# Find an acute angle $\gamma$ such that $\sin \gamma + \cos \gamma= \sqrt{2}$ [closed]

Find an acute angle $\gamma$ such that $$\sin \gamma + \cos \gamma = \sqrt{2}$$

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– user137731
Commented May 29, 2015 at 19:48
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• Commented May 29, 2015 at 19:51

$$\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x=1$$ $$\cos \frac{\pi}{4}\cos x+\sin \frac{\pi}{4}\sin x=1$$ $$\cos (x-\frac{\pi}{4})=1$$ You can do it from here?

Hint:

$\sin \gamma + \cos \gamma$

$= \sqrt2\left(\frac{1}{\sqrt 2}\sin \gamma + \frac{1}{\sqrt 2} \cos \gamma\right)$

$= \sqrt 2 \sin \left(\gamma + \frac{\pi}{4}\right)$

Here is a trick that can be generalized to simplify the sum of a multiple of the sine of an angle with a multiple of the cosine of the same angle.

$$\sin x + \cos x = \sqrt 2$$ $$\sin x\cdot\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2}\cdot\cos x= 1$$ $$\sin x\cdot\cos 45° + \sin 45°\cdot\cos x= 1$$ $$\sin(x+45°)=1$$

You should be able to finish from here.

${1 \over \sqrt 2} \left( \sin\gamma + \cos\gamma\right) = 1$ or equivalently $$\sin\gamma\cos(\pi/4) + \cos\gamma\sin(\pi/4) = 1$$ Thus $\sin(\gamma + \pi/4) = 1$.

Hence $$\gamma + \pi/4 = \pi/2 + 2k\pi, \ k \text{ integer}$$ So if $\gamma$ is acute, the only solution is

$$\gamma = \ ...$$

Given that $$\sin \gamma+\cos\gamma=\sqrt{2}$$ $$\implies \frac{1}{\sqrt{2}}\sin \gamma +\frac{1}{\sqrt{2}}\cos\gamma=1$$ $$\implies \sin \gamma \cos \frac{\pi}{4}+\cos\gamma \sin \frac{\pi}{4}=1$$ $$\implies \sin \left(\gamma+\frac{\pi}{4}\right)=1 =\sin \frac{\pi}{2}$$ writing the general solution of the above equation, we get $$\gamma+\frac{\pi}{4}=2n\pi+\frac{\pi}{2}$$ $$\implies \gamma=2n\pi+\frac{\pi}{4}$$Where $n$ is any integer. As the unknown angle $\gamma$ is acute ($\color{red}{0<\gamma<\frac{\pi}{2}}$) hence $$\gamma=2(0)\pi+\frac{\pi}{4}$$$$\implies \color {blue}{\gamma=\frac{\pi}{4}}$$

we can write $$\sqrt 2 = \sin t + \cos t = \sqrt 2\left(\cos \left(\frac{\pi}4\right) \cos t + \sin \left(\frac{\pi}4\right) \sin t \right) = \sqrt 2 \cos \left(t - \pi/4\right)$$

so that we have $$\cos(t - \pi/4) = 1 \implies t - \pi/4 = 2k\pi$$

that is $$t = 2k\pi + \frac{\pi}4, \text{ where k is any integer.}$$