Proof that sin6 > 1/10 and sin10 > 1/6?
920618
-------------- edit ---------------\n\nThe function\n\n     sin(πx/180) - x/60\n\nis 0 when x = 0 and x = 30, and can be shown, using derivatives, to have a maximum between 0 and 30\n\n∴     sin(πx/180) > x/60 if 0 < x < 30\n\n∴     sin(xº) > x/60 if 0 < x < 30\n\nPutting x = 6 and 10 get\n\nsin 6º > 1/10\nsin 10º > 1/6\n\nand similarly, also get\n\nsin 1º > 1/60\nsin 2º > 1/30\nsin 3º > 1/20\nsin 4º > 1/15\nsin 5º > 1/12\nsin 12º > 1/5\nsin 15º > 1/4\nsin 20º > 1/3\n\n---------------- end of edit ------------------------------\n\nAlternatively, prove the special cases as follows.\n\nProof that sin 6º > 1/10\n-----------------------------\n\nSee Source 1 to calculate:\n\n     sin 36º = ¼√( 2 ( 5 - √5 ) ) \n     cos 36º = ¼√5\n\n     sin 30º = ½  and cos 30º = ½√3\n\nUse difference formula to get\n\n     sin 6º = sin( 36º - 30º ) = sin 36º cos 30º - sin 30º cos 36º\n\nSimplify to get (and check with Source 2):\n\n     sin 6º  =  [√6 √( 5 - √5) - (√5 + 1 ) ] / 8\n\nTo prove this > 1/10 rearrange, square, etc to get\n\n     30 - 6√5 > 206 / 25 + 18 / 5 √5\n\n     √5 < 34 / 15\n\n     1125 < 1156 which is true.\n\nA simpler method is similar to the following for 10º. Use the multiple angle formula for sin (5 × 6º) and show that when s = 1/10\n\n     32 s^5 - 40 s³ + 10 s - 1 < 0\n\nProof that sin 10º > 1/6\n-----------------------------\n\nUse multiple angle formula (Source 3) to get\n\n     sin 30º = 3 cos² 10º sin 10º - sin³ 10º\n\nLet s = sin 10º and simplify to\n\n     f(s) = 8s³ - 6s + 1 = 0\n\nUse calculus to show that this function decreases from 0 to ½ then increases. So sin 10º < 1/6 if f(1/6) > 0 which is true.
535427
To do this, you have to do sin^-1 * decimal. Because Yahoo Answers only displays up to .0145, I'll do sin^-1 * .0145, which would be about 0.8308
121013
if it is in binary, 1 + 1 = 10
926069
suppose it is true for some n . I do not know because you say 22 I can compute and check\n\nSo 1st step is right\n\nlet it be true for some n\n\nwe need to prove for n+1\n\n60*5^n < 6^n\nmultiply both sides by 5\n60*5^(n+1) < 6^n . 5 ...1\n\nbut 6^n. 5 < 6^n . 6 as 6^n > 0\nso 6^n. 5 < 6^(n+1) ....2\n\nfrom 1 and 2\n60.5^(n+1) < 6^(n+1)\n\n\nt QED
690230
(160 x .065) + 160\n\nor\n\n160 x 1.065
738333
(1+sin^2x)dx/\n(1+sin^2x)(1-sin^2x)\n=dx/(1-sin^2x)\ndx/cos^2x\nsec^2xdx\ntanx+C
739421
x<1/2(x+y)<y can be proved by splitting it in two parts and proving those:\nx<1/2(x+y) is true because 1/2(x+y) > 1/2(x+x) = x\n1/2(x+y)<y is true because 1/2(x+y) < 1/2(y+y) = y
978758
There's a mathematical riddle that claims to "prove" 1 = 2, but if you examine it closely enough, it turns out to be dividing by 0.  Sorry I csn't remember the riddle.
876876
6 * [6*(5/6 - 1/6)]\n6 * [6*(4/6)]\n6 * 4\nTough.
319082
RHS cot a - tan b \n= cos a/sina - sinb/cosb \n= cosacosb - sinasinb/sinacosb\n= cos(a+b)/sinacosb\n= LHS (proven)\n\nThe idea is to prove that both sides are the same.
333624
(ln x)^7=(7*(lnx)^6)*(1/x)=7*(lnx)^6/x
168359
1/3 + 1/2 = 2/6 + 3/6 = 5/6
1151870
this is a u sub problem\n\nlet u = sin(2x) than du = 2 cos 2x dx and than du/2 = cos(2x) dx\n\nnow subbing it in we get \n\nint(1/2 u^5 du)\n\nwhich is 1/2 x 1/6 u^6 = 1/12 sin^6(2x) + c
822150
If you're certain that you don't mean sin^2((pi/2)-x) + sin^2 x (= cos^2 x + sin^2 x =1), then no.
626033
I wanted to give a partial answer that wasn't presented elsewhere:\n\ne.g. Sine 10 deg =0.174, while cosine 80 deg= 0.174. how ironic is that?!!\n\nIt's not ironic at all.  If you have a right triangle, you know one angle is 90°, and the sum of all angles must be 180°, so the other two angles must add to 90.  You give sin 10°, so the other angle must be 80°.\nThe sin of any angle must be equal to the cos of its complementary angle (x and 90°-x are complementary), because of the definitions.  The side opposite the x angle is adjacent to the the 90°-x angle.\nThus sin 10° = cos 80°.
144204
-10\n\nand to prove it -10 x -10 x -10 = -1000
338816
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.