Evaluate ${sin}^{4} 15^{\circ} + {cos}^{4} 15^{\circ}$ $sin^4 15^@ + cos^4 15^@$ ?

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Evaluate ${sin}^{4} 15^{\circ} + {cos}^{4} 15^{\circ}$ $sin^4 15^@ + cos^4 15^@$ ?

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Answer 1 · 2018-01-12T00:21:16

The expression gives a result of $\frac{7}{8}$ $7/8$

Explanation:

We can see that

${({sin}^{2} x + {cos}^{2} x)}^{2} = ({sin}^{2} x + {cos}^{2} x) ({sin}^{2} x + {cos}^{2} x) = {sin}^{4} x + {cos}^{4} x + 2 {cos}^{2} x {sin}^{2} x$ $(sin^2x + cos^2x)^2 = (sin^2x + cos^2x)(sin^2x+ cos^2x) = sin^4x + cos^4x + 2cos^2xsin^2x$

Therefore,

${sin}^{4} x + {cos}^{4} x = {({sin}^{2} x + {cos}^{2} x)}^{2} - 2 {cos}^{2} x {sin}^{2} x$ $sin^4x + cos^4x = (sin^2x + cos^2x)^2 - 2cos^2xsin^2x$

We seek to find:

$(sin^2(15˚) + cos^2(15˚))^2 - 2cos^2(15˚)sin^2(15˚) = sin^4(15˚) + cos^4(15˚)$

We know that $sin^2x+ cos^2x = 1$ , therefore,

$1^2 - 2cos^2(15˚)sin^2(15˚) = sin^4(15˚) + cos^4(15˚)$

Now we notice that $sin^2(2x) = 4sin^2xcos^2x$ , therefore,

$1^2 - 1/2sin^2(2(15˚)) = sin^4(15˚) + cos^4(15˚)$

$1^2 - 1/2sin^2(30˚) = sin^4(15˚) + cos^4(15˚)$

$1 - 1/2(1/2)^2 = sin^4(15˚) + cos^4(15˚)$

$7/8 = sin^4(15˚) + cos^4(15˚)$

And if we check by calculator, we see that we do indeed get $7/8$ .

Hopefully this helps!

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