Evaluate the following expression for sine value?

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Evaluate the following expression for sine value?

${sin}^{2} θ - {sin}^{4} θ = \frac{3}{16}$ $sin^2theta -sin^4 theta = 3/16$

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Answer 1 · 2018-08-08T15:31:04

$\pm \frac{1}{2} \pm \frac{1}{2} \sqrt{3}$ $+- 1/2 +- 1/2 sqrt3$

Explanation:

$\frac{3}{16} = {sin}^{2} θ - {sin}^{4} θ = {sin}^{2} θ (1 - {sin}^{2} θ)$ $3/16 = sin^2theta - sin^4theta = sin^2theta ( 1 - sin^2theta)$

$= {sin}^{2} θ {cos}^{2} θ = \frac{1}{4} {sin}^{2} 2 θ$ $= sin^2theta cos^2theta = 1/4 sin^2 2theta$

So,

${sin}^{2} 2 θ = \frac{3}{4} \Rightarrow sin 2 θ = \pm \frac{1}{2} \sqrt{3} = sin (\frac{+_{π}}{3})$ $sin^2 2theta = 3/4 rArr sin 2theta = +-1/2 sqrt 3 = sin ( +_ pi/3)$

$2 θ = k π + {(- 1)}^{k} (\pm \frac{π}{3}) \Rightarrow θ = \frac{k}{2} π + {(- 1)}^{k} (\pm \frac{π}{6})$ $2theta = kpi + (-1)^k ( +-pi/3) rArr theta = k/2pi +(-1)^k(+-pi/6)$

$\Rightarrow sin θ = sin (k \frac{π}{2} + {(- 1)}^{k} (\pm \frac{π}{6})), k = 0, \pm 1, \pm 2, \pm 3,,,$ $rArr sin theta = sin ( kpi/2 + (-1)^k ( +-pi/6)), k =0, +-1, +-2, +-3,,,$

$= r e p e t i t i v e {sin (- \frac{2}{3} π) sin (\frac{1}{6} π) sin (- \frac{1}{6} π) sin (\frac{2}{3} π}$ $= repetitive { sin ( -2/3pi ) sin ( 1/6pi ) sin ( -1/6pi) sin (2/3pi }$

# = { +-1/2 +-sqrt3/2 }

graph{ y-x^2+x^4+3/16=0[-1 1 -.01 .01]}

x-intercepts are the sine values.

Science

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Evaluate the following expression for sine value?

${sin}^{2} θ - {sin}^{4} θ = \frac{3}{16}$ $sin^2theta -sin^4 theta = 3/16$

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

Evaluate the following expression for sine value?

sin2θ−sin4θ=316sin^2theta -sin^4 theta = 3/16

1 Answer

Explanation:

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Impact of this question

${sin}^{2} θ - {sin}^{4} θ = \frac{3}{16}$ $sin^2theta -sin^4 theta = 3/16$