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

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

Answer 2

Here’s two more very straightforward ways to go about it: $${\cos x+\sin x\over\sqrt 2}=\sqrt{1+k\over 2}$$ $$\implies \sin\left(x+\frac\pi4\right) = \sqrt{1+k\over 2}$$ (note that this expression is valid for all $k$ in the range). Hence, $$x=\arcsin\left(\sqrt{1+k\over 2}\right)-\frac \pi4$$

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For the first part, we have $$\sin2x=\sin\left(2\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi2\right)$$

$$=-\cos\left(2\arcsin\left(\sqrt{1+k\over 2}\right)\right)$$ $$={1+k\over 2}+{1+k\over 2}-1$$ $$=k$$

For the second, we have

$$\sin x-\cos x = \sin\left (\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)-\cos\left(\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)$$

$$=\left(\sin\left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)-\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)-\sin \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)-\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)\right)\cdot\frac1{\sqrt 2} $$ $$=-\sqrt 2\cdot\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)$$ $$=-\sqrt2\cdot\mp\sqrt{1-k\over 2}$$

And since $x\in\left(\frac\pi4,\frac\pi2\right)$, $\sqrt{1+k\over 2}>0$ because $k\in(-1,1)$, and because $x = \left(\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)$, $\cos \left(\arcsin\left(\sqrt{1+k\over 2}\right)\right)<0$ and so we have $$\sin x-\cos x = \sqrt{1-k}$$

And finally, $$\tan x= \tan \left(\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4\right)$$ With this and $\tan\left(\alpha-\frac\pi4\right)={1-\tan\alpha\over1+\tan\alpha}$ we get our result that $$\tan x = {1+\sqrt{1-k^2}\over k}$$

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Note that we have used $\cos(\arcsin\alpha) = \sqrt{1-\alpha^2}$ and $\tan(\arcsin\alpha) ={\alpha\over\sqrt{1-\alpha^2}}$ in our simplification. Also note that any part may be solved by steps from the other two similarly to the answer due to Quanto.

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Yet another method picking off from the first line of the first method:

Writing $k$ in terms of $x$ we have:

$$k=2\sin^2\left(x+\frac\pi4\right)-1$$

$$\implies k=-\cos\left(2x+\frac\pi2\right)$$

$$\implies \boxed{k = \sin2x}$$ Next, we have $$\sin x -\cos x = \sqrt{(\sin x +\cos x)^2-4\sin x\cos x}$$ $$=\sqrt{1+k-2k}$$ $$\implies \boxed{\sin x - \cos x = \sqrt{1-k}}$$

Also, I did not see the "Hence" in the last part of the question. So this last step is common to both parts of the solution

$$\tan x = \sin x + \cos x$$ By Componendo et Dividendo

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

$$\implies \sqrt{1+k}\tan x -\sqrt{1+k} = \sqrt{1-k}\tan x +\sqrt{1-k}$$

$$\implies \tan x(\sqrt{1+k}-\sqrt{1-k})=\sqrt{1+k}+\sqrt{1-k}$$

Multiplying both sides by $$\sqrt{1+k}+\sqrt{1-k}\over(\sqrt{1+k}+\sqrt{1-k})\cdot(\sqrt{1+k}-\sqrt{1-k})$$

we have

$$\tan x ={ (\sqrt{1+k}+\sqrt{1-k})^2\over \sqrt{1+k}^2-\sqrt{1-k}^2)}$$ $$ = {1+k+1-k-2\sqrt{1-k^2}\over 1+k-1+k}$$ $$\implies\boxed{\tan x = {{1+\sqrt{1-k^2}\over k}}}$$

The first method may look unwieldy due to the repeated appearance of $\arcsin\left(\sqrt{1+k\over 2}\right)-\frac\pi4$ but it is quite simple. Read it as some angle $\phi$ instead and it’ll make it better.

Answer 3

You did well in determining $\sin2x=k$.

Now, $\cos2x=-\sqrt{1-\sin^22x}=-\sqrt{1-k^2}$, due to $\pi/2<2x<\pi$.

Next, you can recall that $$ \tan x=\frac{1-\cos2x}{\sin2x}=\frac{\sin2x}{1+\cos2x} $$ (a not so well-known bisection formula for the tangent). Using the first formula yields $$ \tan x=\frac{1+\sqrt{1-k^2}}{k} $$ Finally, $(\sin x-\cos x)^2=1-\sin2x=1-k$. The limitation on $x$ ensures that $\sin x>\cos x$, so $$ \sin x-\cos x=\sqrt{1-k} $$