Difference between revisions of "1975 AHSME Problems/Problem 30"
Quantummech (talk | contribs) (Created page with "==Problem 30== Let <math>x=\cos 36^{\circ} - \cos 72^{\circ}</math>. Then <math>x</math> equals <math>\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf...") |
Thechampion (talk | contribs) (→Solution) |
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==Solution== | ==Solution== | ||
+ | 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> | ||
+ | <cmath>x=-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 | ||
+ | <cmath>x=2\cos(90^{\circ}-54^{\circ})\cos(90^{\circ}-18^{\circ}) \implies</cmath> | ||
+ | <cmath>x=2\cos36^{\circ}\cos72^{\circ}.</cmath> | ||
+ | We multiply the entire expression by <math>\sin36^{\circ}</math> and use the double angle identity of sine twice to find | ||
+ | <cmath>x\sin36^{\circ}=2\sin36^{\circ}\cos36^{\circ}\cos72^{\circ} \implies</cmath> | ||
+ | <cmath>x\sin36^{\circ}=\sin72^{\circ}\cos72^{\circ} \implies</cmath> | ||
+ | <cmath>x\sin36^{\circ}=\frac{1}{2}\sin144^{\circ}.</cmath> | ||
+ | Using the property <math>\sin(180^{\circ}-a)=\sin{a}</math>, we find <math>\sin144^{\circ}=\sin36^{\circ}.</math> Substituting this back into the equation, we have | ||
+ | <cmath>x\sin36^{\circ}=\frac{1}{2}\sin36^{\circ}.</cmath> | ||
+ | Dividing both sides by <math>\sin36^{\circ}</math>, we have | ||
+ | <cmath>x=\boxed{\textbf{(B)}\ \frac{1}{2}}</cmath> |
Revision as of 17:29, 14 July 2016
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