Question #a6fe2

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Question #a6fe2

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Answer 1 · 2018-02-06T19:31:29

${sin}^{4} (x) + {cos}^{4} (x) = \frac{7}{8}$ $sin^4(x)+ cos^4(x) = 7/8$

Explanation:

We know that:

${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x) = 1$

Square both sides:

${({sin}^{2} (x) + {cos}^{2} (x))}^{2} = 1^{2}$ $(sin^2(x)+cos^2(x))^2 = 1^2$

Expand the squares and mark as equation [1]:

${sin}^{4} (x) + 2 {sin}^{2} (x) {cos}^{2} (x) + {cos}^{4} (x) = 1 [1]$ $sin^4(x)+ 2sin^2(x)cos^2(x)+cos^4(x) = 1" [1]"$

Given $sin (x) cos (x) = \frac{1}{4}$ $sin(x)cos(x) = 1/4$

Square both sides:

${sin}^{2} (x) {cos}^{2} (x) = \frac{1}{16}$ $sin^2(x)cos^2(x) = 1/16$

Multiply both sides by 2:

$2 {sin}^{2} (x) {cos}^{2} (x) = \frac{2}{16}$ $2sin^2(x)cos^2(x) = 2/16$

Simplify and mark as equation [2]:

$2 {sin}^{2} (x) {cos}^{2} (x) = \frac{1}{8} [2]$ $2sin^2(x)cos^2(x) = 1/8" [2]"$

Subtract equation [2] from equation [1]:

${sin}^{4} (x) + 2 {sin}^{2} (x) {cos}^{2} (x) - 2 {sin}^{2} (x) {cos}^{2} (x) + {cos}^{4} (x) = 1 - \frac{1}{8}$ $sin^4(x)+ 2sin^2(x)cos^2(x)-2sin^2(x)cos^2(x)+cos^4(x) = 1-1/8$

Simplify:

${sin}^{4} (x) + {cos}^{4} (x) = \frac{7}{8}$ $sin^4(x)+ cos^4(x) = 7/8$

Answer 2 · 2018-02-06T19:39:18

$\frac{7}{8}$ $7/8$

Explanation:

Squaring both sides and using the trig identity: ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ we can write the expression in terms of ${sin}^{2} x$ $sin^2x$ :

${sin}^{2} x \times {cos}^{2} x = \frac{1}{16}$ $sin^2xtimescos^2x=1/16$
${sin}^{2} x \times (1 - {sin}^{2} x) = \frac{1}{16}$ $sin^2xtimes(1-sin^2x)=1/16$
${sin}^{2} x - {sin}^{4} x - \frac{1}{16} = 0$ $sin^2x-sin^4x-1/16=0$

Let $y = {sin}^{2} x$ $y=sin^2x$

$y^{2} - y + \frac{1}{16} = 0$ $y^2-y+1/16=0$
$y = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \times 1 \times \frac{1}{16}}}{2 \times 1}$ $y=(-(-1)+-sqrt((-1)^2-4times1times1/16))/(2times1)$
$y = \frac{1}{2} \pm \frac{1}{2} \sqrt{\frac{3}{4}}$ $y=1/2+-1/2sqrt(3/4)$
$y = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$ $y=1/2+-sqrt3/4$

Substitute back:

${sin}^{2} x = \frac{1}{2} + \frac{\sqrt{3}}{4} \lor {sin}^{2} x = \frac{1}{2} - \frac{\sqrt{3}}{4}$ $sin^2x=1/2+sqrt3/4vvsin^2x=1/2-sqrt3/4$

For each possible value of ${sin}^{2} x$ $sin^2x$ we should have a value for the expression ${sin}^{4} x + {cos}^{4} x$ $sin^4x+cos^4x$ .

$if sin^2x=1/2+sqrt3/4:$

$cos^2x=1-(1/2+sqrt3/4)=>cos^4x=(1/2-sqrt3/4)^2=7/16-sqrt3/4$
$sin^4x=(1/2+sqrt3/4)^2=7/16+sqrt3/4$

$sin^4x+cos^4x=7/16+sqrt3/4+7/16-sqrt3/4=7/8$

$if sin^2x=1/2-sqrt3/4:$

$cos^2x=1-(1/2-sqrt3/4)=>cos^4x=(1/2+sqrt3/4)^2=7/16+sqrt3/4$
$sin^4x=(1/2-sqrt3/4)^2=7/16-sqrt3/4$

$sin^4x+cos^4x=7/16-sqrt3/4+7/16+sqrt3/4=7/8$

In both cases, we get the same result, so we can conclude:

$sin^4x+cos^4x=7/8$

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