Infinitely many solutions, but just one value

$\cos{\theta} + \sin{\theta} = k \quad \Rightarrow \sin{\left( \frac{\pi}{2} - \theta \right)} + \cos{\left( \frac{\pi}{2} - \theta \right)} = k$

Therefore, for $\theta \in [0, \frac{\pi}{2} ]$ , there are two $\theta$ 's that satisfy the above equation, $\theta_1$ and $\theta_2 = \frac{\pi}{2} - \theta_1$ , except when $\theta_1 = \frac{\pi}{4}$ then $\theta_2 = \theta_1 = \frac{\pi}{4}$ . Then there is only one solution to the equation and there is only one value for $\cos{\theta} - \sin{\theta} = 0$ .

When $\theta = \frac{\pi}{4}$ , we have:

$\begin{aligned} \cos{\theta} + \sin{\theta} & = k \\ \Rightarrow \sin{\left( \frac{\pi}{4} \right)} + \cos{\left( \frac{\pi}{4} \right)} & = k \\ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} & = k \\ \Rightarrow k & = \boxed{\sqrt{2}} \end{aligned}$

Using the Trigonometric Equations - R method and the substitution $y = \theta + \frac{\pi}{4}$ , we can write the equation as $\sqrt{2} \sin y= k$ . We want to uniquely determine the value of $\sqrt{2} \sin \left(y + \frac{\pi}{2}\right)$ , which is equivalent to $\sqrt{2} \cos y$ .

Using the Pythagorean Identities ,

$\cos y = \pm \sqrt{1 - \sin^2 y}$

A unique value of $\sqrt{2} \cos y$ will be obtained only when $+ \sqrt{1 - \sin^2 y} = - \sqrt{1 - \sin^2 y}$ , i.e. $\cos y = 0$ .

When $\cos y = 0$ , $\sin y = \pm 1$ , so $k = \pm \sqrt{2}$ .

Hence $\boxed{k = \sqrt{2}}$ is the positive value of $k$ which uniquely determines the value of $\cos \theta - \sin \theta$ . $_\square$ .

Let $\cos(\theta)-\sin(\theta)=l$ . Then we have $l^2+k^2=2$ i.e. $l=\pm \sqrt{2-k^2}$ Unless $k^2=2$ , we have either two distinct values of $l$ or no (real) value of $l$ . Hence the only positive value of $k$ with unique value of $l$ is $\sqrt{2}$ .

cos@+sin@=k
[cos@+sin@=k]^2
1+2cos@sin@=k^2
sin2@=k^2-1
@=sin^-1[(k^2-1)]/2

If k=2,
@=[sin^-1(3)]/2=undefined

If k=sqrt(2),
@=..., pi/4, 1.25pi, ...

If k=1,
@=..., 0, pi/2, ...

If k=sqrt(3),
@=[sin^-1(2)]/2=undefined

Now we have 2 possibilities for k=sqrt(2), 1.

Let f(@)=cos@-sin@.

For k=sqrt(2),
... f(0.25pi)=0 f(1.25pi)=0 ... f(@)=0

For k=1,
... f(0)=1 f(0.5pi)=-1 ... f(@)=1,-1

Thus, for cos@-sin@ to be constant, k=sqrt(2).