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$$\frac{1}{\tan (x)(1+\cos(2x))} = \csc(2x)$$

I really don't know what to do with denominator. Sure, I can use the double angle formula for cosine, and get:

$$\frac{1}{\tan(x)(2 - 2\sin^2(x))} = \csc(2x)$$

But what's next?

TravisJ
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Jack
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4 Answers4

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$$\frac{1}{\color{#0F0}{\tan x}(\color{#00F}{1}+\color{#F00}{\cos 2x)}}=\frac{\color{#0f0}{\cos x}}{\color{#0f0}{\sin x}(\color{#00F}{\sin^2 x+\cos^2 x}+\color{#F00}{\cos^2x-\sin^2 x})}$$

$$=\frac{1}{2\sin x\cos x}=\frac{1}{\sin 2x}=\csc 2x$$

user26486
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There's a better form of the double-angle formula for this purpose: remember there are three different forms of the cosine double-angle formula, $$ \cos{2x} = \cos^2{x}-\sin^2{x} = 2\cos^2{x}-1=1-2\sin^2{x}. $$ Hence $$ 1+\cos{2x} = 2\cos^2{x}. $$ Now, you also know $\tan{x}=\frac{\sin{x}}{\cos{x}}$ (right?), so you have $$ \frac{1}{\tan{x}(1+\cos{2x})} = \frac{\cos{x}}{\sin{x} \cdot 2\cos^2{x}} = \frac{1}{2\sin{x}\cos{x}} = \frac{1}{\sin{2x}} = \csc{2x}, $$ using the sine double-angle formula.

Chappers
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We have to prove $$\frac{1}{\tan(x)(1+\cos(2x))}=\csc(2x)=\frac{1}{\sin(2x)}$$ Multiply both sides by $\tan(x)$ and apply $\sin(2x)=2\sin(x)\cos(x)$ and you arrive at $$\frac{1}{2{\cos^2(x)}}=\frac{1}{1+\cos(2x)}$$ raise both sides to the power of -1, and divide both sides by two and arrive at $$\cos^2(x)=\frac{1}{2}+\frac{1}{2}\cos(2x)$$ which is a well-known result that can be easily derived using Euler's formula.

JacksonFitzsimmons
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  • One should not work with both sides of an equality: it is far better practice to proceed by logical steps from one side to the other. If you start with a false premise, you can actually prove anything. – Chappers Jun 23 '15 at 01:36
  • How would one put that principle into practice? For example, how could I start with sin(x)=cos(x) and deduce that 1/x=2^x? – JacksonFitzsimmons Jun 23 '15 at 01:58
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    Well, first you put $x=0$ and discover that $0=\sin{0} = \cos{0} = 1$. Then $2=1+1=0+0=0$, and so $1/x=0/x=0=0^x=(0+0)^x = (1+1)^x=2^x$. It's all so much nonsense, but that's the point. See also this and its linked questions. – Chappers Jun 23 '15 at 02:55
  • Thanks! That's really interesting. I won't make that mistake again. – JacksonFitzsimmons Jun 23 '15 at 02:59
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Here's one more -- this is being a bit "cute", but it is related to another trigonometric relation you should be meeting soon, if you haven't already.

$$ \frac{1}{\tan (x)(1+\cos(2x))} \ = \ \csc(2x) \ \ \Rightarrow \ \ \frac{1}{\tan (x)(1+\cos(2x))} \ = \ \frac{1}{\sin(2x)} $$

$$ \Rightarrow \ \ \tan (x) \ = \ \frac{\sin(2x)}{1+\cos(2x)} \ \ . $$

If we now identify $ \ 2 \ x \ $ as the angle $ \ \theta \ $ , then $ \ x \ = \ \frac{\theta}{2} \ $ and we have

$$ \Rightarrow \ \ \tan \left(\frac{\theta}{2}\right) \ = \ \frac{\sin(\theta)}{1+\cos(\theta)} \ \ , $$

which is one form of the "half-angle formula" for the tangent function. To prove your original relation, you can start from this equation and work backwards.

colormegone
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