π}{3}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{4 π}{3}, \frac{5 π}{3}, \frac{11 π}{6}}$
	Commented by benjo_mathlover last updated on 07/May/21

	$waw . . . master of trigonometry$
	Commented by greg_ed last updated on 07/May/21

	$ok!$
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