Showing that, if $\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$, then $\sqrt2\sin\beta=\sin\alpha-\cos\alpha$

Question

If $$\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$$ then prove that $$\sqrt2\sin\beta=\sin\alpha-\cos\alpha$$

I have been trying to solve this exercise but I don't get it. I need help. Thanks.

Answer 1

You can use the fact that $$ \sin^2\beta=\frac{\sin^2\beta}{\cos^2\beta+\sin^2\beta}=\frac{\tan^2\beta}{1+\tan^2\beta} $$ and this will show that $$ \sin^2\beta=\frac{(\sin\alpha-\cos\alpha)^2}{2}\tag{*} $$ On the other hand, the statement is generally false. Take $\alpha=\pi/3$ and $\beta=\pi+\arctan(2-\sqrt{3})$, after having observed that $$ \frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}=2-\sqrt{3} $$ Then $\sin\beta<0$, but $\sin\alpha-\cos\alpha>0$.

Therefore (*) is the best you can prove.

Answer 2

Writing $\alpha=x,\beta=y,$

As $(\sin x-\cos x)^2+(\sin x+\cos x)^2=2$

Using the given condition, $2=(\sin x-\cos x)^2(1+\cot^2y)$

$\implies(\sin x-\cos x)^2=2\sin^2y$

As we have taken square, we have introduced When do we get extraneous roots?

For example, $$a=b\implies a^2=b^2\implies a=\pm b$$

But actually $a\ne-b$

Answer 3

Avoid squaring whenever possible as it immediately introduces When do we get extraneous roots?

$$\tan\beta=\dfrac{1-\tan\alpha}{1+\tan\alpha}=\tan(\pi/4-\alpha)$$

$$\beta=n\pi+\pi/4-\alpha$$ where $n$ is any integer

$$\implies\sin\beta=\sin(n\pi+\pi/4-\alpha)$$

Now as $\sin(n\pi+x)=(-1)^n\sin x,$

$$\sin\beta=(-1)^n\sin(\pi/4-\alpha)=?$$

Answer 4

Here's a diagrammatic verification for acute $\alpha$ and $\beta$: