avewright/proofwiki-math huggingface dataset:
Theorem: Triple Angle Formulas/Cosine

:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 3 \theta
      | r = \cos \paren {2 \theta + \theta}
}}
{{eqn | r = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta
      | c = Cosine of Sum
}}
{{eqn | r = \paren {\cos^2 \theta - \sin^2 \theta} \cos \theta - \paren {2 \sin \theta \cos \theta} \sin \theta
      | c = Double Angle Formula for Cosine and Double Angle Formula for Sine
}}
{{eqn | r = \cos^3 \theta - \sin^2 \theta \cos \theta - 2 \sin^2 \theta \cos \theta
      | c = 
}}
{{eqn | r = \cos^3 \theta - \paren {1 - \cos^2 \theta} \cos \theta - 2 \paren {1 - \cos^2 \theta} \cos \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \cos^3 \theta - \cos \theta + \cos^3 \theta - 2 \cos \theta + 2 \cos^3 \theta
      | c = multiplying out
}}
{{eqn | r = 4 \cos^3 \theta - 3 \cos \theta
      | c = gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Triple Angle Formulas/Cosine

:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \cos 3 \theta + i \sin 3 \theta
      | r = \paren {\cos \theta + i \sin \theta}^3
      | c = De Moivre's Formula
}}
{{eqn | r = \paren {\cos \theta}^3 + \binom 3 1 \paren {\cos \theta}^2 \paren {i \sin \theta}
      | c = Binomial Theorem
}}
{{eqn | o = 
      | ro=+
      | r = \binom 3 2 \paren {\cos \theta} \paren {i \sin \theta}^2 + \paren {i \sin \theta}^3
}}
{{eqn | r = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta + 3 i^2 \cos \theta \sin^2 \theta + i^3 \sin^3 \theta
      | c = substituting for binomial coefficients
}}
{{eqn | r = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta
      | c = $i^2 = -1$
}}
{{eqn | n = 1
      | r = \cos^3 \theta - 3 \cos \theta \sin^2 \theta 
}}
{{eqn | o = 
      | ro=+
      | r = i \paren {3 \cos^2 \theta \sin \theta - \sin^3 \theta}
      | c = rearranging
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \cos 3 \theta
      | r = \cos^3 \theta - 3 \cos \theta \sin^2 \theta
      | c = equating real parts in $(1)$
}}
{{eqn | r = \cos^3 \theta - 3 \cos \theta \paren {1 - \cos^2 \theta}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 4 \cos^3 \theta - 3 \cos \theta
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Triple Angle Formulas/Cosine

:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 3 \theta
      | r = \cos \paren {2 \theta + \theta}
}}
{{eqn | r = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta
      | c = Cosine of Sum
}}
{{eqn | r = \paren {2 \cos^2 \theta - 1} \cos \theta - \paren {2 \sin \theta \cos \theta} \sin \theta
      | c = Double Angle Formula for Cosine and Double Angle Formula for Sine
}}
{{eqn | r = 2 \cos^3 \theta - \cos \theta - 2 \paren {1 - \cos^2 \theta} \cos \theta
      | c = Sum of Squares of Sine and Cosine after algebra
}}
{{eqn | r = 2 \cos^3 \theta - \cos \theta - 2 \cos \theta + 2 \cos^3 \theta
      | c = multiplying everything out
}}
{{eqn | r = 4 \cos^3 \theta - 3 \cos \theta
      | c = gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Triple Angle Formulas/Tangent

:$\tan 3 \theta = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 3 \theta
      | r = \frac {\sin 3 \theta} {\cos 3 \theta}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}
      | c = Triple Angle Formula for Sine and Triple Angle Formula for Cosine
}}
{{eqn | r = \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}\frac {\cos^3 \theta} {\cos^3 \theta}
      | c = 
}}
{{eqn | r = \frac {\frac {3 \tan \theta} {\cos^2 \theta} - 4 \tan^3 \theta} {4 - \frac {3 \cos \theta} {\cos^3 \theta} }
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {3 \tan \theta \sec^2 \theta - 4 \tan^3 \theta} {4 - 3 \sec^2 \theta}
      | c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {3 \tan \theta \paren {1 + \tan^2 \theta} - 4 \tan^3 \theta} {4 - 3 \paren {1 + \tan^2 \theta} }
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \frac {3 \tan \theta + 3 \tan^3 \theta - 4 \tan^3 \theta} {4 - 3 - 3 \tan^2 \theta}
      | c = multiplying out
}}
{{eqn | r = \frac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}
      | c = gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Triple Angle Formulas/Tangent

