# Difference between revisions of "1975 AHSME Problems/Problem 30"

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==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> |

## Latest revision as of 15:17, 26 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