Trigonometry (Identity)

$\def\D{\displaystyle}$

Example 1 

Prove that $\D \sin^4x+\cos^4x=\frac{1}{4}(3+\cos 4x).$

$\D \begin{array}{|rl|}\hline (x+y)^2&=x^2+y^2+2xy \\ \sin^2 x&=\frac{1-\cos 2x}{2} \\ \sin^2 x+\cos^2 x&=1\\ 2\sin x\cos x&=\sin 2x\\ \hline \end{array} $

Proof: 

\begin{eqnarray*}
\left(\sin^2x+\cos^2x\right)^2&=&\sin^4x+\cos^4x+2\sin^2x\cos^2x\\
1^2&=&\sin^4x+\cos^4x+\frac{1}{2}(2\sin x\cos x)^2\\
1&=&\sin^4x+\cos^4x+\frac{1}{2}\sin^22x\\
&=&\sin^4x+\cos^4x+\frac{1}{2}\times \frac{1-\cos4x}{2}\\
\sin^4x+\cos^4x&=&1-\left( \frac{1}{4}-\frac{\cos4x}{4}\right) \\
&=&\frac{3}{4}+\frac{\cos4x}{4}
\end{eqnarray*}
Hence $\D \sin^4x+\cos^4x=\frac{1}{4}\left(3+\cos 4x\right).$

Example 2 

If $\D \sin x+\cos x=a,$ then show that $\D \sin^6x+\cos^6x=\frac{1}{4}\left(4-3\left(a^2-1\right)^2\right).$

$\D\begin{array}{|rcl|}\hline
(x+y)^3&=&x^3+y^3+3xy(x+y)\\
(x+y)^2&=&x^2+y^2+2xy\\
1&=&\sin^2x+\cos^2x\\ \hline
\end{array}$

Proof: 

\begin{eqnarray*}
a^2&=&\left(\sin x+\cos x\right)^2\\
&=&\sin^2x+\cos^2x+2\sin x\cos x\\
&=&1+2\sin x\cos x\\
\sin x\cos x&=&\frac{a^2-1}{2}
\end{eqnarray*}
\begin{eqnarray*}
\left(\sin^2x+\cos^2x\right)^3
&=&\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\\
&&+3\sin^2x\cos^2x\left(\sin^2x+\cos^2x\right)\\
1^3&=&\sin^6x+\cos^6x+3(\sin x\cos x)^2\times 1\\
1&=&\sin^6x+\cos^6x+3\left(\frac{a^2-1}{2}\right)^2\\
\sin^6x+\cos^6x&=&1-\frac{3}{4}(a^2-1)^2\\
&=&\frac{1}{4}[4-3(a^2-1)^2]
\end{eqnarray*}

Example 3 

If $\D \cos x-\sin x=\sqrt{2}\sin x$, show that $\D cos x+\sin x=\sqrt{2}\cos x.$

Proof: 

$\D \begin{array}{lrll}
&\cos x&=\sin x+\sqrt{2}\sin x&\cdots (1)\\
(1)\times \sqrt{2}:&  \sqrt{2}\cos x&=\sqrt{2}\sin x+2\sin x&\cdots (2)\\
(2)-(1):& \sqrt{2}\cos x-\cos x&=\sin x &
\end{array}$
Hence $\D \sqrt{2}\cos x=\sin x+\cos x.$

Example 4

Prove that \[\frac{\cos 3x+\sin 3x}{\cos x-\sin x}=1+2\sin 2x.\]
$\D \begin{array}{|rl|}\hline
\cos x-\cos y&\D =-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}\\
\sin x+\sin y&=\D \quad 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}\\ \hline
\end{array}$

Proof:

$\D \begin{array}{rll}
\cos 3x-\cos x&=-2\sin \frac{3x+x}{2}\sin \frac{3x-x}{2}\\
&=-2\sin 2x\sin x&\cdots (1)\\
\sin 3x+\sin x&=2\sin \frac{3x+x}{2}\cos\frac{3x-x}{2}\\
&=2\sin2x\cos x&\cdots (2)
\end{array}$

(1)+(2): $\D \cos 3x+\sin 3x-(\cos x-\sin x) =2\sin 2x(\cos x-\sin x).$
\[\div (\cos x-\sin x): \frac{\cos 3x+\sin 3x}{\cos x-\sin x}-1=2\sin x.\]
Therefore \[\frac{\cos 3x+\sin 3x}{\cos x-\sin x}=1+2\sin x.\]