:$\tan 3 \theta = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$

Proof:
Let $\theta$ be such that $\tan 2 \theta$ is defined.
Then:
{{begin-eqn}}
{{eqn | l = \tan 3 \theta
      | r = \dfrac {\tan \theta + \tan 2 \theta} {1 - \tan \theta \tan 2 \theta}
      | c = Tangent of Sum
}}
{{eqn | r = \dfrac {\tan \theta + \dfrac {2 \tan \theta} {1 - \tan^2 \theta} } {1 - \tan \theta \dfrac {2 \tan \theta} {1 - \tan^2 \theta} }
      | c = Double Angle Formula for Tangent
}}
{{eqn | r = \dfrac {\tan \theta \paren {1 - \tan^2 \theta} + 2 \tan \theta} {\paren {1 - \tan^2 \theta} - 2 \tan^2 \theta}
      | c = simplifying
}}
{{eqn | r = \dfrac {\tan \theta - \tan^3 \theta + 2 \tan \theta} {1 - \tan^2 \theta - 2 \tan^2 \theta}
      | c = simplifying
}}
{{eqn | r = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Now suppose $\theta$ is such that $\tan 2 \theta$ is not defined.
Then:
:$2 \theta = \dfrac \pi 2 + n \pi$
for some integer $n$.
Hence:
:$\theta = \dfrac \pi 4 + \dfrac {n \pi} 2$
For $n$ even we will then have:
:$\tan \theta = 1$
and:
:$\tan 3 \theta = -1$
For $n$ odd we will then have:
:$\tan \theta = -1$
and:
:$\tan 3 \theta = 1$
It is then directly verified that the '''Triple Angle Formula for Tangent''' holds for these special cases where $\tan 2 \theta$ is undefined.
{{qed}}
Theorem: Triple Angle Formulas/Tangent

:$\tan 3 \theta = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$

Proof:
From Tangent of Sum of Three Angles:
{{:Tangent of Sum of Three Angles}}
The result follows by setting $\theta = A = B = C$.
{{qed}}
Theorem: Quadruple Angle Formulas/Sine

:$\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \map \sin {4 \theta}
      | r = \map \sin {3 \theta + \theta}
}}
{{eqn | r = \sin 3 \theta \cos \theta + \cos 3 \theta \sin \theta
      | c = Sine of Sum
}}
{{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \cos \theta + \paren {4 \cos^3 \theta - 3 \cos \theta} \sin \theta
      | c = Triple Angle Formula for Sine and Triple Angle Formula for Cosine
}}
{{eqn | r = 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos^3 \theta \sin \theta - 3 \cos \theta \sin \theta
      | c = 
}}
{{eqn | r = 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos \theta \paren {1 - \sin^2 \theta} \sin \theta - 3 \cos \theta \sin \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quadruple Angle Formulas/Sine

:$\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \cos 4 \theta + i \sin 4 \theta
      | r = \paren {\cos \theta + i \sin \theta}^4
      | c = De Moivre's Formula
}}
{{eqn | r = \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2
}}
{{eqn | o = 
      | ro=+
      | r = \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4
      | c = Binomial Theorem
}}
{{eqn | r = \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta
      | c = substituting for binomial coefficients
}}
{{eqn | o = 
      | ro=-
      | r = 4 i \cos \theta \sin^3 \theta + \sin^4 \theta
      | c = and using $i^2 = -1$
}}
{{eqn | n = 1
      | r = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta
}}
{{eqn | o = 
      | ro=+
      | r = 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta
      | c = rearranging
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \sin 4 \theta
      | r = 4 \cos^3 \theta \sin \theta - 4 \cos \theta \sin^3 \theta
      | c = equating imaginary parts in $(1)$
}}
{{eqn | r = 4 \cos \theta \paren {1 - \sin^2 \theta} \sin \theta - 4 \cos \theta \sin^3 \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 4 \cos \theta \sin \theta - 8 \sin^3 \theta \cos \theta
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quadruple Angle Formulas/Sine

