MegpoidBeetle Posted July 3, 2015 Share Posted July 3, 2015 (edited) So am just doing some revisions for my upcoming finals and am stuck on this question Prove the identity (cosec A - sin A)(sec A - cos A) = 1/(tan A+cot A) I got up to 1 - cos^2 A - sin^2 A + sin^2 A cos^2 A and after that imma stuck Somebody help pls, trying not to shoot myself Edited July 3, 2015 by YOUR_MOTHER_HAVE_BANDAIDS Link to comment Share on other sites More sharing options...
Dhrekr Posted July 3, 2015 Share Posted July 3, 2015 First of all, it's "1/(tan A + cot A)". Written as you have it is misleading. Ok. As you know, cosec = 1/sin, sec = 1/cos, tan = sin/cos, cot = cos/sin. Also, sin^2+cos^2 = 1. I'll break it down in smaller units: (cosec A - sin A) = 1/sinA - sinA = (1-sin^2 A)/sin A = cos^2 A / sin A (sec A - cos A) = (1-cos^2 A)/cos A = sin^2 A / cos A So: (cosec A - sin A) * (sec A - cos A) = (cos^2 A / sin A) * (sin^2 A / cos A) = cos A * sin A On the right hand side: 1/(tan A + cot A) = 1/(sinA/cosA + cosA/sinA) = 1/((sin^2 A+cos^2 A)/sinA*cosA) = 1/(1/sinA*cosA) = sin A * cos A The identity is thus proven. Link to comment Share on other sites More sharing options...
Xhos Posted July 3, 2015 Share Posted July 3, 2015 wat Link to comment Share on other sites More sharing options...
Kracken Posted July 3, 2015 Share Posted July 3, 2015 wat don't even bother understanding it ... I'm glad that i'm far past that part of my math history ;p Link to comment Share on other sites More sharing options...
Xhos Posted July 3, 2015 Share Posted July 3, 2015 don't even bother understanding it ... I'm glad that i'm far past that part of my math history ;p lel I haven't even learnt this yet (I'm 14) Link to comment Share on other sites More sharing options...
MegpoidBeetle Posted July 3, 2015 Author Share Posted July 3, 2015 First of all, it's "1/(tan A + cot A)". Written as you have it is misleading. Ok. As you know, cosec = 1/sin, sec = 1/cos, tan = sin/cos, cot = cos/sin. Also, sin^2+cos^2 = 1. I'll break it down in smaller units: (cosec A - sin A) = 1/sinA - sinA = (1-sin^2 A)/sin A = cos^2 A / sin A (sec A - cos A) = (1-cos^2 A)/cos A = sin^2 A / cos A So: (cosec A - sin A) * (sec A - cos A) = (cos^2 A / sin A) * (sin^2 A / cos A) = cos A * sin A On the right hand side: 1/(tan A + cot A) = 1/(sinA/cosA + cosA/sinA) = 1/((sin^2 A+cos^2 A)/sinA*cosA) = 1/(1/sinA*cosA) = sin A * cos A The identity is thus proven. Thanks, got it at last! :D Will mail you a potato once finals are over :3 lel I haven't even learnt this yet (I'm 14) IT'S NEVER TOO YOUNG TO LEARN Link to comment Share on other sites More sharing options...
Kracken Posted July 3, 2015 Share Posted July 3, 2015 IT'S NEVER TOO YOUNG TO LEARN I believe that is university level of math ;p he still have time xD Link to comment Share on other sites More sharing options...
MegpoidBeetle Posted July 3, 2015 Author Share Posted July 3, 2015 I believe that is university level of math ;p he still have time xD Not at all, I learned it when I was 16 and this is just a question from my first semester of foundations :B Link to comment Share on other sites More sharing options...
JefPlays Posted July 3, 2015 Share Posted July 3, 2015 (edited) I believe that is university level of math ;p he still have time xD University? Hardly. If y'all want some actual university math to chew on, I'm sure I can dig something up... Edited July 3, 2015 by Jeffrey94 Link to comment Share on other sites More sharing options...
Kracken Posted July 3, 2015 Share Posted July 3, 2015 If y'all want some actual university math to chew on, I'm sure I can dig something up... what about simplified laplace transformation of a healthy human heartbeat ? ;p (that was the last question of my math exam) Link to comment Share on other sites More sharing options...
JefPlays Posted July 3, 2015 Share Posted July 3, 2015 what about simplified laplace transformation of a healthy human heartbeat ? ;p (that was the last question of my math exam) Not being a math major myself, I actually don't know that myself. What I've been doing lately is combinatorics and graph theory, which often is less reliant on algebra but tends to be harder for it. Link to comment Share on other sites More sharing options...
