maths expert
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4sin2t + 3cos2t = 5cos(2t - 53.1 degrees)
i forget how to do...
neone knows? -
4sin2t + 3cos2t use R-formula?Originally posted by valentinoo:4sin2t + 3cos2t = 5cos(2t - 53.1 degrees)
i forget how to do...
neone knows? -
mi o level maths no teach leh
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the british and americans dunno how to solve this ...i wanna know if singaporeans can....too bad i forgot..all give back
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square root of 4^2 + 3^2 = 5
lidat = 5sin(2t - tan^-1 3/4)
no sure correct anot -
den find using general solution?
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man it looks easy i know i can do it if im stil studying...lol...after few yrs..cannot remember..i only remember 1+1..kindergarten teach one hahahah
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i don't know about the pommies but the yankees have sent people to space, you say they know how to solve or not?Originally posted by valentinoo:the british and americans dunno how to solve this ...i wanna know if singaporeans can....too bad i forgot..all give back -
they got give u formula for the sine cosine and tangent de.. juz sub in
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they dunno lah..british also dunno...Originally posted by vito_corleone:i don't know about the pommies but the yankees have sent people to space, you say they know how to solve or not? -
there r two approaches. both use gotta r-forumale..
(1) Start with cosine
R = (4²+3²)^0.5=5
3cos2t + 4sin2t
= R cos ( 2t - b )
= R cos2tsinb + Rsin2tcosb
where b: basic angle
comparing coef,
Rsinb = 3
Rcosb = 4
so b = arc tan (3/4) = 36.9
=> 4sin2t + 3cos2t = 5cos (2t - 36.9)
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(2) Start with sine
4sin2t + 3cos2t
= Rsin (2t + b)
= Rsin2tcosb + Rcos2tsinb
tanb = Rsin2t/Rcos2t = 4/3
b = 53.1
4sin2t + 3cos2t
= 5sin (2t + 53.1)
= 5cos (90+2t-53.1)
= 5cos (2t - 36.9)
got the same Ans for both approaches perhaps u can double check? =p -
i know who know..let's ask guatamela.Originally posted by valentinoo:they dunno lah..british also dunno... -
linear interpolation, use numerical methods + computer to solveOriginally posted by valentinoo:4sin2t + 3cos2t = 5cos(2t - 53.1 degrees)
i forget how to do...
neone knows? -
dude u r a genius u came out wid sometthin... lol...Originally posted by papercut87:there r two approaches. both use gotta r-forumale..
(1) Start with cosine
R = (4²+3²)^0.5=5
3cos2t + 4sin2t
= R cos ( 2t - b )
= R cos2tsinb + Rsin2tcosb
where b: basic angle
comparing coef,
Rsinb = 3
Rcosb = 4
so b = arc tan (3/4) = 36.9
=> 4sin2t + 3cos2t = 5cos (2t - 36.9)
----
(2) Start with sine
4sin2t + 3cos2t
= Rsin (2t + b)
= Rsin2tcosb + Rcos2tsinb
tanb = Rsin2t/Rcos2t = 4/3
b = 53.1
4sin2t + 3cos2t
= 5sin (2t + 53.1)
= 5cos (90+2t-53.1)
= 5cos (2t - 36.9)
got the same Ans for both approaches perhaps u can double check? =p -
oh man, i'm invisibleOriginally posted by valentinoo:dude u r a genius u came out wid sometthin... lol... -
Use 'R' formula? Die...... exam so close, I forgot liao!
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im helping u all to refresh...
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u how old? how come still need to solve this?Originally posted by valentinoo:im helping u all to refresh... -
1)try this. find the last three digits of 9^100 -1(without the use of calculator)
2)find the smallest three digit number n such that if the three digits are a, b, and c, then n= a + b + c + ab +bc + ac +abc -
What level is this ar?
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I try hor:Originally posted by Ray87:1)try this. find the last three digits of 9^100 -1(without the use of calculator)
2)find the smallest three digit number n such that if the three digits are a, b, and c, then n= a + b + c + ab +bc + ac +abc
i)146?
ii)107? -
i don't think so.Originally posted by CoolMyth:I try hor:
i)146?
ii)107?
for i), 9^2=081, 9^4=561, 9^6=...441, 9^8=...721, 9^10=...401, 9^12=...481
......................................................................9^20=...?01, 9^22=...?81
......................................................................9^30=...?01, 9^32=...?81
therefore for 9^100, last 2 digits must be 01, and for (9^100)-1, last 2 digits must be 00.
although a more intelligent method is needed to find last 3 digits...
for (ii) im quite sure: ans=199. isit? -
I used the following method to solve, don't know correct or not:
i) 3^200 =(actually triple lines) x (mod 1000)
Find x, then the answer is x - 1.
ii)I factorise until it becomes
n = (c+1)(a + ab + b) + c
Then, I just did some trial and errors.
Correct or not har? -
Your arguments are sound, all three digits are obtained by pattern recognition. i got 400 as my answer, and a couple of my friends has gotten that too. Just list out the first ten powers and observe the pattern. Actually im not completely sure of my answer because i dont have the answer sheet. Its from math olympiad.Originally posted by crushedhearts:i don't think so.
for i), 9^2=081, 9^4=561, 9^6=...441, 9^8=...721, 9^10=...401, 9^12=...481
......................................................................9^20=...?01, 9^22=...?81
......................................................................9^30=...?01, 9^32=...?81
therefore for 9^100, last 2 digits must be 01, and for (9^100)-1, last 2 digits must be 00.
although a more intelligent method is needed to find last 3 digits...
for (ii) im quite sure: ans=199. isit?
2nd ans should be correct. Haha, cause i had that answer too. fun eh? trial and error to get answer... -
Wasn't these from the maths olympiad senior section this year?Originally posted by Ray87:1)try this. find the last three digits of 9^100 -1(without the use of calculator)
2)find the smallest three digit number n such that if the three digits are a, b, and c, then n= a + b + c + ab +bc + ac +abc