:$\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \sin {4 \theta}
      | r = \map \sin {2 \times 2 \theta}
}}
{{eqn | r = 2 \sin 2 \theta \cos 2 \theta
      | c = Double Angle Formula for Sine
}}
{{eqn | r = 2 \paren {2 \sin \theta \cos \theta} \paren {\cos^2 \theta - \sin^2 \theta}
      | c = Double Angle Formula for Sine, Double Angle Formula for Cosine
}}
{{eqn | r = 4 \sin \theta \cos^3 \theta - 4 \sin^3 \theta \cos \theta
      | c = Distributive Laws of Arithmetic
}}
{{eqn | r = 4 \sin \theta \paren {1 - \sin^2 \theta} \cos \theta - 4 \sin^3 \theta \cos \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Quadruple Angle Formulas/Cosine

:$\cos 4 \theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 1$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 4 \theta
      | r = \cos \paren {2 \theta + 2 \theta}
}}
{{eqn | r = \cos 2 \theta \cos 2 \theta - \sin 2 \theta \sin 2 \theta
      | c = Cosine of Sum
}}
{{eqn | r = \paren {\cos^2 \theta - \sin^2 \theta} \paren {\cos^2 \theta - \sin^2 \theta} - \paren {2 \sin \theta \cos \theta} \paren {2 \sin \theta \cos \theta}
      | c = Double Angle Formulas
}}
{{eqn | r = \cos^4 \theta - 2 \cos^2 \theta \sin^2 \theta + \sin^4 \theta - 4 \cos^2 \theta \sin^2 \theta
      | c = multiplying out
}}
{{eqn | r = \cos^4 \theta - 2 \cos^2 \theta \paren {1 - \cos^2 \theta} + \paren {1 - \cos^2 \theta}^2 - 4 \cos^2 \theta \paren {1 - \cos^2 \theta}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 8 \cos^4 \theta - 8 \cos^2 \theta + 1
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quadruple Angle Formulas/Cosine

:$\cos 4 \theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 1$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \cos 4 \theta + i \sin 4 \theta
      | r = \paren {\cos \theta + i \sin \theta}^4
      | c = De Moivre's Formula
}}
{{eqn | r = \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2
}}
{{eqn | o = 
      | ro= +
      | r = \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4
      | c = Binomial Theorem
}}
{{eqn | r = \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta
      | c = substituting for binomial coefficients
}}
{{eqn | o = 
      | ro= -
      | r = 4 i \cos \theta \sin^3 \theta + \sin^4 \theta
      | c = and using $i^2 = -1$
}}
{{eqn | n = 1
      | r = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta
}}
{{eqn | o = 
      | ro= +
      | r = 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta
      | c = rearranging
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \cos 4 \theta
      | r = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta
      | c = equating real parts in $(1)$
}}
{{eqn | r = \cos^4 \theta - 6 \cos^2 \theta \paren {1 - \cos^2 \theta} + \paren {1 - \cos^2 \theta}^2 
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 8 \cos^4 \theta - 8 \cos^2 \theta + 1
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quadruple Angle Formulas/Tangent

:$\tan 4 \theta = \dfrac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 4 \theta
      | r = \frac {\sin 4 \theta} {\cos 4 \theta}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta} {8 \cos^4 \theta - 8 \cos^2 \theta + 1}
      | c = Quadruple Angle Formula for Sine and Quadruple Angle Formula for Cosine
}}
{{eqn | r = \frac {4 \frac {\tan \theta} {\cos^2 \theta} - 8 \tan^3 \theta} {8 - \frac 8 {\cos^2 \theta} + \frac 1 {\cos^4 \theta} }
      | c = dividing top and bottom by $\cos^4 \theta$
}}
{{eqn | r =  \frac {4 \tan \theta \sec^2 \theta - 8 \tan^3 \theta} {8 - 8 \sec^2 \theta + \sec^4 \theta}
      | c = Secant is Reciprocal of Cosine
}}
{{eqn | r =  \frac {4 \tan \theta \paren {1 + \tan^2 \theta} - 8 \tan^3 \theta} {8 - 8 \paren {1 + \tan^2 \theta} + \paren {1 + \tan^2 \theta}^2}
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r =  \frac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quintuple Angle Formulas/Sine

