How to solve $y = \cos(x) + \sin(x)$

Question

How do I solve $y=\cos(x) + \cos(90-x)$ where $0 \leq x \leq 45$ degrees? If I have some value $Y$, I need to find a value for $x$ such that this equation is true. Thanks for the help

Answer 1

Hint:

It implies: $$y^2=(\cos x+\sin x)^2=\cos^2x+\sin^2x+2\sin x\cos x=1+\sin 2x$$

Then further you can work with $\arcsin$.

Answer 2

From your question, it looks you are finding range of $y$ for $x\in [0,\tfrac{\pi}{4}]$. You can try $$\sin(x)+\cos(x) = \sqrt{2} \sin(x+\tfrac{\pi}{4})$$

And since $x\in [0,\tfrac{\pi}{4}], (x+\tfrac{\pi}{4}) \in [\tfrac{\pi}{4}, \tfrac{\pi}{2}]$. So in this range, $\sin(x+\tfrac{\pi}{4})$ is increasing and so the range of $y$ is $[\sin(\tfrac{\pi}{4}), \sin(\tfrac{\pi}{2})]$ that is $y\in [1, \sqrt2]$

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Alternatively, $y' = \cos(x)-\sin(x)$ and in $[0,\tfrac{\pi}{4}]$ we have $\cos(x) \ge \sin(x)$. So again $y$ is strictly increasing in given domain and so range is simply $[y(0), y(\tfrac{\pi}{4})] = [1,\sqrt{2}]$

Answer 3

Let use sum to product formula

$$\cos \theta + \cos \varphi = 2 \cos\left( \frac{\theta + \varphi} {2} \right) \cos\left( \frac{\theta - \varphi}{2} \right)$$

that is

$$\cos x+ \cos (90-x) = 2 \cos\left( 45° \right) \cos\left(45°-2x\right)$$

Answer 4

$$y=cos(x) + cos(90-x) = cos(x) + sin(x)$$

Note that $$ \sin (\pi /4) = \cos (\pi /4)= \sqrt 2/2 $$

Thus $$cos(x) + sin(x)= \sqrt 2 (\sqrt 2/2 \cos (x) +\sqrt 2/2 sin(x))= $$

$$\sqrt 2 ( \sin (\pi /4) \cos (x) + \cos (\pi /4) sin(x))=$$

$$ \sqrt 2 ( \sin (\pi /4 +x)$$

$$ y=\sqrt 2 \sin (\pi /4 +x) \implies $$

$$\pi /4 +x = \sin ^{-1} (y/{\sqrt 2}) \implies $$

$$x= \sin ^{-1} (y/{\sqrt 2}-\pi /4)$$