Can all trigonometric expressions be written in terms of sine and cosine?












6












$begingroup$


I know that sine and cosine can be rewritten in terms of the real and complex parts of the exponential function as a result of Euler's formula.



My question is, can every trigonometric expression be written in terms of elementary trigonometric functions ($sin$, $cos$)? If not, why couldn't they be?



I would think that they could, although I understand that sometimes it may be prohibitive to do so since most trig identities can be derived from Euler's formula. Are there any cases when a trig expression absolutely cannot be written in terms of the elementary functions?



The only potential counterexamples I could think of would include some non trigonometric terms or factors.



I know that hyperbolic sine and cosine can be rewritten in terms of sine and cosine in the complex plane.










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hyperbolic sin and cos can be written in terms if you are working in the complex plane... And yes, since all trig functions can be written in terms of sin and cos since the remaining 4 trig function are defined in terms of sin and cos.
    $endgroup$
    – Eleven-Eleven
    6 hours ago






  • 1




    $begingroup$
    If you allow complex numbers, the hyperbolic functions are the same as their trigonometric counterparts with argument multiplied by $i$
    $endgroup$
    – Andrei
    6 hours ago






  • 1




    $begingroup$
    It comes down to what is "every trigonometric expression" and what is "in terms of"? Is $x^2 sin(x)$ a trig expression and is it already "in terms of" the sine function, or does the $x^2$ factor spoil that?
    $endgroup$
    – David K
    6 hours ago
















6












$begingroup$


I know that sine and cosine can be rewritten in terms of the real and complex parts of the exponential function as a result of Euler's formula.



My question is, can every trigonometric expression be written in terms of elementary trigonometric functions ($sin$, $cos$)? If not, why couldn't they be?



I would think that they could, although I understand that sometimes it may be prohibitive to do so since most trig identities can be derived from Euler's formula. Are there any cases when a trig expression absolutely cannot be written in terms of the elementary functions?



The only potential counterexamples I could think of would include some non trigonometric terms or factors.



I know that hyperbolic sine and cosine can be rewritten in terms of sine and cosine in the complex plane.










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hyperbolic sin and cos can be written in terms if you are working in the complex plane... And yes, since all trig functions can be written in terms of sin and cos since the remaining 4 trig function are defined in terms of sin and cos.
    $endgroup$
    – Eleven-Eleven
    6 hours ago






  • 1




    $begingroup$
    If you allow complex numbers, the hyperbolic functions are the same as their trigonometric counterparts with argument multiplied by $i$
    $endgroup$
    – Andrei
    6 hours ago






  • 1




    $begingroup$
    It comes down to what is "every trigonometric expression" and what is "in terms of"? Is $x^2 sin(x)$ a trig expression and is it already "in terms of" the sine function, or does the $x^2$ factor spoil that?
    $endgroup$
    – David K
    6 hours ago














6












6








6





$begingroup$


I know that sine and cosine can be rewritten in terms of the real and complex parts of the exponential function as a result of Euler's formula.



My question is, can every trigonometric expression be written in terms of elementary trigonometric functions ($sin$, $cos$)? If not, why couldn't they be?



I would think that they could, although I understand that sometimes it may be prohibitive to do so since most trig identities can be derived from Euler's formula. Are there any cases when a trig expression absolutely cannot be written in terms of the elementary functions?



The only potential counterexamples I could think of would include some non trigonometric terms or factors.



I know that hyperbolic sine and cosine can be rewritten in terms of sine and cosine in the complex plane.










share|cite|improve this question











$endgroup$




I know that sine and cosine can be rewritten in terms of the real and complex parts of the exponential function as a result of Euler's formula.



My question is, can every trigonometric expression be written in terms of elementary trigonometric functions ($sin$, $cos$)? If not, why couldn't they be?



I would think that they could, although I understand that sometimes it may be prohibitive to do so since most trig identities can be derived from Euler's formula. Are there any cases when a trig expression absolutely cannot be written in terms of the elementary functions?



The only potential counterexamples I could think of would include some non trigonometric terms or factors.