:$\sin 5 \theta = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 5 \theta
      | r = \map \sin {3 \theta + 2 \theta}
}}
{{eqn | r = \sin 3 \theta \cos 2 \theta + \cos 3 \theta \sin 2 \theta
      | c = Sine of Sum
}}
{{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \cos 2 \theta + \paren {4 \cos^3 \theta - 3 \cos \theta} \sin 2 \theta
      | c = Triple Angle Formulas
}}
{{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {\cos^2 \theta - \sin^2 \theta} + \paren {4 \cos^3 \theta - 3 \cos \theta} 2 \sin \theta \cos \theta
      | c = Double Angle Formulas
}}
{{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {\cos^2 \theta - \sin^2 \theta} + 8 \cos^4 \theta \sin \theta - 6 \cos^2 \theta \sin \theta
      | c = multiplying out
}}
{{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {1 - 2 \sin^2 \theta} + 8 \paren {1 - \sin^2 \theta}^2 \sin \theta - 6 \paren {1 - \sin^2 \theta} \sin \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin ^5 \theta
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quintuple Angle Formulas/Sine

:$\sin 5 \theta = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \cos 5 \theta + i \sin 5 \theta
      | r = \paren {\cos \theta + i \sin \theta}^5
      | c = De Moivre's Formula
}}
{{eqn | r = \paren {\cos \theta}^5 + \binom 5 1 \paren {\cos \theta}^4 \paren {i \sin \theta} + \binom 5 2 \paren {\cos \theta}^3 \paren {i \sin \theta}^2
}}
{{eqn | o = 
      | ro=+
      | r = \binom 5 3 \paren {\cos \theta}^2 \paren {i \sin \theta}^3 + \binom 5 4 \paren {\cos \theta} \paren {i \sin \theta}^4 + \paren {i \sin \theta}^5
      | c = Binomial Theorem
}}
{{eqn | r = \cos^5 \theta + 5 i \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta
      | c = substituting for binomial coefficients
}}
{{eqn | o = 
      | ro=-
      | r = 10 i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta
      | c = and using $i^2 = -1$
}}
{{eqn | n = 1
      | r = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta
}}
{{eqn | o = 
      | ro=+
      | r = i \paren {5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta}
      | c = rearranging
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \sin 5 \theta
      | r = 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta
      | c = equating imaginary parts in $(1)$
}}
{{eqn | r = 5 \paren {1 - \sin^2 \theta}^2 \sin \theta - 10 \paren {1 - \sin^2 \theta} \sin^3 \theta + \sin^5 \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quintuple Angle Formulas/Cosine

:$\cos 5 \theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 5 \theta
      | r = \map \cos {4 \theta + \theta}
}}
{{eqn | r = \cos 4 \theta \cos \theta - \sin 4 \theta \sin \theta
      | c = Cosine of Sum
}}
{{eqn | r = \paren {8 \cos^4 \theta - 8 \cos^2 \theta + 1} \cos \theta - \paren {4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta} \sin \theta
      | c = Quadruple Angle Formulas
}}
{{eqn | r = 8 \cos^5 \theta - 8 \cos^3 \theta + \cos \theta - 4 \sin^2 \theta \cos \theta - 8 \sin^4 \theta \cos \theta
      | c = multiplying out
}}
{{eqn | r = 8 \cos^5 \theta - 8 \cos^3 \theta + \cos \theta - 4 \paren {1 - \cos^2 \theta} \cos \theta - 8 \paren {1 - \cos^2 \theta}^2 \cos \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quintuple Angle Formulas/Cosine

