If $cos(x+a)=cos(x+y+z)$, then can we deduce that $a=y+z$?
$begingroup$
If I have the following equation:
$$cos(x+a)=cos(x+y+z)$$
Can I take the inverse cos of both sides?
$$begin{align}
cos^{-1}(cos(x + a))&=cos^{-1}(cos(x+y+z)) \
x+a&=x+y+z \
a&=y+z
end{align}$$
Is this correct?
algebra-precalculus trigonometry
edited Dec 10 '18 at 0:28
Blue
49.1k870156
asked Dec 10 '18 at 0:08
user367640user367640
268
$endgroup$
add a comment |
$begingroup$
If I have the following equation:
$$cos(x+a)=cos(x+y+z)$$
Can I take the inverse cos of both sides?
$$begin{align}
cos^{-1}(cos(x + a))&=cos^{-1}(cos(x+y+z)) \
x+a&=x+y+z \
a&=y+z
end{align}$$
Is this correct?
algebra-precalculus trigonometry
edited Dec 10 '18 at 0:28
Blue
49.1k870156
asked Dec 10 '18 at 0:08
user367640user367640
268
$endgroup$
1
$begingroup$
No, you can hardly ever take inverse cosine of both sides, and there appears to be no reason for this to be one of those.
$endgroup$
– Saucy O'Path
Dec 10 '18 at 0:10
add a comment |
$begingroup$
If I have the following equation:
$$cos(x+a)=cos(x+y+z)$$
Can I take the inverse cos of both sides?
$$begin{align}
cos^{-1}(cos(x + a))&=cos^{-1}(cos(x+y+z)) \
x+a&=x+y+z \
a&=y+z
end{align}$$
Is this correct?
algebra-precalculus trigonometry
edited Dec 10 '18 at 0:28
Blue
49.1k870156
asked Dec 10 '18 at 0:08
user367640user367640
268
$endgroup$
If I have the following equation:
$$cos(x+a)=cos(x+y+z)$$
Can I take the inverse cos of both sides?
$$begin{align}
cos^{-1}(cos(x + a))&=cos^{-1}(cos(x+y+z)) \
x+a&=x+y+z \
a&=y+z
end{align}$$
Is this correct?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Dec 10 '18 at 0:28
Blue
49.1k870156
asked Dec 10 '18 at 0:08
user367640user367640
268
edited Dec 10 '18 at 0:28
Blue
49.1k870156
asked Dec 10 '18 at 0:08
user367640user367640
268
edited Dec 10 '18 at 0:28
Blue
49.1k870156
edited Dec 10 '18 at 0:28
Blue
49.1k870156
edited Dec 10 '18 at 0:28
Blue
49.1k870156
49.1k870156
asked Dec 10 '18 at 0:08
user367640user367640
268
asked Dec 10 '18 at 0:08
user367640user367640
268
asked Dec 10 '18 at 0:08
user367640user367640
268
268
1
$begingroup$
No, you can hardly ever take inverse cosine of both sides, and there appears to be no reason for this to be one of those.
$endgroup$
– Saucy O'Path
Dec 10 '18 at 0:10
add a comment |
1
$begingroup$
No, you can hardly ever take inverse cosine of both sides, and there appears to be no reason for this to be one of those.
$endgroup$
– Saucy O'Path
Dec 10 '18 at 0:10
1
1
$begingroup$
No, you can hardly ever take inverse cosine of both sides, and there appears to be no reason for this to be one of those.
$endgroup$
– Saucy O'Path
Dec 10 '18 at 0:10
$begingroup$
No, you can hardly ever take inverse cosine of both sides, and there appears to be no reason for this to be one of those.
$endgroup$
– Saucy O'Path
Dec 10 '18 at 0:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No it'is wrong, let instead consider the trigonometric circle to derive that
$$cos theta = cos alpha implies theta = alpha+2kpi quad lor quad theta = -alpha+2kpi$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
$endgroup$
1
$begingroup$
Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because: $$cos(x) = cos(x + 2pi) = cos(x + 4pi) = ... = cos(x + 2kpi)$$ for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2pi$. We also have this because cosine is an even function, i.e. $$cos(-x) = cos(x)$$
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:16
$begingroup$
@EeveeTrainer That's not complete we also have $$cos(x) = cos(-x + 2pi) = cos(-x + 4pi) = ... = cos(-x + 2kpi)$$
$endgroup$
– gimusi
Dec 10 '18 at 0:17
2
$begingroup$
I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities.
$endgroup$
– gimusi
Dec 10 '18 at 0:18
$begingroup$
The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero.
$endgroup$
– user367640
Dec 10 '18 at 0:18
$begingroup$
Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now.
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:19
|
show 5 more comments
$begingroup$
Notice that due to $costheta$ having a period of $2pi$, for any $lambda, muinBbb Z, thetain Bbb R$, we have:
$$cos(theta+2lambdapi)=cos(theta+2mupi)$$
$lambda=mu$ is not a necessity to this end.
