Difference between revisions of "1975 AHSME Problems/Problem 30"
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Dividing both sides by <math>\sin36^{\circ}</math>, we have | Dividing both sides by <math>\sin36^{\circ}</math>, we have | ||
<cmath>x=\boxed{\textbf{(B)}\ \frac{1}{2}}</cmath> | <cmath>x=\boxed{\textbf{(B)}\ \frac{1}{2}}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We observe that <math>72^{\circ}</math> is twice of <math>36^{\circ}</math>, so evaluating <math>\cos 36^{\circ}</math> and using the double angle identify is enough to find <math>\cos 36^{\circ} - \cos 72^{\circ}</math> | ||
+ | |||
+ | We first let <math>n = \cos 36^{\circ}</math>. | ||
+ | Notice that <math>36^{\circ}/4 = 9^{\circ}</math> and <math>36^{\circ} = 45^{\circ} - 9^{\circ}</math>. | ||
+ | Rewriting <math>\cos 36^{\circ}</math> as <math>\cos (45^{\circ} - 9^{\circ})</math>, we then use the difference of angles identify to find | ||
+ | <cmath>n = \cos 36^{\circ} = \cos (45^{\circ} - 9^{\circ}) = \cos 45^{\circ}\cos 9^{\circ} - \sin 45^{\circ}\sin 9^{\circ} = \frac{\sqrt2}{2}(\cos 9^{\circ} - \sin 9^{\circ})</cmath> | ||
+ | Rearranged to get | ||
+ | <cmath>\sqrt2 n = \cos 9^{\circ} - \sin 9^{\circ}</cmath> | ||
+ | Using the half-angle identify twice, and noticing that <math>\cos 9^{\circ}</math> is positive, we get | ||
+ | <cmath> \cos 9^{\circ} = \cos \frac{18^{\circ}}{2} = \sqrt \frac{1 + \cos 18^{\circ}}{2} = \sqrt \frac{1 + \sqrt \frac{1 + \cos 36^{\circ}}{2}}{2}</cmath> | ||
+ | Also, | ||
+ | <cmath>\sin 9^{\circ} = \sqrt {1 - \cos^2 9} = \sqrt {1 - \frac{1 + \sqrt{\frac{1 + 36^{\circ}}{2}}}{2}} = \sqrt {\frac{1 - \sqrt{\frac{1 + 36^{\circ}}{2}}}{2}}</cmath> | ||
+ | |||
+ | Thus we can get | ||
+ | <cmath> \sqrt2 n = \sqrt \frac{1 + \sqrt \frac{1 + \cos 36^{\circ}}{2}}{2} - \sqrt {\frac{1 - \sqrt{\frac{1 + 36^{\circ}}{2}}}{2}}</cmath> | ||
+ | Squaring both sides and simplifying, this results in | ||
+ | <cmath>2n^2 = 1 - \sqrt{\frac{1-n}{2}}</cmath> | ||
+ | From which we get | ||
+ | <cmath> 2n^2 - 1 = - \sqrt{\frac{1-n}{2}}</cmath> | ||
+ | Squaring both sides again, | ||
+ | <cmath> 4n^4 - 4n^2 +1 = \frac{1-n}{2}</cmath> | ||
+ | Multiplying both sides by <math>2</math> and rearranging, | ||
+ | <cmath>8n^4 - 8n^2 + n + 1 = 0</cmath> | ||
+ | Factoring out <math>8n^2</math> from the first two terms and applying difference of squares, | ||
+ | <cmath>8n^2(n+1)(n-1) + n+1 = 0</cmath> | ||
+ | We know that <math>n = \cos 36^{\circ}</math> which is not equal to <math>-1</math>, so we divide both sides by <math>n+1</math> | ||
+ | <cmath> 8n^3 -8n^2 +1 = 0</cmath> | ||
+ | Using the rational root test, we find that <math>n = \frac{1}{2}</math> is a solution of this equation. However, we also know that <math>\cos 36^{\circ}</math> is not equal to <math>\frac{1}{2}</math>, so we divide both sides by <math>n- \frac{1}{2}</math> to get | ||
+ | <cmath> 8n^2 - 4n - 2 = 0</cmath> | ||
+ | Using the quadratic formula, we find that | ||
+ | <cmath>n = \frac{1 \pm \sqrt{5}}{4}</cmath> | ||
+ | We know <math>n</math> must be positive, which gives us | ||
+ | <cmath>n = \frac{1 + \sqrt{5}}{4}</cmath> | ||
+ | |||
+ | Thus, using the double angle identify, we can find | ||
+ | <cmath>x = \cos 36^{\circ} - \cos 72^{\circ} = \cos 36^{\circ} - 2\cos^2 36^{\circ} + 1 = n - 2n^2 + 1 = \boxed{\textbf{(B)} \frac{1}{2}} </cmath> | ||
+ | |||
+ | -SaraLiu24 | ||
==See Also== | ==See Also== | ||
− | {{AHSME box|year=1975|num-b=28|Last Problem}} | + | {{AHSME box|year=1975|num-b=28|after=Last Problem}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:11, 15 July 2024
Contents
Problem 30
Let . Then
equals
Solution
Using the difference to product identity, we find that
is equivalent to
Since sine is an odd function, we find that
, and thus
. Using the property
, we find
We multiply the entire expression by
and use the double angle identity of sine twice to find
Using the property
, we find
Substituting this back into the equation, we have
Dividing both sides by
, we have
Solution 2
We observe that is twice of
, so evaluating
and using the double angle identify is enough to find
We first let .
Notice that
and
.
Rewriting
as
, we then use the difference of angles identify to find
Rearranged to get
Using the half-angle identify twice, and noticing that
is positive, we get
Also,
Thus we can get
Squaring both sides and simplifying, this results in
From which we get
Squaring both sides again,
Multiplying both sides by
and rearranging,
Factoring out
from the first two terms and applying difference of squares,
We know that
which is not equal to
, so we divide both sides by
Using the rational root test, we find that
is a solution of this equation. However, we also know that
is not equal to
, so we divide both sides by
to get
Using the quadratic formula, we find that
We know
must be positive, which gives us
Thus, using the double angle identify, we can find
-SaraLiu24
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.