:$\cos 5 \theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \cos 5 \theta + i \sin 5 \theta
      | r = \paren {\cos \theta + i \sin \theta}^5
      | c = De Moivre's Formula
}}
{{eqn | r = \paren {\cos \theta}^5 + \binom 5 1 \paren {\cos \theta}^4 \paren {i \sin \theta} + \binom 5 2 \paren {\cos \theta}^3 \paren {i \sin \theta}^2
      | c = Binomial Theorem
}}
{{eqn | o = 
      | ro=+
      | r = \binom 5 3 \paren {\cos \theta}^2 \paren {i \sin \theta}^3 + \binom 5 4 \paren {\cos \theta} \paren {i \sin \theta}^4 + \paren {i \sin \theta}^5
}}
{{eqn | r = \cos^5 \theta + 5 i \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta
      | c = substituting for binomial coefficients
}}
{{eqn | o = 
      | ro=-
      | r = 10 i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta
      | c = and using $i^2 = -1$
}}
{{eqn | n = 1
      | r = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta
}}
{{eqn | o = 
      | ro=+
      | r = i \paren {5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta}
      | c = rearranging
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \cos 5 \theta
      | r = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta
      | c = equating real parts in $(1)$
}}
{{eqn | r = \cos^5 \theta - 10 \cos^3 \theta \paren {1 - \cos^2 \theta} + 5 \cos \theta \paren {1 - \cos^2 \theta}^2
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Quintuple Angle Formulas/Tangent

:$\tan 5 \theta = \dfrac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 5 \theta
      | r = \frac {\sin 5 \theta} {\cos 5 \theta}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta} {16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta}
      | c = Quintuple Angle Formulas
}}
{{eqn | r = \frac {5 \frac {\tan \theta} {\cos^4 \theta} - 20 \frac {\tan^3 \theta} {\cos^2 \theta} + 16 \tan^5 \theta} {16 \cos \theta - \frac {20} {\cos^2 \theta} + \frac 5 {\cos^4 \theta} }
      | c = dividing top and bottom by $\cos^5 \theta$
}}
{{eqn | r = \frac {5 \tan \theta \sec^4 \theta - 20 \tan^3 \theta \sec^2 \theta + 16 \tan^5 \theta} {16 - 20 \sec^2 \theta + 5 \sec^4 \theta}
      | c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {5 \tan \theta \paren {1 + \tan^2 \theta}^2 - 20 \tan^3 \theta \paren {1 + \tan^2 \theta} + 16 \tan^5 \theta} {16 - 20 \paren {1 + \tan^2 \theta} + 5 \paren {1 + \tan^2 \theta}^2}
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r =  \frac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}
      | c = multiplying out and gathering terms
}}
{{end-eqn}}
{{qed}}
Theorem: Power Reduction Formulas/Sine Squared

:$\sin^2 x = \dfrac {1 - \cos 2 x} 2$

Proof:
{{begin-eqn}}
{{eqn | l = 1 - 2 \sin^2 x
      | r = \cos 2 x
      | c = {{Corollary|Double Angle Formula for Cosine|2}}
}}
{{eqn | ll= \leadsto
      | l = \sin^2 x
      | r = \frac {\cos 2 x - 1} {-2}
      | c = solving for $\sin^2x$
}}
{{eqn | r = \frac {1 - \cos 2 x} 2
      | c = multiplying top and bottom by $-1$ and rearranging terms
}}
{{end-eqn}}
{{qed}}
Theorem: Power Reduction Formulas/Cosine Squared

:$\cos^2 x = \dfrac {1 + \cos 2 x} 2$

Proof:
{{begin-eqn}}
{{eqn | l = 2 \cos^2 x - 1
      | r = \cos 2 x
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | l = \cos^2 x
      | r = \frac {1 + \cos 2 x} 2
      | c = solving for $\cos^2 x$
}}
{{end-eqn}}
{{qed}}
Theorem: Power Reduction Formulas/Cosine Squared

:$\cos^2 x = \dfrac {1 + \cos 2 x} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {1 + \cos 2 x} 2
      | r = \dfrac 1 2 \paren {1 + \dfrac {e^{2 i x} + e^{-2 i x} } 2}
      | c = Euler's Cosine Identity
}}
{{eqn | r = \dfrac 1 4 \paren {e^{2 i x} + 2 + e^{-2 i x} }
      | c = simplifying
}}
{{eqn | r = \dfrac 1 4 \paren {e^{2 i x} + 2 \paren {e^{i x} } \paren {e^{-i x} } + e^{-2 i x} }
      | c = 
}}
{{eqn | r = \paren {\dfrac {e^{i x} + e^{-i x} } 2}^2
      | c = Square of Sum
}}
{{eqn | r = \cos^2 x
      | c = Euler's Cosine Identity
}}
{{end-eqn}}
{{qed}}
Theorem: Power Reduction Formulas/Tangent Squared

