If $cos(x+a)=cos(x+y+z)$, then can we deduce that $a=y+z$?












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 0:28









Blue

49.1k870156




49.1k870156










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
















  • 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












2 Answers
2






active

oldest

votes


















2












$begingroup$

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 $cos(3x) = cos(2x)$






share|cite|improve this answer











$endgroup$









  • 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



















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $cos(3x) = cos(2x)$






share|cite|improve this answer











$endgroup$









  • 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
















2












$begingroup$

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 $cos(3x) = cos(2x)$






share|cite|improve this answer











$endgroup$









  • 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














2












2








2





$begingroup$

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 $cos(3x) = cos(2x)$






share|cite|improve this answer











$endgroup$



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 $cos(3x) = cos(2x)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 0:24

























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














  • 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











0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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