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Show that $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$$

I have substituted the expansions for $\cos2x$ and $\sin2x$ and gotten, after simplification:

$$\frac{1-\sin x\cos x + 2\sin^2x}{1+\sin x\cos x-2\sin^2x}$$ I'm not sure how to carry on. I factored out the $\sin x$, but ended up with $$\frac{1+\sin x}{1-\sin x}$$

I haven't been taught that as equal to $tan x$.

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\begin{align} \frac{1-\cos 2x + \sin 2x}{1+\cos 2x + \sin 2x} &=\frac{1-(1-2\sin^2 x) + 2\sin x \cos x}{1+(2\cos^2 x -1) + 2\sin x \cos x} \\ &=\frac{2\sin^2x +2\sin x \cos x}{2\cos^2x + 2\sin x \cos x} \end{align}

Can you simplify from here?

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  • $\begingroup$ Okay. I factor out the two. Then, I factor out sin from the top and cos from the bottom. How do I get rid of $(1+cosx)$ in the numerator and $1+sinx$ in the denominator? $\endgroup$
    – Alpha
    Oct 29 '20 at 9:20
  • $\begingroup$ After getting rid of $2$ and if you factor out $\sin x$ form the numerator, you should be left with $\sin x + \cos x$. Try to check that. Also verify your working for the denominator as well. $\endgroup$ Oct 29 '20 at 9:22
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Hint:

Use $$\cos2x=2\cos^2x-1=1-2\sin^2x\text{ and }\sin2x=2\sin x\cos x$$

Or use Weierstrass substitution

Or $$\dfrac{1+\sin2x}{\cos2x}=\dfrac{(\cos x+\sin x)^2}{\cos^2x-\sin^2x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}\text { assuming }\cos x+\sin x\ne0$$

Apply Componendo et Dividendo

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By Tangent half-angle formulas with $t=\tan x$

  • $\cos 2x=\frac{1-t^2}{1+t^2}$
  • $\sin 2x=\frac{2t}{1+t^2}$

we have

$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\frac{1-\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}=\frac{1+t^2-1+t^2+2t}{1+t^2+1-t^2+2t}=\frac{2t+2t^2}{2+2t}=t$$

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$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \frac{1- (2\cos^2x-1)+2\sin x .\cos x}{1+ 2\cos^2x-1+2\sin x .\cos x}$$ $$ = \frac{2- 2\cos^2 x +2\sin x .\cos x}{ 2\cos^2 x +2\sin x .\cos x} = \frac{1- \cos^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x}$$ $$ = \frac{\sin^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x} = \frac{\sin x (\sin x + \cos x )}{ \cos x (\sin x + \cos x )} = \frac{\sin x }{ \cos x } = \tan x$$

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  • $\begingroup$ Please pay attention to your formatting when you write an answer. $\endgroup$
    – Toby Mak
    Oct 29 '20 at 9:35

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