Triggered Trigonometry #1

We note that the RHS is $>0$ and let us assume that a solution exists for $\sin {x}$ and $\cos {x}$ are positive. Cauchy-Schwartz inequality implies:

$(\sin{x} + \cos{x})^2 \le (1^2+1^2)(\sin^2{x} + \cos^2{x}) = 2$

$\Rightarrow$ LHS $= \sin{x} + \cos{x} \le \sqrt{2}$ and we note that RHS $= \sqrt{y+\dfrac{1}{y}} \ge \sqrt{2}$ .

This implies that $\sin{x} + \cos{x} = \sqrt{y+\dfrac{1}{y}} = \sqrt{2}$ and equality happens when $\sin{x} = \cos{x} = \dfrac {1}{\sqrt{2}}\Rightarrow \boxed{x = \dfrac {\pi}{4}}$ and $\boxed{y = \dfrac {1}{y} = 1}$ .

Here I will use a calculus approach to find the maximum of the given expression.

Let ${f}$ ( ${x}$ ) = sin( ${x}$ ) + cos( ${x}$ ). Then: $f'(x) = cosx -sinx = 0 \Rightarrow\ cosx = sinx = \frac{1}{\sqrt{2}}$ Differentiating again to fond the nature of the stationary point: $f''(x) = -sinx - cosx = -\sqrt{2}\ < 0 \therefore\ sinx + cosx \leq\sqrt{2}$ Equality occurs when $x = \frac{\pi}{4}\ , y =1$

Note that $(\sin x+\cos x)^2=1+\sin 2x$ , so we have $1+\sin 2x=y+\frac 1y$ . By AM-GM, $y+\frac 1y\geq2$ , and obviously $1+\sin 2x\leq2$ . Therefore, they must both be 2, which happens at $x=\frac \pi4,\ y=1$

Here , i will use the value of x = π/4, y = 1 and this x and y value satisfied equation sin⁡〖θ+ cos⁡〖θ= √(y+ 1/y〗 〗)