I know that hyperbolic sine and cosine can be rewritten in terms of sine and cosine in the complex plane.







trigonometry elementary-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







Gnumbertester

















asked 6 hours ago









GnumbertesterGnumbertester

35918




35918








  • 4




    $begingroup$
    Hyperbolic sin and cos can be written in terms if you are working in the complex plane... And yes, since all trig functions can be written in terms of sin and cos since the remaining 4 trig function are defined in terms of sin and cos.
    $endgroup$
    – Eleven-Eleven
    6 hours ago






  • 1




    $begingroup$
    If you allow complex numbers, the hyperbolic functions are the same as their trigonometric counterparts with argument multiplied by $i$
    $endgroup$
    – Andrei
    6 hours ago






  • 1




    $begingroup$
    It comes down to what is "every trigonometric expression" and what is "in terms of"? Is $x^2 sin(x)$ a trig expression and is it already "in terms of" the sine function, or does the $x^2$ factor spoil that?
    $endgroup$
    – David K
    6 hours ago














  • 4




    $begingroup$
    Hyperbolic sin and cos can be written in terms if you are working in the complex plane... And yes, since all trig functions can be written in terms of sin and cos since the remaining 4 trig function are defined in terms of sin and cos.
    $endgroup$
    – Eleven-Eleven
    6 hours ago






  • 1




    $begingroup$
    If you allow complex numbers, the hyperbolic functions are the same as their trigonometric counterparts with argument multiplied by $i$
    $endgroup$
    – Andrei
    6 hours ago






  • 1




    $begingroup$
    It comes down to what is "every trigonometric expression" and what is "in terms of"? Is $x^2 sin(x)$ a trig expression and is it already "in terms of" the sine function, or does the $x^2$ factor spoil that?
    $endgroup$
    – David K
    6 hours ago








4




4




$begingroup$
Hyperbolic sin and cos can be written in terms if you are working in the complex plane... And yes, since all trig functions can be written in terms of sin and cos since the remaining 4 trig function are defined in terms of sin and cos.
$endgroup$
– Eleven-Eleven
6 hours ago




$begingroup$
Hyperbolic sin and cos can be written in terms if you are working in the complex plane... And yes, since all trig functions can be written in terms of sin and cos since the remaining 4 trig function are defined in terms of sin and cos.
$endgroup$
– Eleven-Eleven
6 hours ago




1




1




$begingroup$
If you allow complex numbers, the hyperbolic functions are the same as their trigonometric counterparts with argument multiplied by $i$
$endgroup$
– Andrei
6 hours ago




$begingroup$
If you allow complex numbers, the hyperbolic functions are the same as their trigonometric counterparts with argument multiplied by $i$
$endgroup$
– Andrei
6 hours ago




1




1




$begingroup$
It comes down to what is "every trigonometric expression" and what is "in terms of"? Is $x^2 sin(x)$ a trig expression and is it already "in terms of" the sine function, or does the $x^2$ factor spoil that?
$endgroup$
– David K
6 hours ago




$begingroup$
It comes down to what is "every trigonometric expression" and what is "in terms of"? Is $x^2 sin(x)$ a trig expression and is it already "in terms of" the sine function, or does the $x^2$ factor spoil that?
$endgroup$
– David K
6 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

$$tantheta=frac{sintheta}{costheta}$$



$$cottheta=frac{costheta}{sintheta}$$



$$sectheta=frac{1}{costheta}$$



$$csctheta=frac{1}{sintheta}$$



and since in the complex plane, we have



$$begin{align}
coshtheta&=phantom{-i}cos{itheta} \
sinhtheta&=-isin{itheta} \
tanhtheta&=-itan{itheta} \
coththeta&=phantom{-}icot{itheta} \
operatorname{sech}theta&=phantom{-i}sec{itheta} \
operatorname{csch}theta&=phantom{-}icsc{itheta}
end{align}$$



