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Prove by writing: $$\theta = 2A$$ that: $$\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1+\sin\theta}{\cos\theta}$$

First I subbed in the $2A$ such that:

$$\frac{\sin 2A-\cos 2A+1}{\sin 2A+\cos 2A-1}=\frac{1+\sin 2A}{\cos 2A}$$

Then I considered the following trigonometric formulae:

$$\begin{align} \sin 2A &= 2\sin A\cos A \\ \cos 2A &= \cos^{2}A - \sin^{2}A \\ \cos 2A &= 2\cos^{2}A-1 \\ \cos 2A &= 1-2\sin^{2}A \end{align}$$

I have taken various approaches but keep ending up with:

$$\frac{\cos A+\sin A}{\cos A-\sin A}$$ on the left hand side. But I can't seem to get the right hand side to agree with this. I end up with something like: $$\frac{1+2\sin A\cos A}{1-2\sin^{2}A}$$ I've spent a few hours on this and can't find the solution. Help appreciated.

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marked as duplicate by Shailesh, Watson, colormegone, John B, Antonios-Alexandros Robotis Mar 16 '16 at 0:32

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You're almost there.

You say you end up with $1+2\sin A \cos A$ in the numerator, but by the double angle formula, $2\sin A \cos A = \sin (2A)$ so you get the numerator of the RHS.

In the denominator you end up with $1-2\sin^2 A$, use a suitable double angle formula for the cosine ($1-2\sin^2 A = \cos(2A)$).

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Here is another simple proof, $$LHS=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$$ $$=\frac{(\sin\theta-\cos\theta+1)(\sin\theta+\cos\theta+1)}{(\sin\theta+\cos\theta-1)(\sin\theta+\cos\theta+1)}$$ $$=\frac{(\sin\theta+1)^2-(\cos\theta)^2}{(\sin\theta+\cos\theta)^2-(1)^2}$$ $$=\frac{\sin^2\theta+2\sin\theta+1-\cos^2\theta}{\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1}$$ $$=\frac{\sin^2\theta+2\sin\theta+\sin^2\theta}{1+2\sin\theta\cos\theta-1}$$ $$=\frac{2\sin^2\theta+2\sin\theta}{2\sin\theta\cos\theta}$$ $$=\frac{2\sin\theta(1+\sin\theta)}{2\sin\theta\cos\theta}$$$$=\frac{1+\sin\theta}{\cos\theta}=RHS$$

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Thanks all. $$\frac{1+2sinAcosA}{cos^2{A}-sin^2{A}}$$ $$=\frac{cos^{2}A+sin^{2}A+2sinAcosA}{cos^{2}A-sin^{2}A}$$ $$=\frac{(cosA+sinA)(cosA+sinA)}{(cosA+sinA)(cosA-sinA)}$$ $$=\frac{cosA+sinA}{cosA-sinA}$$ Time for tea.

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    $\begingroup$ Actually you should use \sin and \cos to get the properly upright spaced trigonometric functions, but I was too lazy to edit all of them... $\endgroup$ – user21820 Jul 30 '16 at 6:22

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