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The NCERT Math Textbook for Grade 11 mentions these two general solutions for Cosine Trigonometric function:

$\cos x = 0$ gives $x=(2n+1)\pi/2$, where $n\in Z$
$\cos x = \cos y$ gives $x=2n\pi \pm y$, where $n\in Z$

So if I have to solve

$\cos x = 0$

why can't I simplify it as

$\cos x = \cos \pi/2$

and use the second formula to say

$x=2n\pi \pm \pi/2$

I understand that both the solutions cover all the odd multiples of $\pi/2$ for different values of $n$, but I've not come across even one example that solves $\cos x = 0$ using the second solution

Alekhya Vellanki
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  • Going through the trouble of writing a specific value ($0$) as a specific cosine ($\cos(\pi/2)$) just adds an unnecessary extra step. However, there are situations where the $\cos x=\cos y$ approach arises. For instance, in this answer showing that $\cos(\sin x)\neq \sin(\cos x)$ starts by rewriting the latter expression as a cosine ($\cos(\pi/2-\cos x)$), then invoking the $\cos x=\cos y$ solution to move the argument along. – Blue Sep 23 '19 at 17:38
  • @Blue Thanks for the response. Here's a sample problem that could help us discuss better. They use both the solutions to solve the problem. My question is around why they haven't used the second (the more generic) one even for cos2x=0 case – Alekhya Vellanki Sep 23 '19 at 17:59
  • Trigonometry offers a huge number of equations, from the basic $sin^2+cos^2=1$ to some "twisted". The trick to deal with a specific case is to know the basis and "smell" which equation would fit in. There is no best aproximation to any problem. Smelling inproves with practice. – Ripi2 Sep 23 '19 at 18:03

1 Answers1

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What do you mean? Just factor out $π/2,$ to get $$x=π/2(4n\pm 1).$$ The last factor contains all odd numbers since $4n+3$ can be taken instead of $4n-1$ (this is just a shift), and all odd numbers are of either of the forms, since $4n,4n+2$ can never be odd.

So the solutions are actually equivalent, only apparently different.

Allawonder
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  • Like I mentioned in the question, I understand that both the formulae give the same values and cover all odd multiples of $\pi/2$. But I haven't come across any accepted solutions (at least NCERT ones) that use the second one. All of them always seem to "prefer" the first one. Totally confused as to why that preference... – Alekhya Vellanki Sep 23 '19 at 17:50
  • @AlekhyaVellanki Oh, I see. I think they prefer that because given an equation of the form $\cos x=a,$ you can immediately write down the solutions after finding one $\phi$ so that $a=\cos\phi.$ However, another way to do this is to first write $$\cos x-\cos\phi=-2\sin\frac12(x+\phi)\sin\frac12(x-\phi)=0.$$ You can see which is longer, although I generally prefer the second. It reduces further to the basic forms $\cos y=0,\sin y=0,$ which I personally find easier to remember. – Allawonder Sep 23 '19 at 17:58