I have become stuck while solving a trig identity. It is:


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!

  • $\begingroup$ try using the angle sum formula again to get rid of those pesky $2x$ arguments. $\endgroup$ – The Count May 31 '18 at 21:27

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 $\cos (3x)=4\cos^3x-3\cos x$

$$\iff1-4\sin^2(x)=\frac{4\cos^3x-3\cos x}{\cos x}\iff1-4\sin^2(x)=4\cos^2x-3$$


  • 1
    $\begingroup$ You are allowed to cross multiply if the two fractions are equal to each other, but isn't that what you have to show in the first place? I though you had to work on one side and get that expression equal to the other.PS, no downvote from me! $\endgroup$ – imranfat May 31 '18 at 21:30
  • $\begingroup$ @imranfat $\frac A B = \frac C D \iff AD=BC$ for $B\neq 0$ and $D\neq 0$. $\endgroup$ – gimusi May 31 '18 at 21:32
  • $\begingroup$ Thanks for your help, although could you please show how you got to your final equation from the second one please? I just don't understand how you got to the last equation $\endgroup$ – Owen Johnstone May 31 '18 at 21:33
  • $\begingroup$ Just run it in reverse. $\endgroup$ – marty cohen May 31 '18 at 21:34
  • $\begingroup$ @OwenJohnstone Simply note that $$(1-2\sin(x))(1+2\sin(x))=1-4sin^2 x$$ $\endgroup$ – gimusi May 31 '18 at 21:34

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} $


$\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.

  • $\begingroup$ I'm not sure what you mean by "run this in reverse". $\endgroup$ – Owen Johnstone May 31 '18 at 21:39
  • $\begingroup$ @OwenJohnstone From that formula for $\cos 2x$, compute $\cos 3x$. $\endgroup$ – J.G. May 31 '18 at 21:40

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


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}


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