Proving identity $cos^2 a= sinh^2 b$ if $sin(a+ib)=cos b + isin b$

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This is my first question on StackExchange. I have some trouble with proving this identity with the following condition:




If: $sin(a+ib)= cos b + i sin b$



Prove: $cos^2 (a) = sinh^2 (b)$




I tried using different methods, but the furthest I've come to is this:



$sinh^2 (b) = sin^2 (b)/cos^2 (a)$



I expanded $sin(a+ib)$ and rearranged the equation to isolate $sinh(b)$. I then squared it (to get $sinh^2 (b)$) and got the above statement.



I would like to know whether I'm on the right track, or if there are any mistakes. Thanks a lot.




trigonometry complex-numbers




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  • Welcome here. Use mathjax please.
    – user 108128
    Aug 2 at 17:05











  • Thanks. I will try.
    – Ra1pid
    Aug 2 at 17:06














up vote
1
down vote

favorite












This is my first question on StackExchange. I have some trouble with proving this identity with the following condition:




If: $sin(a+ib)= cos b + i sin b$



Prove: $cos^2 (a) = sinh^2 (b)$




I tried using different methods, but the furthest I've come to is this:



$sinh^2 (b) = sin^2 (b)/cos^2 (a)$



I expanded $sin(a+ib)$ and rearranged the equation to isolate $sinh(b)$. I then squared it (to get $sinh^2 (b)$) and got the above statement.



I would like to know whether I'm on the right track, or if there are any mistakes. Thanks a lot.




trigonometry complex-numbers




share|cite|improve this question





















  • Welcome here. Use mathjax please.
    – user 108128
    Aug 2 at 17:05











  • Thanks. I will try.
    – Ra1pid
    Aug 2 at 17:06












up vote
1
down vote

favorite









up vote
1
down vote

favorite











This is my first question on StackExchange. I have some trouble with proving this identity with the following condition:




If: $sin(a+ib)= cos b + i sin b$



Prove: $cos^2 (a) = sinh^2 (b)$




I tried using different methods, but the furthest I've come to is this:



$sinh^2 (b) = sin^2 (b)/cos^2 (a)$



I expanded $sin(a+ib)$ and rearranged the equation to isolate $sinh(b)$. I then squared it (to get $sinh^2 (b)$) and got the above statement.



I would like to know whether I'm on the right track, or if there are any mistakes. Thanks a lot.




trigonometry complex-numbers




share|cite|improve this question













This is my first question on StackExchange. I have some trouble with proving this identity with the following condition:




If: $sin(a+ib)= cos b + i sin b$



Prove: $cos^2 (a) = sinh^2 (b)$




I tried using different methods, but the furthest I've come to is this:



$sinh^2 (b) = sin^2 (b)/cos^2 (a)$



I expanded $sin(a+ib)$ and rearranged the equation to isolate $sinh(b)$. I then squared it (to get $sinh^2 (b)$) and got the above statement.



I would like to know whether I'm on the right track, or if there are any mistakes. Thanks a lot.





trigonometry complex-numbers





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 17:12









zipirovich

9,89911630




9,89911630









asked Aug 2 at 17:03









Ra1pid

113




113











  • Welcome here. Use mathjax please.
    – user 108128
    Aug 2 at 17:05











  • Thanks. I will try.
    – Ra1pid
    Aug 2 at 17:06
















  • Welcome here. Use mathjax please.
    – user 108128
    Aug 2 at 17:05











  • Thanks. I will try.
    – Ra1pid
    Aug 2 at 17:06















Welcome here. Use mathjax please.
– user 108128
Aug 2 at 17:05





Welcome here. Use mathjax please.
– user 108128
Aug 2 at 17:05













Thanks. I will try.
– Ra1pid
Aug 2 at 17:06




Thanks. I will try.
– Ra1pid
Aug 2 at 17:06










2 Answers
2






active

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up vote
2
down vote



accepted










beginalign
sin(a+ib)
&=sin acos ib+cos asin ib\
&=sin acosh b+icos asinh b\
&= cos b + i sin b
endalign
so take real part and imaginary of sides give us
$$sin acosh b= cos b$$
$$cos asinh b= sin b$$
then squaring two equations and adding concludes
$$sin^2 acosh^2 b+cos^2 asinh^2 b=1$$
$$(1-cos^2 a)(1+sinh^2 b)+cos^2 asinh^2 b=1$$
which gives $colorbluecos^2 a = sinh^2 b$.






