It is known that f (x) = asin (π x + α) + bcos (π X - β), where α, β, a and B are non-zero real numbers. If f (2012) = - 1, then f (2013) =? [the answer is 1]

It is known that f (x) = asin (π x + α) + bcos (π X - β), where α, β, a and B are non-zero real numbers. If f (2012) = - 1, then f (2013) =? [the answer is 1]

f﹙2012﹚
＝asin﹙2012π＋α﹚＋bcos﹙2012π－β﹚
＝asinα＋bcosβ
＝－1
∴f﹙2013﹚
＝asin﹙2013π＋α﹚＋bcos﹙2013π－β﹚
＝－asinα－bcosβ
＝1

Let f (x) = asin (π x + a) + bcos (π + X + b), a and B are all real numbers. If f (2013) = 1, find f (2014)=
Thank you very much

Because f (x) = asin (π x + a) + bcos (π x + b)
f(2013)=asin(2013π+a)+bcos(2013π+b)=1
So f (2014) = asin (2014 π + a) + bcos (2014 π + b)
=asin(2013π+a+π)+bcos(2013π+b+π)
=-asin(2013π+a)-bcos(2013π+b)
=-[asin(2013π+a)+bcos(2013π+b)]
=-1
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It is known that f (x) = asin (π x + α) + bcos (π X - β), where α. β. A.B are all non-zero real numbers. If f (2010) = - 1, then f (2011)=

f﹙2010﹚＝asin﹙2010π＋α﹚＋bcos﹙2010π－β﹚
＝asinα＋bcosβ＝－1
∴f﹙2011﹚＝asin﹙2011π＋α﹚＋bcos﹙2011π－β﹚
＝－asinα－bcosβ＝1

Let f (x) = asin (π x + a) + bcos (π x + k), where a, b.a.k are all non-zero real numbers and satisfy f (2004) = - 1, find the value of F (2008)

f(2004)=asin(2004π+α)+bcos(2004π+k)
= asinα + bcosk
That is - asin α - bcosk = 1;
Then asin α + bcosk = - 1;
Then: F (2008) = asin (2008 π + α) + bcos (2008 π + k) = asin α + bcosk
=-1

Let f (x) = asin (π x + α) + bcos (π x + β) + 7, where α, β, a and B are real numbers
If f (2001) = 6, find the value of F (2008)

f(2001)=asin(2001π+α)+bcos(2001π+β)+7=6
asin(π+α)+bcos(π+β)+7=6
So - asin α - bcos β + 7 = 6
asinα+bcosβ=1
f(2001)=asin(2008π+α)+bcos(2008π+β)+7
=asinα+bcosβ+7
=8

Let f (x) = asin (π x + Q) + bcos (π x + b) + 4, and f (2003) = 5, then f (2004) =

If f (2003) = asin (2003 π + Q) + bcos (2003 π + Q) + 4 = 5, then asin (2003 π + Q) + bcos (2003 π + Q) = 1F (2004) = asin (2003 π + Q + π) + bcos (2003 π + Q + π) + 4 = - asin (2003 π + Q) - bcos (2003 π + Q) + 4 = 4 - [asin (2003 π + Q) + bcos (2003 π + Q)] = 4-1 = 3

Given that f (x) = asin (π x + α) + bcos (π x + β) + 7, if f (2001) = 6, find the value of F (2008)

f(2001)=aSin(2001π+α)+bCos(2001π+β)+7=aSin(2000π+π+α)+bCos(2000π+π+β)+7=aSin(π+α)+bCos(π+β)+7=-aSinα-bCosβ+7=6-aSinα-bCosβ+7=6-aSinα-bCosβ=-1aSinα+bCosβ=1f(2008)=aSin(2008π+α)+...

Given f (x) = asin (π x + α) + bcos (π x + β), if f (2007) = 5, find the value of F (2008)
Where ab α β is a nonzero real number

f(2007)=asin(2007π+α)+bcos(2007π+β) =asin(π+α)+bcos(π+β) =-asinα-bcosβ=5 f(2008)=asin(πx+α)+bcos(πx+β) =asin(2008π+α)+bcos(2008π+β) =asinα+bcosβ =-5

Let f (x) = asin (π x + α) + bcos (π x + β) (where a, B, α and β are non-zero real numbers), if f (2006) = 5,
Find the value of F (2007)

F (x) = asin (π x + α) + bcos (π x + β), f (2006) = 5, that is, asin (2006 π + α) + bcos (2006 π + β) = 5. By using the induction formula, we can get: asin (α) + bcos (β) = 5

Let s be a set satisfying the following conditions: if a ∈ s, then 1 / (1-A) ∈ s, and 1 &; s
1. If 2 ∈ a, find other elements in set a;
2. Prove: if a ∈ a, then 1-1 / a ∈ a

1. Because 2 ∈ a, so - 1 = 1 / (1-2) ∈ a, so 1 / 2 = 1 / (1 + 1) ∈ a
So the other elements in a are - 1,1 / 2
2. If a ∈ a, then 1 / (1-A) ∈ a, and then 1 / (1-1 / (1-A)) = 1-1 / a ∈ a