Proving $cos^5(a)=cos(a)-2sin^2(a)cos(a)$, using only Pythagorean identities












0














I came across a problem which said prove using only Pythagorean Identities and I got stumped. Some Insight would be helpful.



$$cos^5(a)=cos(a)-2sin^2(a)cos(a)$$



What I first did was factor the right side and used the main pythagorean identity to try and simplify but I don't really think it got me anywhere. My final step somehow got me to;



$$cos^2(a)-sin^2(a)=cos^4(a)$$



Where do I go from there or what could I have done differently in the start?






trigonometry







share|cite|improve this question




















  • 8




    The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
    – Blue
    19 hours ago










  • the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
    – Will Jagy
    19 hours ago










  • So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
    – Savvas Nicolaou
    19 hours ago










  • Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
    – Blue
    19 hours ago






  • 7




    If you were being asked to solve, then replacing $sin^2a$ with $1-cos^2a$ would allow you to rearrange and factor, getting this equation $$cos a,left(1-cos aright)^2left(1+cos aright)^2 = 0$$ From this, we can deduce that the relation only holds for integer multiples of $90^circ$.
    – Blue
    19 hours ago
















0














I came across a problem which said prove using only Pythagorean Identities and I got stumped. Some Insight would be helpful.



$$cos^5(a)=cos(a)-2sin^2(a)cos(a)$$



What I first did was factor the right side and used the main pythagorean identity to try and simplify but I don't really think it got me anywhere. My final step somehow got me to;



$$cos^2(a)-sin^2(a)=cos^4(a)$$



Where do I go from there or what could I have done differently in the start?






trigonometry







share|cite|improve this question




















  • 8




    The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
    – Blue
    19 hours ago










  • the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
    – Will Jagy
    19 hours ago










  • So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
    – Savvas Nicolaou
    19 hours ago










  • Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
    – Blue
    19 hours ago






  • 7




    If you were being asked to solve, then replacing $sin^2a$ with $1-cos^2a$ would allow you to rearrange and factor, getting this equation $$cos a,left(1-cos aright)^2left(1+cos aright)^2 = 0$$ From this, we can deduce that the relation only holds for integer multiples of $90^circ$.
    – Blue
    19 hours ago














0












0








0







I came across a problem which said prove using only Pythagorean Identities and I got stumped. Some Insight would be helpful.



$$cos^5(a)=cos(a)-2sin^2(a)cos(a)$$



What I first did was factor the right side and used the main pythagorean identity to try and simplify but I don't really think it got me anywhere. My final step somehow got me to;



$$cos^2(a)-sin^2(a)=cos^4(a)$$



Where do I go from there or what could I have done differently in the start?






trigonometry







share|cite|improve this question















I came across a problem which said prove using only Pythagorean Identities and I got stumped. Some Insight would be helpful.



$$cos^5(a)=cos(a)-2sin^2(a)cos(a)$$



What I first did was factor the right side and used the main pythagorean identity to try and simplify but I don't really think it got me anywhere. My final step somehow got me to;



$$cos^2(a)-sin^2(a)=cos^4(a)$$



Where do I go from there or what could I have done differently in the start?






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 19 hours ago









Blue

47.6k870151




47.6k870151










asked 20 hours ago









Savvas Nicolaou

866




866








  • 8




    The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
    – Blue
    19 hours ago










  • the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
    – Will Jagy
    19 hours ago










  • So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
    – Savvas Nicolaou
    19 hours ago










  • Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
    – Blue
    19 hours ago






  • 7




    If you were being asked to solve, then replacing $sin^2a$ with $1-cos^2a$ would allow you to rearrange and factor, getting this equation $$cos a,left(1-cos aright)^2left(1+cos aright)^2 = 0$$ From this, we can deduce that the relation only holds for integer multiples of $90^circ$.
    – Blue
    19 hours ago














  • 8




    The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
    – Blue
    19 hours ago










  • the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
    – Will Jagy
    19 hours ago










  • So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
    – Savvas Nicolaou
    19 hours ago










  • Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
    – Blue
    19 hours ago






  • 7




    If you were being asked to solve, then replacing $sin^2a$ with $1-cos^2a$ would allow you to rearrange and factor, getting this equation $$cos a,left(1-cos aright)^2left(1+cos aright)^2 = 0$$ From this, we can deduce that the relation only holds for integer multiples of $90^circ$.
    – Blue
    19 hours ago








8




8




The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
– Blue
19 hours ago




The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
– Blue
19 hours ago












the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
– Will Jagy
19 hours ago




the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
– Will Jagy
19 hours ago












So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
– Savvas Nicolaou
19 hours ago




So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
– Savvas Nicolaou
19 hours ago












Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
– Blue
19 hours ago




Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
– Blue
19 hours ago




7




7




If you were being asked to solve, then replacing $sin^2a$ with $1-cos^2a$ would allow you to rearrange and factor, getting this equation $$cos a,left(1-cos aright)^2left(1+cos aright)^2 = 0$$ From this, we can deduce that the relation only holds for integer multiples of $90^circ$.
– Blue
19 hours ago




If you were being asked to solve, then replacing $sin^2a$ with $1-cos^2a$ would allow you to rearrange and factor, getting this equation $$cos a,left(1-cos aright)^2left(1+cos aright)^2 = 0$$ From this, we can deduce that the relation only holds for integer multiples of $90^circ$.
– Blue
19 hours ago










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