Solve $\sin x + \cos x = \sqrt{1+k}$ for $\sin 2x$, $\sin x-\cos x$, and $\tan x$ in terms of $k$

Question

Given that $\sin x + \cos x = \sqrt{1+k}$, $-1 \le k \le 1$

• Find the value of $\sin 2x$ in terms of k

• Given that $x \in (45^{\circ}, 90^{\circ})$ deduce that $\sin x – \cos x = \sqrt{1-k}$

• Hence, show that $\tan x = \dfrac{1 + \sqrt{1-k^2}}{k}$

Okay, I have figured out that

$$ \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1+k \implies \sin 2x = k $$

Now, I am not sure what approach I should use to deduce the second one.

And can somebody give me a hint on how to start with the third one?

Answer 1

Try this for the second one,

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

For the third one, solve the equations below for $\sin x$ and $\cos x$,

$$ \sin x + \cos x = \sqrt{1+k}$$ $$ \sin x- \cos x = \sqrt{1-k}$$

and then plug them into

$$\tan x= \frac{\sin x}{\cos x}$$