Proving $cos^5(a)=cos(a)-2sin^2(a)cos(a)$, using only Pythagorean identities
0
$begingroup$
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
edited Jan 3 at 1:52
Blue
48.8k870156
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
$endgroup$
|
show 3 more comments
0
$begingroup$
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
edited Jan 3 at 1:52
Blue
48.8k870156
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
$endgroup$
8
$begingroup$
The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
$endgroup$
– Blue
Jan 3 at 1:55
$begingroup$
the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
$endgroup$
– Will Jagy
Jan 3 at 1:56
$begingroup$
So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
$endgroup$
– Savvas Nicolaou
Jan 3 at 1:57
$begingroup$
Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
$endgroup$
– Blue
Jan 3 at 1:58
7
$begingroup$
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 $90^circ$.
$endgroup$
– Blue
Jan 3 at 2:05
|
show 3 more comments
0
0
0
$begingroup$
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
edited Jan 3 at 1:52
Blue
48.8k870156
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
$endgroup$
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
edited Jan 3 at 1:52
Blue
48.8k870156
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
edited Jan 3 at 1:52
Blue
48.8k870156
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
edited Jan 3 at 1:52
Blue
48.8k870156
edited Jan 3 at 1:52
Blue
48.8k870156
edited Jan 3 at 1:52
Blue
48.8k870156
48.8k870156
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
asked Jan 3 at 1:48
Savvas NicolaouSavvas Nicolaou
867
867
8
$begingroup$
The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
$endgroup$
– Blue
Jan 3 at 1:55
$begingroup$
the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
$endgroup$
– Will Jagy
Jan 3 at 1:56
$begingroup$
So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
$endgroup$
– Savvas Nicolaou
Jan 3 at 1:57
$begingroup$
Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
$endgroup$
– Blue
Jan 3 at 1:58
7
$begingroup$
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 $90^circ$.
$endgroup$
– Blue
Jan 3 at 2:05
|
show 3 more comments
8
$begingroup$
The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
$endgroup$
– Blue
Jan 3 at 1:55
$begingroup$
the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
$endgroup$
– Will Jagy
Jan 3 at 1:56
$begingroup$
So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
$endgroup$
– Savvas Nicolaou
Jan 3 at 1:57
$begingroup$
Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
$endgroup$
– Blue
Jan 3 at 1:58
7
$begingroup$
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 $90^circ$.
$endgroup$
– Blue
Jan 3 at 2:05
8
8
$begingroup$
The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
$endgroup$
– Blue
Jan 3 at 1:55
$begingroup$
The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
$endgroup$
– Blue
Jan 3 at 1:55
$begingroup$
the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
$endgroup$
– Will Jagy
Jan 3 at 1:56
$begingroup$
the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
$endgroup$
– Will Jagy
Jan 3 at 1:56
$begingroup$
So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
$endgroup$
– Savvas Nicolaou
Jan 3 at 1:57
$begingroup$
So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
$endgroup$
– Savvas Nicolaou
Jan 3 at 1:57
$begingroup$
Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
$endgroup$
– Blue
Jan 3 at 1:58
$begingroup$
Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
$endgroup$
– Blue
Jan 3 at 1:58
7
7
$begingroup$
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 $90^circ$.
$endgroup$
– Blue
Jan 3 at 2:05
$begingroup$
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 $90^circ$.
$endgroup$
– Blue
Jan 3 at 2:05
|
show 3 more comments
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8
$begingroup$
The relation doesn't hold for, say, $a = 45^circ$. (The right-hand side vanishes, but the left-hand side does not.)
$endgroup$
– Blue
Jan 3 at 1:55
$begingroup$
the thing you wish to prove is false. Let $a = pi/4,$ so that $cos a = sin a = 1/sqrt 2,$ especially $2 sin^2 a = 1$
$endgroup$
– Will Jagy
Jan 3 at 1:56
$begingroup$
So, If the book say's prove, It can be unprovable? A bit of a useless problem. Perhaps there is a typo.
$endgroup$
– Savvas Nicolaou
Jan 3 at 1:57
$begingroup$
Are you being asked to prove that the equation holds for all $a$, or to solve it for values of $a$ that work?
$endgroup$
– Blue
Jan 3 at 1:58
7
$begingroup$
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 $90^circ$.
$endgroup$
– Blue
Jan 3 at 2:05