Suppose $a\sin x + b\cos x = c $ . My teacher told me that with changing variable we can solve it by this : $(c + b) t^2 - 2at + (c-b) = 0$ where $t = \tan \frac{x}{2}$ . I thought it is true always but today found a problem . When we define $\sin x$ and $\cos x$ with $\tan \frac{x}{2}$ , we should consider $\cos \frac{x}{2} \not = 0$ and $x \not = 2k\pi + \pi$ but it can be the answer of the main equation (i.e. $a\sin x + b\cos x = c $ ) ! So , that formula is incomplete ?
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1You can either check for those few points or allow $t=\pm\infty$. No big deal. – Hellen Jul 21 '17 at 12:48
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1If you find something ambiguous in your teacher's saying, then you can use different substitutions , as $$sin(x)=t , cos(x)=\sqrt{1-t^2}$$ which will lead you to a quadratic equation which is easier to solve – Atul Mishra Jul 21 '17 at 12:53
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2observe that $$\cos(x)=\pm\sqrt{1-t^2}$$ – Dr. Sonnhard Graubner Jul 21 '17 at 12:57
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...and in particular, you have to make BOTH substitutions for $\cos x$ to get all solutions. – John Hughes Jul 21 '17 at 13:06
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@AtulMishra yes I agree with your teacher and my below solution is completely same to what your teacher indicates you to do. But fair enough I have elaborated it by speaking the steps I followed in that method of solving. – Aditya Guha Roy Jul 21 '17 at 13:17
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Oops, this is a duplicate : https://math.stackexchange.com/questions/213545/solving-trigonometric-equations-of-the-form-a-sin-x-b-cos-x-c?rq=1 – Aditya Guha Roy Jul 21 '17 at 13:18
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1For reference, the approach your teacher suggests is discussed on Wikipedia: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution. – Semiclassical Jul 21 '17 at 13:19
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@Semiclassical wow it has a name. Good to know !! – Aditya Guha Roy Jul 21 '17 at 13:21
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Can't we also use http://www.mash.dept.shef.ac.uk/Resources/web-rcostheta-alphaetc.pdf ? – Quantaliinuxite Jul 21 '17 at 13:24
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use that $$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2}}\cos(x)\right)=c$$ and $$\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}}$$ $$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$ therefore we get $$\sin(x+\phi)=\frac{c}{\sqrt{a^2+b^2}}$$ or you write $$2\,{\frac {a\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}+{\frac {b \left( 1- \left( \tan \left( x/2 \right) \right) ^{2} \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}=c $$ with $$\tan(x/2)=t$$ as your teacher said
Mariuslp
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Dr. Sonnhard Graubner
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why should you lose answers? i have never squared the given equation – Dr. Sonnhard Graubner Jul 21 '17 at 12:55
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You lose answers because $\tan(x/2)$ could be "infinite", i.e., because $x/2$ might not be in the domain of $\tan$, and yet $x$ might still be a solution. Squaring is not the only way to reduce a domain; the problem arises in the same way that a negative number might not be the square of anything, but still could be a solution to some equation. – John Hughes Jul 21 '17 at 12:57
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@JohnHughes So if I check $x = 2k\pi + \pi $ , then I can use that formula ? – S.H.W Jul 21 '17 at 13:00
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@Dr.SonnhardGraubner How we can define $\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}}$ ? – S.H.W Jul 21 '17 at 13:03
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yes you can use this formula, but you must check that case, at the end or in the beginning – Dr. Sonnhard Graubner Jul 21 '17 at 13:04
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@S.H.W If $ a \ne 0 $ or $ b \ne 0 $, we have $ \sqrt{a^2 + b^2} \ge \lvert {a} \rvert \implies \lvert {{a} \over {\sqrt{a^2 + b^2}}}\rvert \le 1 \implies 1 \ge {{a} \over {\sqrt{a^2 + b^2}}} \ge -1 $ – hamster on wheels Jul 21 '17 at 13:28
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NOte that:$${\color{Red}{a \sin x+ b \cos x=c \\ \frac{|a|}{a} \sqrt{a^2+b^2} \sin (x+\alpha)=c\\ \tan \alpha=\frac{b}{a}} }\\\sin (x+\alpha)=\frac{|a|}{a}.\frac{c}{\sqrt{a^2+b^2}}$$ This is general form
Khosrotash
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Why you are using $\frac{|a|}{a}$ ? I think the above answer is the same . – S.H.W Jul 21 '17 at 21:23