Trouble proving the trigonometric identity $\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$ I have become stuck while solving a trig identity. It is:
$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$
I have simplified the left side as far as I can:
\begin{align}
\frac{1-2\sin(x)}{\sec(x)}
&=\frac{1-2\sin(x)}{1/\cos(x)}=(1-2\sin(x))\cos(x)\\
&=\cos(x)-2\sin(x)\cos(x)=\cos(x)-\sin(2x)
\end{align}
However, I'm not sure what to do on the right side. I know I can use a compound angle formula to break $\cos(3x)$ into $\cos(2x)\cos(x)-\sin(2x)\sin(x)$; however, I do not know where to go after that. My main problem is with the denominator of the right side, I can't figure out how to get rid of it, either by multiplying, or by using a trig identity. Any help in solving this identity would be greatly appreciated!
 A: We have that for $\cos x\neq 0$ and $\sin x \neq -\frac12$
$$ \frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}\iff(1-2\sin(x))(1+2\sin(x))=\frac{\cos (3x)}{\cos x}$$
then recall that $\cos (3x)=4\cos^3x-3\cos x$
$$\iff1-4\sin^2(x)=\frac{4\cos^3x-3\cos x}{\cos x}\iff1-4\sin^2(x)=4\cos^2x-3$$
$$\iff4=4(\cos^2x+\sin^2x)\iff4=4$$
A: First, in case you don't know how, I'll expand $\cos{3x}$.
$$\require{cancel}\begin{aligned}\cos{3x}&=\cos\left(x+\left(x+x\right)\right)\\&=\cos x\cos\left(x+x\right)-\sin x\sin\left(x+x\right)\\&=\cos{x}\left(\cos^{2}x-\sin^{2}x\right)-\sin{x}\left(2\sin{x}\cos{x}\right)\\&=\cos^3 x-\sin^{2}{x}\cos{x}-2\sin^{2}{x}\cos{x}\\&=\cos^3 x-3\sin^{2}{x}\cos{x}\\&=\cos^2 x\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}\left(1-\sin^2{x}\right)-3\sin^{2}{x}\cos{x}\\&=\cos{x}-\sin^2{x}\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}-4\sin^2{x}\cos{x}\end{aligned}$$
Let's rock!
$$\begin{aligned}\frac{1-2\sin{x}}{\sec{x}}&=\frac{1-2\sin{x}}{\frac{1}{\cos x}}\\&=\cos{x}\left(1-2\sin x\right)\\&=\cos x-2\sin{x}\cos{x}\\&=\cos x-2\sin{x}\cos{x}\cdot\frac{1+2\sin x}{1+2\sin x}\\&=\frac{\cos x\cancel{+2\sin{x}\cos{x}}\cancel{-2\sin{x}\cos{x}}-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos x-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos{3x}}{1+2\sin{x}}\end{aligned}$$
A: I would cross-multiply,
not worrying about
where functions are zero,
to get
$\begin{array}\\
\cos(x)(1-4\sin^2(x))
&=cos(3x)\\
&=\cos(2x)\cos(x)-\sin(2x)\sin(x)\\
&=\cos(2x)\cos(x)-2\cos(x)\sin^2(x)\\
\end{array}
$
or
$\begin{array}\\
1-4\sin^2(x)
&=\cos(2x)-2\sin^2(x)\\
\text{or}\\
1-2\sin^2(x)
&=\cos(2x)\\
\end{array}
$
which is well known.
Then run this in reverse
to get the original
equation.
A: Working on one side and finally obtaining the other side:
$$\frac{1-2sinx}{secx}$$
$$cosx-2sinxcosx$$
$$\frac{(cosx-2sinxcosx)(1+2sinx)}{1+2sinx}$$
$$\frac{cosx+2sinxcosx-2sinxcosx-4sin^2xcosx}{1+2sinx}$$
$$\frac{cosx-4(1-cos^2x)cosx}{1+2sinx}$$
Numerator is standard identity for the numerator of the RHS
A: The identity is equivalent to
$$
\cos3x=\cos x(1-4\sin^2x)
$$
(except for the values where the denominators vanish). The right-hand side can be rewritten as
$$
\cos x(\cos^2x+\sin^2x-4\sin^2x)=\cos^3x-3\cos x\sin^2x
$$
which is known to be the same as $\cos3x$: by De Moivre
\begin{align}
\cos 3x+i\sin3x
&=(\cos x+i\sin x)^3 \\
&=\cos^3x+3i\cos^2x\sin x+3i^2\cos x\sin^2x+i^3\sin^3x\\
&=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2x\sin x-\sin^3x)
\end{align}
Of course you can also use
\begin{align}
\cos3x
&=\cos(2x+x)\\
&=\cos2x\cos x-\sin2x\sin x\\
&=(\cos^2x-\sin^2x)\cos x-2\cos x\sin^2x\\
&=\cos^3x-3\cos x\sin^2x
\end{align}