share|cite|improve this answer























  • Wow thanks this really seems to make sense!
    – Ra1pid
    Aug 2 at 17:17










  • Wait for people verify the answer, and give more solutions, then decide about it!
    – user 108128
    Aug 2 at 17:20

















up vote
0
down vote













Use that $$sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$$






share|cite|improve this answer





















  • I used $sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$ to expand $sin(a+bi)$, however I got the issue after isolating $sinh(b)$.
    – Ra1pid
    Aug 2 at 17:13











  • So your problem is solved? That's nice!
    – Dr. Sonnhard Graubner
    Aug 2 at 17:18










  • Thanks Dr. Sonnhard Graubner
    – Ra1pid
    Aug 2 at 17:20











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










beginalign
sin(a+ib)
&=sin acos ib+cos asin ib\
&=sin acosh b+icos asinh b\
&= cos b + i sin b
endalign
so take real part and imaginary of sides give us
$$sin acosh b= cos b$$
$$cos asinh b= sin b$$
then squaring two equations and adding concludes
$$sin^2 acosh^2 b+cos^2 asinh^2 b=1$$
$$(1-cos^2 a)(1+sinh^2 b)+cos^2 asinh^2 b=1$$
which gives $colorbluecos^2 a = sinh^2 b$.






share|cite|improve this answer























  • Wow thanks this really seems to make sense!
    – Ra1pid
    Aug 2 at 17:17










  • Wait for people verify the answer, and give more solutions, then decide about it!
    – user 108128
    Aug 2 at 17:20














up vote
2
down vote



accepted










beginalign
sin(a+ib)
&=sin acos ib+cos asin ib\
&=sin acosh b+icos asinh b\
&= cos b + i sin b
endalign
so take real part and imaginary of sides give us
$$sin acosh b= cos b$$
$$cos asinh b= sin b$$
then squaring two equations and adding concludes
$$sin^2 acosh^2 b+cos^2 asinh^2 b=1$$
$$(1-cos^2 a)(1+sinh^2 b)+cos^2 asinh^2 b=1$$
which gives $colorbluecos^2 a = sinh^2 b$.






share|cite|improve this answer























  • Wow thanks this really seems to make sense!
    – Ra1pid
    Aug 2 at 17:17










  • Wait for people verify the answer, and give more solutions, then decide about it!
    – user 108128
    Aug 2 at 17:20












up vote
2
down vote



accepted







up vote
2
down vote



accepted






beginalign
sin(a+ib)
&=sin acos ib+cos asin ib\
&=sin acosh b+icos asinh b\
&= cos b + i sin b
endalign
so take real part and imaginary of sides give us
$$sin acosh b= cos b$$
$$cos asinh b= sin b$$
then squaring two equations and adding concludes
$$sin^2 acosh^2 b+cos^2 asinh^2 b=1$$
$$(1-cos^2 a)(1+sinh^2 b)+cos^2 asinh^2 b=1$$
which gives $colorbluecos^2 a = sinh^2 b$.






