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| Math Central | Quandaries & Queries |
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Question from citra, a student: sin54 cos36/cos18 - 2cos36 -2sin18 = ... |
Hi Citra.
Did you notice that all of those numbers are multiples of 18?
So if I can show you how to find sin 18°, you should be able to use that in conjunction with double angle identities and sum identities to find the whole thing.
[ credit for the following: http://www.andrews.edu/~calkins/math/webtexts/NUMB18.HTM#SIN18 ]
| sin 72° = 2 sin 36° cos 36° | by the double angle relationship. |
| sin 72° = 4 sin 18° cos 18° (1 - 2sin2 18°) | by the double angle relationship, again. |
| cos 18° = 4 sin 18° cos 18° (1 - 2sin2 18°) | by the cofunction properties: sin 72° = cos 18°. |
| 1 = 4 sin 18° (1 - 2sin2 18°) | Let x = sin 18°, this is known as |
| 1 = 4x(1-2x2) | substitution, a useful technique in calculus. |
| 8x3-4x+1 = 0 | A product is zero only when one of its factors is zero. |
| 8x3-4x+1 = (2x-1)(4x2+2x-1)=0 | (2x-1)=0 implies x= ½=sin 30° > sin 18° ; |
| Since we know sin is increasing on [0°,90°]. | |
| x = (-2 ± √(4 + 4•4•1))/8 | So we must solve the other factor, |
| = (-2 ± √20)/8 | using the quadratic formula. |
| = (-2 ± √4 √5)/8 | |
| = (-1 ± √5)/4 But the sin 18° > 0, | so it cannot be negative. |
| sin 18° = (√ 5 - 1) / 4 | Hence the middle root is the one we want. |
If you consider the unit circle, this means sin 18° is shown on a triangle of radius 1, with the vertical side going up to (√ 5 - 1) / 4. Pythagoras can tell you what the horizontal side is, which is cos 18°.
Now try using double angle formulas and addition (3*18 = 2*18 + 18) formulas to expand:
sin(3*18°) cos(2*18°) / cos(18°) - 2 cos(2*18°) - 2 sin(18°).
Then you'll have the exact value of the expression you sent to us.
Cheers,
Stephen La Rocque.
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