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...") |
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(e) <math>\frac{26}{25}</math> | (e) <math>\frac{26}{25}</math> | ||
| − | ==Solution == | + | ==Solution== |
| + | 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 16:11, 8 January 2018
Problem 5
Compute 
, knowing that 
.
(a) 
(b) 
(c) 
(d) 
(e) 
Solution
We know that 
, so:
But remember that:
Thus:
Again:
