Trigonometry Crush 5

$\begin{aligned} \color{#3D99F6}{A} & = \sin{45^\circ} + \cos{45^\circ} = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} \color{#3D99F6}{= \sqrt{2}} \\ B & = \sin{44^\circ} + \cos{44^\circ} \\ & = \sin{(45-1)^\circ} + \cos{(45-1)^\circ} \\ & = \sin{45^\circ} \cos{1^\circ} - \cos{45^\circ} \sin{1^\circ} + \cos{45^\circ} \cos{1^\circ} + \sin{45^\circ} \sin{1^\circ} \\ & = \dfrac{1}{\sqrt{2}} \cos{1^\circ} - \dfrac{1}{\sqrt{2}} \sin{1^\circ} + \dfrac{1}{\sqrt{2}} \cos{1^\circ} + \dfrac{1}{\sqrt{2}} \sin{1^\circ} \\ & = \sqrt{2} \cos{1^\circ} \color{#D61F06}{<} \color{#3D99F6} {\sqrt{2} = A} \quad \quad \small \color{#D61F06} { [\cos{1^\circ} < 1]} \end{aligned}$

Therefore, $\boxed {A>B}$ .

Let us consider SinA + CosA.

The maximum value of this expression can be found by differentiation.

f'(x) = cosA-sinA = 0 gives the maxima .

that implies that sinA=cosA which occurs at $2n\pi +\alpha$ .(where n is an integer and alpha is equal to 45 degrees).

So, the principal solution(alpha) is 45 degrees.

So, SinA + CosA is maximum at 45 degrees.

To know for sure that this is the maxima, differentiate the function twice.

We get,

f''(x) = -sinA-cosA = negative at 45 degrees implies at 45 degrees, it is maxima.

So, Sin44+Cos44 < Sin45+Cos45.

Please correct me if I am wrong.

sin45+cos45 ? sin44+cos44

sin45-cos44 ? sin44-cos45

2 sin(1/2) cos(89/2) ? 2sin(-1/2) cos(89/2) sin(1/2) ? sin(-1/2) , as (2 cos(89/2) > 0)

sin(1/2) > sin(-1/2) A>B