I was required to solve the function for it domain and range, $$y=\log_2(\sin x-\cos x+3\sqrt{2})-\log_2\sqrt{2}$$
Here I'm not able to solve due my inability to manipulate $\sin x-\cos x$. Are there any important ones that I should try to remember?
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Do you mean $\log(2(\cdots))$ (the natural log of two-times something) or $\log_2(\cdots)$ (the base-$2$ log of something)? Also, I don't see an inequality. – Blue Apr 13 '20 at 12:17
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1this isn't an inequity, this is a function. – Noa Even Apr 13 '20 at 12:18
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$a\sin x+b\cos x+c=A\cos (x+\varphi)$ (added angle formula for example), into $\log_2 (\ldots)$. – Sebastiano Apr 13 '20 at 12:21
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1This question is about the relation you probably need: "Why does $A\sin k(x+c)=a\sin kx+b\cos kx$ imply that $A=\sqrt{a^2+b^2}$ and $\tan c=−b/a$?" (Take $k=1$, $a=1$, and $b=-1$.) – Blue Apr 13 '20 at 12:21
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Where is that inequality ? – Apr 13 '20 at 12:44
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im sorry, i have edited the question now. – MrKhonsu Apr 13 '20 at 13:03
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You have to introduce what we call normalizer factor that is defined as: $$b=\sqrt{a^2+b^2}$$ where $a$ and $b$ are the coefficent of sine and cosine function (in this case both $1$). So we multiply and divide by $\sqrt2$, obtaining: $$sin(x)-\cos(x)=\sqrt2(\frac{\sqrt2}{2}\sin(x)-\frac{\sqrt2}{2}\cos(x))$$ Now we can notice the expansion of $\sin(x+\pi/4)$, so all becomes: $$\sqrt2\sin(x-\pi/4)$$
Matteo
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@MrKhonsu: if this answer has been useful, maybe a $+1$ or a favourite answer? – Matteo Apr 13 '20 at 20:34
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Presumably, you want to solve
$$\sin x-\cos x=a.$$
You can address this by squaring, while you recall that the LHS must have the sign of $a$:
$$(\sin x-\cos x)^2=\sin^2x-2\sin x\cos x+\cos^2x=1-\sin 2x=a^2$$
so that
$$\sin 2x=1-a^2.$$