:$\tan^2x = \dfrac {1 - \cos2x} {1 + \cos2x}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan^2 x
      | r = \frac {\sin^2 x} {\cos^2 x}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\frac {1 - \cos 2 x} 2} {\frac {\cos 2 x + 1} 2}
      | c = Square of Sine and Square of Cosine
}}
{{eqn | r = \frac {1 - \cos 2 x} {1 + \cos 2 x}
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Category:Tangent Function
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Theorem: Power Reduction Formulas/Sine Cubed

:$\sin^3 x = \dfrac {3 \sin x - \sin 3 x} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 3 x
      | r = 3 \sin x - 4 \sin^3 x
      | c = Triple Angle Formula for Sine
}}
{{eqn | ll= \leadsto
      | l = 4 \sin^3 x
      | r = 3 \sin x - \sin 3 x
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \sin^3 x
      | r = \frac {3 \sin x - \sin 3 x} 4
      | c = dividing both sides by $4$
}}
{{end-eqn}}
{{qed}}
Theorem: Power Reduction Formulas/Sine Cubed

:$\sin^3 x = \dfrac {3 \sin x - \sin 3 x} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin^3 x
      | r = \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^3
      | c = Euler's Sine Identity
}}
{{eqn | r = \frac {\paren {e^{i x} - e^{-i x} }^3} {8 i^3}
      | c = rearranging
}}
{{eqn | r = -\frac 1 {8 i} \paren {\paren {e^{i x} }^3 - 3 \paren {e^{i x} }^2 \paren {e^{-i x} } + 3 \paren {e^{i x} } \paren {e^{-i x} }^2 - \paren {e^{-i x} }^3}
      | c = multiplying out
}}
{{eqn | r = -\frac 1 {8 i} \paren {e^{3 i x} - 3 e^{i x} + 3 e^{-i x} - e^{-3 i x} }
      | c = simplifying
}}
{{eqn | r = \frac 3 4 \paren {\frac {e^{i x} - e^{-i x} } {2 i} } - \frac 1 4 \paren {\frac {e^{3 i x} - e^{-3 i x} } {2 i} }
      | c = gathering terms
}}
{{eqn | r = \frac {3 \sin x - \sin 3 x} 4
      | c = Euler's Sine Identity
}}
{{end-eqn}}
{{qed}}
Theorem: Power Reduction Formulas/Cosine Cubed

:$\cos^3 x = \dfrac {3 \cos x + \cos 3 x} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 3 x
      | r = 4 \cos^3 x - 3 \cos x
      | c = Triple Angle Formula for Cosine
}}
{{eqn | ll= \leadsto
      | l = 4 \cos^3 x
      | r = 3 \cos x + \cos 3 x
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \cos^3 x
      | r = \dfrac {3 \cos x + \cos 3 x} 4
      | c = dividing both sides by $4$
}}
{{end-eqn}}
{{qed}}
Theorem: Power Reduction Formulas/Sine to 4th

:$\sin^4 x = \dfrac {3 - 4 \cos 2 x + \cos 4 x} 8$

Proof:
{{begin-eqn}}
{{eqn | l = \sin^4 x
      | r = \paren {\sin^2 x}^2
}}
{{eqn | r = \paren {\frac {1 - \cos 2 x} 2}^2
      | c = Square of Sine
}}
{{eqn | r = \frac {1 - 2 \cos 2 x + \cos^2 2 x} 4
      | c = multiplying out
}}
{{eqn | r = \frac {1 - 2 \cos 2 x + \frac {1 + \cos 4 x} 2} 4
      | c = Square of Cosine
}}
{{eqn | r = \frac {2 - 4 \cos 2 x + 1 + \cos 4 x} 8
      | c = multiplying top and bottom by $2$
}}
{{eqn | r = \frac {3 - 4 \cos 2 x + \cos 4 x} 8
      | c = rearrangement
}}
{{end-eqn}}
{{qed}}