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

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6 Answers 6

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

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Hint:

$\sin \gamma + \cos \gamma $

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

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

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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.

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

$$\gamma = \ ...$$

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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 $$\gamma=2(0)\pi+\frac{\pi}{4}$$$$\implies \color {blue}{\gamma=\frac{\pi}{4}}$$

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

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