share|cite|improve this answer















beginalign
sin(a+ib)
&=sin acos ib+cos asin ib\
&=sin acosh b+icos asinh b\
&= cos b + i sin b
endalign
so take real part and imaginary of sides give us
$$sin acosh b= cos b$$
$$cos asinh b= sin b$$
then squaring two equations and adding concludes
$$sin^2 acosh^2 b+cos^2 asinh^2 b=1$$
$$(1-cos^2 a)(1+sinh^2 b)+cos^2 asinh^2 b=1$$
which gives $colorbluecos^2 a = sinh^2 b$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Aug 2 at 17:19


























answered Aug 2 at 17:15









user 108128

18.8k41544




18.8k41544











  • Wow thanks this really seems to make sense!
    – Ra1pid
    Aug 2 at 17:17










  • Wait for people verify the answer, and give more solutions, then decide about it!
    – user 108128
    Aug 2 at 17:20
















  • Wow thanks this really seems to make sense!
    – Ra1pid
    Aug 2 at 17:17










  • Wait for people verify the answer, and give more solutions, then decide about it!
    – user 108128
    Aug 2 at 17:20















Wow thanks this really seems to make sense!
– Ra1pid
Aug 2 at 17:17




Wow thanks this really seems to make sense!
– Ra1pid
Aug 2 at 17:17












Wait for people verify the answer, and give more solutions, then decide about it!
– user 108128
Aug 2 at 17:20




Wait for people verify the answer, and give more solutions, then decide about it!
– user 108128
Aug 2 at 17:20










up vote
0
down vote













Use that $$sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$$






share|cite|improve this answer





















  • I used $sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$ to expand $sin(a+bi)$, however I got the issue after isolating $sinh(b)$.
    – Ra1pid
    Aug 2 at 17:13











  • So your problem is solved? That's nice!
    – Dr. Sonnhard Graubner
    Aug 2 at 17:18










  • Thanks Dr. Sonnhard Graubner
    – Ra1pid
    Aug 2 at 17:20















up vote
0
down vote













Use that $$sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$$






share|cite|improve this answer





















  • I used $sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$ to expand $sin(a+bi)$, however I got the issue after isolating $sinh(b)$.
    – Ra1pid
    Aug 2 at 17:13











  • So your problem is solved? That's nice!
    – Dr. Sonnhard Graubner
    Aug 2 at 17:18










  • Thanks Dr. Sonnhard Graubner
    – Ra1pid
    Aug 2 at 17:20













up vote
0
down vote










up vote
0
down vote









Use that $$sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$$






share|cite|improve this answer













Use that $$sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 2 at 17:08









Dr. Sonnhard Graubner

66.6k32659




66.6k32659











  • I used $sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$ to expand $sin(a+bi)$, however I got the issue after isolating $sinh(b)$.
    – Ra1pid
    Aug 2 at 17:13











  • So your problem is solved? That's nice!
    – Dr. Sonnhard Graubner
    Aug 2 at 17:18










  • Thanks Dr. Sonnhard Graubner
    – Ra1pid
    Aug 2 at 17:20

















  • I used $sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$ to expand $sin(a+bi)$, however I got the issue after isolating $sinh(b)$.
    – Ra1pid
    Aug 2 at 17:13











  • So your problem is solved? That's nice!
    – Dr. Sonnhard Graubner
    Aug 2 at 17:18










  • Thanks Dr. Sonnhard Graubner
    – Ra1pid
    Aug 2 at 17:20
















I used $sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$ to expand $sin(a+bi)$, however I got the issue after isolating $sinh(b)$.
– Ra1pid
Aug 2 at 17:13





I used $sin(a+bi)=sin(a)cosh(b)+icos(a)sinh(b)$ to expand $sin(a+bi)$, however I got the issue after isolating $sinh(b)$.
– Ra1pid
Aug 2 at 17:13













So your problem is solved? That's nice!
– Dr. Sonnhard Graubner
Aug 2 at 17:18




So your problem is solved? That's nice!
– Dr. Sonnhard Graubner
Aug 2 at 17:18












Thanks Dr. Sonnhard Graubner
– Ra1pid
Aug 2 at 17:20





Thanks Dr. Sonnhard Graubner
– Ra1pid
Aug 2 at 17:20













 

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