# Equation with sine and cosine - coefficients

I have some trouble with the conceptual understanding of the way we solve this kind of equations. Let's say we have: $$(3-3b^2)\sin(bx)+3a\cos(2x)=6\cos(2x)$$ The method employed on classes was equating coefficients on both sides of the equation. For example there's no sine function on the right hand side, so $(3-3b^2)=0$. Therefore we're left with $3a\cos(2x)=6\cos(2x)$ so obviously $3a=6$. From that it follows that $a=2, b=1$ or $-1$. This is logical to me but I still can't understand why the sine and the cosine on LHS can't "coexist" (not sure how to phrase that) and give together the result on RHS. Example of what I think if that's not clear:

$$\underbrace{(3-3b^2)\sin(bx)}_n+\underbrace{3a\cos(2x)}_m=\underbrace{6\cos(2x)}_{n+m}$$

• Perhaps you can express your sin term in purely cosine terms. – David G. Stork Jul 2 '15 at 16:04
• If the arguments for the trigonometric functions were just integer multiples one might argue that $\cos kx$ and $\sin kx$ are orthogonal for some suitable scalar product on the function space and thus linear independent, so you could not express one as linear combination of the others. – mvw Jul 2 '15 at 16:37

You're being quite vague in a way that large numbers of students are. They see something like $$(3-3b^2)\sin(bx)+3a\cos(2x)=6\cos(2x) \tag 1$$ and think that this specifies an problem to be solved. But the words you wrote after that make it clear that whoever posed this said more than what you've told us.

A set of exercises in a textbook may say "Solve the following equations for $x$." It it had said that, that would be a very different problem from what you write about in your paragraph beginning with "The method employed". There is no reason why the coefficient of the sine term should not be $0$ if it says "Solve this equation for $x$."

But if it said that $(1)$ is an identity true of all values of $x$, and asks you to find the coefficients, then, because of the word all, we can conclude that the coefficient of the sine term is $0$.

If $x=0$ then the sine term vanishes and the cosine terms are equal to $1$, and we get $3a=6$ so $a=2$. Then if $x$ has some value that makes the sine nonzero, we get $$(3-3b^2)\cdot(\text{some number other than 0}) + 6\cos(2x) = 6\cos(2x).$$ Subtracting $6\cos(2x)$ from both sides, we get $$(3-3b^2)\cdot(\text{some number other than 0}) = 0,$$ and from that we can conclude that the coefficient is $0$.

• The other option is that $b=0$, so the coefficient is $3$ but since $\sin(0x)=0$ for all $x$ the coefficient is irrelevant. I suspect that in the examples actually worked in class, all had constant coefficients of $x$ inside the sine and cosine functions, so this particular loophole never came up. – David K Jul 2 '15 at 16:24
• Wow, I haven't noticed that. So if I added $b=0$ to the list of the solutions, that would be good as well? And I'm not sure what you mean by "constant coefficients" - don't you mean nonzero coefficients? Zero is constant too. – Richard Smith Jul 2 '15 at 16:42
• Usually one might take "constant" to mean "not depending on $x$", but in this case he appears to mean "not depending on the coefficients of the sine and cosine functions". "Constant" always means not depending on something. Probably just what the "something" is should be stated explicitly more often than it is. – Michael Hardy Jul 2 '15 at 16:48

You probably want to find $a$ and $b$ such that, for all $x$, $$(3-3b^2)\sin(bx)+3a\cos(2x)=6\cos(2x)$$ If this is the problem, then you can plug in whatever value of $x$; for instance, when $x=0$, we get $$(3-3b^2)\sin(b\cdot0)+3a\cos(2\cdot0)=6\cos(2\cdot0)$$ that becomes $$3a=6$$ so $a=2$. With this information we remain with $$(3-3b^2)\sin(bx)=0$$ for all $x$. If $b=0$, the equality surely holds. If $b^2\ne1$ and $b\ne0$, there surely are values of $x$ such that $\sin(bx)\ne0$, for instance $x=\frac{\pi}{2b}$. So we have the equality holding for all $x$ if and only if $b^2=1$ or $b=0$.