Effectively in your expression we have that $a=y+z+2lambdapi, lambda in Bbb Z$.
$lambda=0$ is a possibility, and there $a=y+z$, but there are infinitely many other possibilities as well.
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
$endgroup$
$begingroup$
Here is a graph showing why this is the case. $f(x)=cos^{-1}(x)$ is a one-to-many function.
$endgroup$
– Rhys Hughes
Dec 10 '18 at 0:46
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No it'is wrong, let instead consider the trigonometric circle to derive that
$$cos theta = cos alpha implies theta = alpha+2kpi quad lor quad theta = -alpha+2kpi$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
$endgroup$
1
$begingroup$
Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because: $$cos(x) = cos(x + 2pi) = cos(x + 4pi) = ... = cos(x + 2kpi)$$ for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2pi$. We also have this because cosine is an even function, i.e. $$cos(-x) = cos(x)$$
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:16
$begingroup$
@EeveeTrainer That's not complete we also have $$cos(x) = cos(-x + 2pi) = cos(-x + 4pi) = ... = cos(-x + 2kpi)$$
$endgroup$
– gimusi
Dec 10 '18 at 0:17
2
$begingroup$
I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities.
$endgroup$
– gimusi
Dec 10 '18 at 0:18
$begingroup$
The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero.
$endgroup$
– user367640
Dec 10 '18 at 0:18
$begingroup$
Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now.
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:19
|
show 5 more comments
$begingroup$
No it'is wrong, let instead consider the trigonometric circle to derive that
$$cos theta = cos alpha implies theta = alpha+2kpi quad lor quad theta = -alpha+2kpi$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
$endgroup$
1
$begingroup$
Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because: $$cos(x) = cos(x + 2pi) = cos(x + 4pi) = ... = cos(x + 2kpi)$$ for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2pi$. We also have this because cosine is an even function, i.e. $$cos(-x) = cos(x)$$
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:16
$begingroup$
@EeveeTrainer That's not complete we also have $$cos(x) = cos(-x + 2pi) = cos(-x + 4pi) = ... = cos(-x + 2kpi)$$
$endgroup$
– gimusi
Dec 10 '18 at 0:17
2
$begingroup$
I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities.
$endgroup$
– gimusi
Dec 10 '18 at 0:18
$begingroup$
The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero.
$endgroup$
– user367640
Dec 10 '18 at 0:18
$begingroup$
Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now.
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:19
|
show 5 more comments
$begingroup$
No it'is wrong, let instead consider the trigonometric circle to derive that
$$cos theta = cos alpha implies theta = alpha+2kpi quad lor quad theta = -alpha+2kpi$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
$endgroup$
No it'is wrong, let instead consider the trigonometric circle to derive that
$$cos theta = cos alpha implies theta = alpha+2kpi quad lor quad theta = -alpha+2kpi$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
edited Dec 10 '18 at 0:24
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
answered Dec 10 '18 at 0:13
gimusigimusi
93k84594
93k84594
1
$begingroup$
Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because: $$cos(x) = cos(x + 2pi) = cos(x + 4pi) = ... = cos(x + 2kpi)$$ for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2pi$. We also have this because cosine is an even function, i.e. $$cos(-x) = cos(x)$$
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:16
$begingroup$
@EeveeTrainer That's not complete we also have $$cos(x) = cos(-x + 2pi) = cos(-x + 4pi) = ... = cos(-x + 2kpi)$$
$endgroup$
– gimusi
Dec 10 '18 at 0:17
2
$begingroup$
I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities.
$endgroup$
– gimusi
Dec 10 '18 at 0:18
$begingroup$
The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero.
$endgroup$
– user367640
Dec 10 '18 at 0:18
$begingroup$
Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now.
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:19
|
show 5 more comments
1
$begingroup$
Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because: $$cos(x) = cos(x + 2pi) = cos(x + 4pi) = ... = cos(x + 2kpi)$$ for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2pi$. We also have this because cosine is an even function, i.e. $$cos(-x) = cos(x)$$
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:16
$begingroup$
@EeveeTrainer That's not complete we also have $$cos(x) = cos(-x + 2pi) = cos(-x + 4pi) = ... = cos(-x + 2kpi)$$
$endgroup$
– gimusi
Dec 10 '18 at 0:17
2
$begingroup$
I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities.
$endgroup$
– gimusi
Dec 10 '18 at 0:18
$begingroup$
The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero.
$endgroup$
– user367640
Dec 10 '18 at 0:18
$begingroup$
Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now.
