Proving that $cosleft(ax-frac{bpi}{2}right)=sin(ax)$ for all $a$, $b$
$begingroup$
Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.
I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?
trigonometry
$endgroup$
add a comment |
$begingroup$
Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.
I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?
trigonometry
$endgroup$
4
$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06
3
$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09
2
$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14
add a comment |
$begingroup$
Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.
I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?
trigonometry
$endgroup$
Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.
I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?
trigonometry
trigonometry
edited Jan 7 at 6:32
Blue
47.8k870152
47.8k870152
asked Jan 7 at 6:04
Anson PangAnson Pang
6215
6215
4
$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06
3
$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09
2
$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14
add a comment |
4
$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06
3
$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09
2
$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14
4
4
$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06
$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06
3
3
$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09
$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09
2
2
$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14
$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].
$endgroup$
$begingroup$
If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
$endgroup$
– John Omielan
Jan 7 at 6:14
$begingroup$
@JohnOmielan Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:15
add a comment |
$begingroup$
If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$
$ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer
$iff2ax-b=4mpm(1-2ax)$
$-implies2ax-b=4m-(1-2x)iff b=1-4m$
$+implies2ax-b=4m+1-2axiff4ax-b=4m+1$
which is dependent on $x$ hence can not be held for all $a,b$
$endgroup$
add a comment |
$begingroup$
If
$$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
$$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.
So, consider small values of $x$ and use series expansion to get
$$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
b}{2}right)right)+Oleft(x^4right)$$
This would imply
$$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].
$endgroup$
$begingroup$
If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
$endgroup$
– John Omielan
Jan 7 at 6:14
$begingroup$
@JohnOmielan Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:15
add a comment |
$begingroup$
The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].
$endgroup$
$begingroup$
If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
$endgroup$
– John Omielan
Jan 7 at 6:14
$begingroup$
@JohnOmielan Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:15
add a comment |
$begingroup$
The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].
$endgroup$
The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].
edited Jan 7 at 6:15
answered Jan 7 at 6:11
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
54.4k32055
$begingroup$
If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
$endgroup$
– John Omielan
Jan 7 at 6:14
$begingroup$
@JohnOmielan Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:15
add a comment |
$begingroup$
If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
$endgroup$
– John Omielan
Jan 7 at 6:14
$begingroup$
@JohnOmielan Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:15
$begingroup$
If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
$endgroup$
– John Omielan
Jan 7 at 6:14
$begingroup$
If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
$endgroup$
– John Omielan
Jan 7 at 6:14
$begingroup$
@JohnOmielan Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:15
$begingroup$
@JohnOmielan Very true. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:15
add a comment |
$begingroup$
If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$
$ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer
$iff2ax-b=4mpm(1-2ax)$
$-implies2ax-b=4m-(1-2x)iff b=1-4m$
$+implies2ax-b=4m+1-2axiff4ax-b=4m+1$
which is dependent on $x$ hence can not be held for all $a,b$
$endgroup$
add a comment |
$begingroup$
If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$
$ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer
$iff2ax-b=4mpm(1-2ax)$
$-implies2ax-b=4m-(1-2x)iff b=1-4m$
$+implies2ax-b=4m+1-2axiff4ax-b=4m+1$
which is dependent on $x$ hence can not be held for all $a,b$
$endgroup$
add a comment |
$begingroup$
If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$
$ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer
$iff2ax-b=4mpm(1-2ax)$
$-implies2ax-b=4m-(1-2x)iff b=1-4m$
$+implies2ax-b=4m+1-2axiff4ax-b=4m+1$
which is dependent on $x$ hence can not be held for all $a,b$
$endgroup$
If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$
$ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer
$iff2ax-b=4mpm(1-2ax)$
$-implies2ax-b=4m-(1-2x)iff b=1-4m$
$+implies2ax-b=4m+1-2axiff4ax-b=4m+1$
which is dependent on $x$ hence can not be held for all $a,b$
answered Jan 7 at 7:32
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$begingroup$
If
$$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
$$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.
So, consider small values of $x$ and use series expansion to get
$$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
b}{2}right)right)+Oleft(x^4right)$$
This would imply
$$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.
$endgroup$
add a comment |
$begingroup$
If
$$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
$$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.
So, consider small values of $x$ and use series expansion to get
$$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
b}{2}right)right)+Oleft(x^4right)$$
This would imply
$$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.
$endgroup$
add a comment |
$begingroup$
If
$$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
$$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.
So, consider small values of $x$ and use series expansion to get
$$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
b}{2}right)right)+Oleft(x^4right)$$
This would imply
$$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.
$endgroup$
If
$$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
$$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.
So, consider small values of $x$ and use series expansion to get
$$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
b}{2}right)right)+Oleft(x^4right)$$
This would imply
$$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.
answered Jan 7 at 7:36
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
add a comment |
add a comment |
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4
$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06
3
$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09
2
$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14