Difference between revisions of "1975 AHSME Problems/Problem 30"
Thechampion (talk | contribs) (→Solution) |
(→See Also) |
||
(3 intermediate revisions by 2 users not shown) | |||
Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
Using the difference to product identity, we find that | Using the difference to product identity, we find that | ||
− | <math>x=\cos 36^{\circ} - \cos 72^{\circ}</math> is equivalent to <cmath>x=-2\sin{\frac{(36^{\circ}+72^{\circ})}{2}}\sin{\frac{(36^{\circ}-72^{\circ})}{2}} \implies</cmath> | + | <math>x=\cos 36^{\circ} - \cos 72^{\circ}</math> is equivalent to <cmath>x=\text{-}2\sin{\frac{(36^{\circ}+72^{\circ})}{2}}\sin{\frac{(36^{\circ}-72^{\circ})}{2}} \implies</cmath> |
− | <cmath>x=-2\sin54^{\circ}\sin(\text{-}18^{\circ}).</cmath> | + | <cmath>x=\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ}).</cmath> |
− | Since sine is an odd function, we find that <math>\sin{(\text{-}18^{\circ})}= - \sin{18^{\circ}}</math>, and thus <math>-2\sin54^{\circ}\sin(\text{-}18^{\circ})=2\sin54^{\circ}\sin18^{\circ}</math>. Using the property <math>\sin{(90^{\circ}-a)}=\cos{a}</math>, we find | + | Since sine is an odd function, we find that <math>\sin{(\text{-}18^{\circ})}= \text{-} \sin{18^{\circ}}</math>, and thus <math>\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ})=2\sin54^{\circ}\sin18^{\circ}</math>. Using the property <math>\sin{(90^{\circ}-a)}=\cos{a}</math>, we find |
<cmath>x=2\cos(90^{\circ}-54^{\circ})\cos(90^{\circ}-18^{\circ}) \implies</cmath> | <cmath>x=2\cos(90^{\circ}-54^{\circ})\cos(90^{\circ}-18^{\circ}) \implies</cmath> | ||
<cmath>x=2\cos36^{\circ}\cos72^{\circ}.</cmath> | <cmath>x=2\cos36^{\circ}\cos72^{\circ}.</cmath> | ||
Line 19: | Line 19: | ||
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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1975|num-b=28|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Revision as of 22:47, 15 May 2022
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
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.