And thus you can continue the definitions of the hyperbolic functions by substituting the sine and cosine definitions of the the remaining 4 trigonometric functions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not sure why sech and csch don't render right?
    $endgroup$
    – Eleven-Eleven
    6 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

$$tantheta=frac{sintheta}{costheta}$$



$$cottheta=frac{costheta}{sintheta}$$



$$sectheta=frac{1}{costheta}$$



$$csctheta=frac{1}{sintheta}$$



and since in the complex plane, we have



$$begin{align}
coshtheta&=phantom{-i}cos{itheta} \
sinhtheta&=-isin{itheta} \
tanhtheta&=-itan{itheta} \
coththeta&=phantom{-}icot{itheta} \
operatorname{sech}theta&=phantom{-i}sec{itheta} \
operatorname{csch}theta&=phantom{-}icsc{itheta}
end{align}$$



And thus you can continue the definitions of the hyperbolic functions by substituting the sine and cosine definitions of the the remaining 4 trigonometric functions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not sure why sech and csch don't render right?
    $endgroup$
    – Eleven-Eleven
    6 hours ago
















5












$begingroup$

$$tantheta=frac{sintheta}{costheta}$$



$$cottheta=frac{costheta}{sintheta}$$



$$sectheta=frac{1}{costheta}$$



$$csctheta=frac{1}{sintheta}$$



and since in the complex plane, we have



$$begin{align}
coshtheta&=phantom{-i}cos{itheta} \
sinhtheta&=-isin{itheta} \
tanhtheta&=-itan{itheta} \
coththeta&=phantom{-}icot{itheta} \
operatorname{sech}theta&=phantom{-i}sec{itheta} \
operatorname{csch}theta&=phantom{-}icsc{itheta}
end{align}$$



And thus you can continue the definitions of the hyperbolic functions by substituting the sine and cosine definitions of the the remaining 4 trigonometric functions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not sure why sech and csch don't render right?
    $endgroup$
    – Eleven-Eleven
    6 hours ago














5












5








5





$begingroup$

$$tantheta=frac{sintheta}{costheta}$$



$$cottheta=frac{costheta}{sintheta}$$



$$sectheta=frac{1}{costheta}$$



$$csctheta=frac{1}{sintheta}$$



and since in the complex plane, we have



$$begin{align}
coshtheta&=phantom{-i}cos{itheta} \
sinhtheta&=-isin{itheta} \
tanhtheta&=-itan{itheta} \
coththeta&=phantom{-}icot{itheta} \
operatorname{sech}theta&=phantom{-i}sec{itheta} \
operatorname{csch}theta&=phantom{-}icsc{itheta}
end{align}$$



And thus you can continue the definitions of the hyperbolic functions by substituting the sine and cosine definitions of the the remaining 4 trigonometric functions.






share|cite|improve this answer











$endgroup$



$$tantheta=frac{sintheta}{costheta}$$



$$cottheta=frac{costheta}{sintheta}$$



$$sectheta=frac{1}{costheta}$$



$$csctheta=frac{1}{sintheta}$$



and since in the complex plane, we have



$$begin{align}
coshtheta&=phantom{-i}cos{itheta} \
sinhtheta&=-isin{itheta} \
tanhtheta&=-itan{itheta} \
coththeta&=phantom{-}icot{itheta} \
operatorname{sech}theta&=phantom{-i}sec{itheta} \
operatorname{csch}theta&=phantom{-}icsc{itheta}
end{align}$$



And thus you can continue the definitions of the hyperbolic functions by substituting the sine and cosine definitions of the the remaining 4 trigonometric functions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 6 hours ago









Blue

47.9k870152




47.9k870152










answered 6 hours ago









Eleven-ElevenEleven-Eleven

5,48072659




5,48072659












  • $begingroup$
    Not sure why sech and csch don't render right?
    $endgroup$
    – Eleven-Eleven
    6 hours ago


















  • $begingroup$
    Not sure why sech and csch don't render right?
    $endgroup$
    – Eleven-Eleven
    6 hours ago
















$begingroup$
Not sure why sech and csch don't render right?
$endgroup$
– Eleven-Eleven
6 hours ago




$begingroup$
Not sure why sech and csch don't render right?
$endgroup$
– Eleven-Eleven
6 hours ago


















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