If cos(x-a),cosx,cos(x+a) are in h.p then prove that cos2x=1+cosa
Here cos2x is cos square x not cos2x
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The first of these expressions is ambiguous. When we write cos2(x) we really mean (cos(x))2 i.e. find the cosine of x and square the result. When we write cos(x2) we mean squre x first, and then take the cosine. The parentheses make the order of operations clear.
We often omit some of the parentheses when the intertion is clear, sometimes it helps to leave a space, e.g. cos2x, but does cosx2 mean (cos(x))2 or (cos(x2))?
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No, cos^2(x) is not equal to cos(x^2).
Do you remember the domain and range ?
Example if I talk about y = cos x
x is domain and cos x is range.
So, here in first case domain is x and range is cos^2(x) but in second case x^2 is dimain and cos(x^2) is range.
You might get confuse the domain set is Real no. for both and also Range set is Real no. then how they differ?
Two functions are only equal whey they give equal output for equal input.
So when we put x=2
cos^2(2) =y=(cos(2))^2
cos(4)=y'
y is not equal to y' .
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As far as I am concerned (and I have been doing engineering-style math for a long time), these expressions are ambiguous, notwithstanding that there may be some notational convention which is applicable. It is not that hard to add brackets and/or parentheses so as to avoid these ambiguities, and notational conventions are not necessarily uniform across disciplines.
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No, (cos(x^2)) is different from cos^2(x).
Proof: if a verification is wrong for one given variable of x, then it’s definitively wrong.
We consider x=1 , cosx^2=cos1. (1)
and cos^2(x)=cos^2(1)=cos1*cos1. (2)
(1) and (2) are equal if and only if (1)-(2)=0
In fact (1)-(2)=cos1(1-cos1). (3)
I consider (3)=0, => 1-cos1=0=>cos1=1 impossible because the fonction cosinus is a bijection between [0,90]→[0,1]; cos0=1; 1€[0,90]. The unique zero has it’s unique image 1. cos1 is now then different of 1. The proposition (3) is incorrect therefore (1)&(2) are not equal ie cos1#cos^2(1).
Finally cos^2(x)#cos(x^2).
We often omit some of the parentheses when the intertion is clear, sometimes it helps to leave a space, e.g. cos2x, but does cosx2 mean (cos(x))2 or (cos(x2))?
9k Views ·
Your feedback is private.
Is this answer still relevant and up to date?
No, cos^2(x) is not equal to cos(x^2).
Do you remember the domain and range ?
Example if I talk about y = cos x
x is domain and cos x is range.
So, here in first case domain is x and range is cos^2(x) but in second case x^2 is dimain and cos(x^2) is range.
You might get confuse the domain set is Real no. for both and also Range set is Real no. then how they differ?
Two functions are only equal whey they give equal output for equal input.
So when we put x=2
cos^2(2) =y=(cos(2))^2
cos(4)=y'
y is not equal to y' .
1k Views
As far as I am concerned (and I have been doing engineering-style math for a long time), these expressions are ambiguous, notwithstanding that there may be some notational convention which is applicable. It is not that hard to add brackets and/or parentheses so as to avoid these ambiguities, and notational conventions are not necessarily uniform across disciplines.
8.7k Views · · Answer requested by
Your feedback is private.
Is this answer still relevant and up to date?
No, (cos(x^2)) is different from cos^2(x).
Proof: if a verification is wrong for one given variable of x, then it’s definitively wrong.
We consider x=1 , cosx^2=cos1. (1)
and cos^2(x)=cos^2(1)=cos1*cos1. (2)
(1) and (2) are equal if and only if (1)-(2)=0
In fact (1)-(2)=cos1(1-cos1). (3)
I consider (3)=0, => 1-cos1=0=>cos1=1 impossible because the fonction cosinus is a bijection between [0,90]→[0,1]; cos0=1; 1€[0,90]. The unique zero has it’s unique image 1. cos1 is now then different of 1. The proposition (3) is incorrect therefore (1)&(2) are not equal ie cos1#cos^2(1).
Finally cos^2(x)#cos(x^2).
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