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:19
1
1
$begingroup$
Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because: $$cos(x) = cos(x + 2pi) = cos(x + 4pi) = ... = cos(x + 2kpi)$$ for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2pi$. We also have this because cosine is an even function, i.e. $$cos(-x) = cos(x)$$
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:16
$begingroup$
Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because: $$cos(x) = cos(x + 2pi) = cos(x + 4pi) = ... = cos(x + 2kpi)$$ for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2pi$. We also have this because cosine is an even function, i.e. $$cos(-x) = cos(x)$$
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:16
$begingroup$
@EeveeTrainer That's not complete we also have $$cos(x) = cos(-x + 2pi) = cos(-x + 4pi) = ... = cos(-x + 2kpi)$$
$endgroup$
– gimusi
Dec 10 '18 at 0:17
$begingroup$
@EeveeTrainer That's not complete we also have $$cos(x) = cos(-x + 2pi) = cos(-x + 4pi) = ... = cos(-x + 2kpi)$$
$endgroup$
– gimusi
Dec 10 '18 at 0:17
2
2
$begingroup$
I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities.
$endgroup$
– gimusi
Dec 10 '18 at 0:18
$begingroup$
I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities.
$endgroup$
– gimusi
Dec 10 '18 at 0:18
$begingroup$
The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero.
$endgroup$
– user367640
Dec 10 '18 at 0:18
$begingroup$
The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero.
$endgroup$
– user367640
Dec 10 '18 at 0:18
$begingroup$
Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now.
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:19
$begingroup$
Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now.
$endgroup$
– Eevee Trainer
Dec 10 '18 at 0:19
|
show 5 more comments
$begingroup$
Notice that due to $costheta$ having a period of $2pi$, for any $lambda, muinBbb Z, thetain Bbb R$, we have:
$$cos(theta+2lambdapi)=cos(theta+2mupi)$$
$lambda=mu$ is not a necessity to this end.
Effectively in your expression we have that $a=y+z+2lambdapi, lambda in Bbb Z$.
$lambda=0$ is a possibility, and there $a=y+z$, but there are infinitely many other possibilities as well.
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
$endgroup$
$begingroup$
Here is a graph showing why this is the case. $f(x)=cos^{-1}(x)$ is a one-to-many function.
$endgroup$
– Rhys Hughes
Dec 10 '18 at 0:46
add a comment |
$begingroup$
Notice that due to $costheta$ having a period of $2pi$, for any $lambda, muinBbb Z, thetain Bbb R$, we have:
$$cos(theta+2lambdapi)=cos(theta+2mupi)$$
$lambda=mu$ is not a necessity to this end.
Effectively in your expression we have that $a=y+z+2lambdapi, lambda in Bbb Z$.
$lambda=0$ is a possibility, and there $a=y+z$, but there are infinitely many other possibilities as well.
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
$endgroup$
$begingroup$
Here is a graph showing why this is the case. $f(x)=cos^{-1}(x)$ is a one-to-many function.
$endgroup$
– Rhys Hughes
Dec 10 '18 at 0:46
add a comment |
$begingroup$
Notice that due to $costheta$ having a period of $2pi$, for any $lambda, muinBbb Z, thetain Bbb R$, we have:
$$cos(theta+2lambdapi)=cos(theta+2mupi)$$
$lambda=mu$ is not a necessity to this end.
Effectively in your expression we have that $a=y+z+2lambdapi, lambda in Bbb Z$.
$lambda=0$ is a possibility, and there $a=y+z$, but there are infinitely many other possibilities as well.
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
$endgroup$
Notice that due to $costheta$ having a period of $2pi$, for any $lambda, muinBbb Z, thetain Bbb R$, we have:
$$cos(theta+2lambdapi)=cos(theta+2mupi)$$
$lambda=mu$ is not a necessity to this end.
Effectively in your expression we have that $a=y+z+2lambdapi, lambda in Bbb Z$.
$lambda=0$ is a possibility, and there $a=y+z$, but there are infinitely many other possibilities as well.
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
answered Dec 10 '18 at 0:42
Rhys HughesRhys Hughes
7,0351630
7,0351630
$begingroup$
Here is a graph showing why this is the case. $f(x)=cos^{-1}(x)$ is a one-to-many function.
$endgroup$
– Rhys Hughes
Dec 10 '18 at 0:46
add a comment |
$begingroup$
Here is a graph showing why this is the case. $f(x)=cos^{-1}(x)$ is a one-to-many function.
$endgroup$
– Rhys Hughes
Dec 10 '18 at 0:46
$begingroup$
Here is a graph showing why this is the case. $f(x)=cos^{-1}(x)$ is a one-to-many function.
$endgroup$
– Rhys Hughes
Dec 10 '18 at 0:46
$begingroup$
Here is a graph showing why this is the case. $f(x)=cos^{-1}(x)$ is a one-to-many function.
$endgroup$
– Rhys Hughes
Dec 10 '18 at 0:46
add a comment |
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$begingroup$
No, you can hardly ever take inverse cosine of both sides, and there appears to be no reason for this to be one of those.
$endgroup$
– Saucy O'Path
Dec 10 '18 at 0:10