MegpoidBeetle Posted July 3, 2015 Author Share Posted July 3, 2015 Not being a math major myself, I actually don't know that myself. What I've been doing lately is combinatorics and graph theory, which often is less reliant on algebra but tends to be harder for it. Gimme couple of years and imma be throwing engineering math at y'all Link to comment Share on other sites More sharing options...
Ghost333 Posted July 3, 2015 Share Posted July 3, 2015 It's like trying to read the Black Speech of Cthulhu in programming language, except someone decided to write it in katakana, and they had a really shaky hand while doing it. Won't drive you insane, but it'll confuse the hell out of anyone who doesn't know what you are saying (like me.) Link to comment Share on other sites More sharing options...
MegpoidBeetle Posted July 3, 2015 Author Share Posted July 3, 2015 It's like trying to read the Black Speech of Cthulhu in programming language, except someone decided to write it in katakana, and they had a really shaky hand while doing it. Won't drive you insane, but it'll confuse the hell out of anyone who doesn't know what you are saying (like me.) Bob has 69 watermelons Link to comment Share on other sites More sharing options...
JefPlays Posted July 3, 2015 Share Posted July 3, 2015 (edited) Bob has 69 watermelons ( ͡° ͜ʖ ͡°) On topic though, here's a fun one (I know how to calculate this, but let's see if you lot can pull it off) How many binary strings are there without 11000 as a substring? (A binary string being any sequence of 1's and 0's) Edited July 3, 2015 by Jeffrey94 Link to comment Share on other sites More sharing options...
MegpoidBeetle Posted July 3, 2015 Author Share Posted July 3, 2015 ( ͡° ͜ʖ ͡°) On topic though, here's a fun one (I know how to calculate this, but let's see if you lot can pull it off) How many binary strings are there without 11000 as a substring? (A binary string being any sequence of 1's and 0's) Equals to me dialing the suicide hotline HAVE MERCY Link to comment Share on other sites More sharing options...
Xhos Posted July 3, 2015 Share Posted July 3, 2015 ( ͡° ͜ʖ ͡°) On topic though, here's a fun one (I know how to calculate this, but let's see if you lot can pull it off) How many binary strings are there without 11000 as a substring? (A binary string being any sequence of 1's and 0's) Lemme just go make a deal with Hermaeus Mora... I'll just go read a Black Book and come back with an answer... hopefully. Link to comment Share on other sites More sharing options...
MegpoidBeetle Posted July 3, 2015 Author Share Posted July 3, 2015 Lemme just go make a deal with Hermaeus Mora... I'll just go read a Black Book and come back with an answer... hopefully. You probably gonna get violated by those tentacles Link to comment Share on other sites More sharing options...
Dhrekr Posted July 13, 2015 Share Posted July 13, 2015 ( ͡° ͜ʖ ͡°) On topic though, here's a fun one (I know how to calculate this, but let's see if you lot can pull it off) How many binary strings are there without 11000 as a substring? (A binary string being any sequence of 1's and 0's) Well, that is actually an interesting question. You don't say how long the strings should be, so I'll assume it's an 8-bits string. With more than 10 strings, the problem becomes a bit too county for the 5 minutes that I have. We can start by knowing that there are 2^8 = 256 individual strings, so the number of strings without 11000 as substring is 256 minus the number of strings that contain 11000, which is a much smaller number. So we are looking for the number of 8-bit strings containing 11000 - that is, 11000 appears at "some" point of the string. Let the other three bits be called ABC. Trivially there are 2^3 = 8 individual such strings. There are 4 points where 11000 can start: 11000ABC, A11000BC, AB11000C, and ABC11000. So this means 4*8 = 32 strings of 8 digits that contain 11000 as a substring. Note that none of them can be equivalent to the other, so there is no problem with double-counting (which there would be if the substring was, say, 10101). Hence there are 32 8-bit strings that contain 11000 as substring, and 224 strings that do not contain 11000. Link to comment Share on other sites More sharing options...
Mirakid Posted August 17, 2015 Share Posted August 17, 2015 Not at all, I learned it when I was 16 and this is just a question from my first semester of foundations :BI learned this in the top set of Year 8... It's easy once you get used to it. Link to comment Share on other sites More sharing options...
Renegade343 Posted August 17, 2015 Share Posted August 17, 2015 ( ͡° ͜ʖ ͡°) On topic though, here's a fun one (I know how to calculate this, but let's see if you lot can pull it off) How many binary strings are there without 11000 as a substring? (A binary string being any sequence of 1's and 0's) Assuming that "substring" equates to any binary string > 5, there are: 2n - (2n-5 x (n-4)) binary strings that do not contain 11000 as a substring, where n = length of a binary string, and n > 5. Link to comment Share on other sites More sharing options...
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