Difference between revisions of "MIE 2016/Day 1/Problem 5"

(Created page with " ===Problem 5=== Compute <math>\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}</math>, knowing that <math>\sin\alpha\cos\alpha=\frac{1}{5}</math>. (a) <math>\fra...")
 
(Added solution)
 
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(e) <math>\frac{26}{25}</math>
 
(e) <math>\frac{26}{25}</math>
  
==Solution ==
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==Solution==
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We know that <math>\sin^2\alpha+\cos^2\alpha=1</math>, so:
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<math>(\sin^2\alpha+\cos^2\alpha)^2=1^2</math>
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<math>\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha=1</math>
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 +
 
 +
But remember that:
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<math>\sin \alpha \cos \alpha =\frac{1}{5}</math>
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<math>\left(\sin\alpha\cos\alpha\right)^2=\left(\frac{1}{5}\right)^2</math>
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<math>2\sin^2\alpha\cos^2\alpha=\frac{2}{25}</math>
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Thus:
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<math>\sin^4\alpha+\frac{2}{25}+\cos^4\alpha=1</math>
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<math>\boxed{\sin^4\alpha+\cos^4\alpha=\frac{23}{25}}</math>
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Again:
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<math>(\sin^2\alpha+\cos^2\alpha)^3=1^3</math>
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<math>\sin^6\alpha+3\sin^4\alpha\cos^2\alpha+3\sin^2\alpha\cos^4\alpha+\cos^6\alpha=1</math>
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<math>\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha\cos^2\alpha(\sin^2\alpha+\cos^2\alpha)=1</math>
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<math>\sin^6\alpha+\cos^6\alpha+3\cdot\frac{1}{25}\cdot1=1</math>
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<math>\boxed{\sin^6\alpha+\cos^6\alpha=\frac{22}{25}}</math>
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<math>\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}\implies\boxed{\frac{23}{22}}\to\boxed{B}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 15:11, 8 January 2018

Problem 5

Compute $\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}$, knowing that $\sin\alpha\cos\alpha=\frac{1}{5}$.


(a) $\frac{22}{21}$

(b) $\frac{23}{22}$

(c) $\frac{25}{23}$

(d) $\frac{13}{12}$

(e) $\frac{26}{25}$

Solution

We know that $\sin^2\alpha+\cos^2\alpha=1$, so:


$(\sin^2\alpha+\cos^2\alpha)^2=1^2$

$\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha=1$


But remember that:


$\sin \alpha \cos \alpha =\frac{1}{5}$

$\left(\sin\alpha\cos\alpha\right)^2=\left(\frac{1}{5}\right)^2$

$2\sin^2\alpha\cos^2\alpha=\frac{2}{25}$


Thus:


$\sin^4\alpha+\frac{2}{25}+\cos^4\alpha=1$

$\boxed{\sin^4\alpha+\cos^4\alpha=\frac{23}{25}}$


Again:


$(\sin^2\alpha+\cos^2\alpha)^3=1^3$

$\sin^6\alpha+3\sin^4\alpha\cos^2\alpha+3\sin^2\alpha\cos^4\alpha+\cos^6\alpha=1$

$\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha\cos^2\alpha(\sin^2\alpha+\cos^2\alpha)=1$

$\sin^6\alpha+\cos^6\alpha+3\cdot\frac{1}{25}\cdot1=1$

$\boxed{\sin^6\alpha+\cos^6\alpha=\frac{22}{25}}$


$\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}\implies\boxed{\frac{23}{22}}\to\boxed{B}$

See Also

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