Proving $cos^5(a)=cos(a)-2sin^2(a)cos(a)$, using only Pythagorean identities
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 19 hours ago
Blue
47.6k870151
asked 20 hours ago
Savvas Nicolaou
866
|
show 3 more comments
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 19 hours ago
Blue
47.6k870151
asked 20 hours ago
Savvas Nicolaou
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
|
show 3 more comments
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 19 hours ago
Blue
47.6k870151
asked 20 hours ago
Savvas Nicolaou
866
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 19 hours ago
Blue
47.6k870151
asked 20 hours ago
Savvas Nicolaou
866
edited 19 hours ago
Blue
47.6k870151
asked 20 hours ago
Savvas Nicolaou
866
edited 19 hours ago
Blue
47.6k870151
edited 19 hours ago
Blue
47.6k870151
edited 19 hours ago
Blue
47.6k870151
47.6k870151
asked 20 hours ago
Savvas Nicolaou
866
asked 20 hours ago
Savvas Nicolaou
866
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
|
show 3 more comments
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
|
show 3 more comments
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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