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I tried the following:

$$\begin{aligned}a\sin\alpha +2\sin\alpha + 2a\cos\alpha - \cos\alpha &= 2a+1\\ a(\sin\alpha +2\cos\alpha)+(2\sin\alpha-\cos\alpha)&=2a+1\end{aligned}$$

Therefore, $$\sin\alpha +2 \cos\alpha=2$$ $$2\sin\alpha - \cos\alpha=1$$

From these two equations, we get

$$\sin\alpha=\frac{4}{5},\cos\alpha=\frac{3}{5}$$

Therefore,

$$\tan\alpha = \frac{\sin\alpha} {\cos\alpha} = \frac{4} {3}$$

Is this a correct method to solve the question? Since $a$ is a constant, it does not seem necessary to me that its coefficients on the two sides of the equation be equal. Should I find $\sin\alpha$ and $\cos\alpha$ using some other method? Are there specific cases where this method of equating the coefficients will break?

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    This is just a particular solution, you can have $\cos x=1$ and $\sin x = 0$(I've used $x$ instead of $\alpha$ because it's easier to type). – kingW3 Oct 05 '20 at 11:08
  • Also it should be $2a-1$ on the right side, which gives the solution I mentioned above. I think that for example $\sin x+2\cos x =1$ and $2\sin x-\cos x =a-1$ might be possible. Also a question what is $a$ is it a parameter or is the equation supposed to work for every $a$? – kingW3 Oct 05 '20 at 11:18
  • @kingW3 There was a typo in the title, it should be $2a + 1$ on the RHS. Apologies! –  Oct 05 '20 at 11:20

3 Answers3

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The OP finds one constant root that applies for all $a$. But there is a second root for most specific values of $a$, which is a function of $a$. The full answer is $\tan\alpha\in\{4/3,2a/(a^2-1)\}$.

Properly, the given equation should be combined with the identity $\sin^2\alpha+\cos^2\alpha=1$. There are two ways to do this:

Method 1

Isolate one of the trigonometric function, square the resulting equations and substitute to get a quadratic equation for the remaining function. Choosing to isolate the cosine we then have

$(2a-1)\cos\alpha=(2a+1)-(a+2)\sin\alpha$

$(2a-1)^2\cos^2\alpha=(2a+1)^2-2(2a+1)(a+2)\sin\alpha+(a+2)^2\sin^2\alpha$

$(4a^2-4a+1)-(4a^2-4a+1)\sin^2\alpha=(4a^2+4a+1)-(4a^2+10a+4)\sin\alpha+(a^2+4a+4)\sin^2\alpha$

$(5a^2+5)\sin^2\alpha-(4a^2+10a+4)\sin\alpha+8a=0$

The quadratic equation looks like a mouthful, but its discriminant is a squared quantity, to wit $(4a^2-10a+4)^2$, thus we get the two roots

$\sin\alpha=\dfrac{(4a^2+10a+4)\pm(4a^2-10a+4)}{2(5a^2+5)}\in\{4/5,2a/(a^2+1)\}$

For each root of $\sin\alpha$ the previous equation with $\cos\alpha$ isolated is used to assure the proper sign of that function:

$(2a-1)\cos\alpha=(2a+1)-(a+2)(4/5); \cos\alpha=3/5$

$(2a-1)\cos\alpha=(2a+1)-(a+2)(2a/(a^2+1)); \cos\alpha=(a^2-1)/(a^2+1)$

Correspondingly $\tan\alpha\in\{4/3,2a/(a^2-1)\}$.

Thus the answer given by the OP is correct for one root that applies for all $a$, but in most cases there will be a second root for any specific value of $a$ (the only exceptions being $a=2$ where there is one doubly degenerate root instead, and $a=\pm 1$ where the second root fails to give a defined value for $\tan\alpha$; also $a=-1/2$ gives the second root with $\tan\alpha=4/3$ but different values for the sine and cosine). The existence of two roots ultimately comes from the fact that except for $\pm 1$, any value of the sine or cosine corresponds to two different arguments within any fundamental period.

Method 2

In this method, we use a trick by combining the original equation with one involving the orthogonal linear combination. To get the orthogonal combination, switch the coefficients $a+2$ and $2a-1$ and then reverse the sign before the second term.

$(a+2)\sin\alpha+(2a-1)\cos\alpha=(2a+1)$

$(2a-1)\sin\alpha-(a+2)\cos\alpha=x$

Square both sides and add them together getting:

$(5a^2+5)(\sin^2\alpha+\cos^2\alpha)=5a^2+5=(2a+1)^2+x^2$

Thus $x=\pm\sqrt{5a^2+5-(2a+1)^2}=\pm(a-2)$. We then have a linear system to solve for $\sin\alpha$ and $\cos\alpha$ for each root of $x$. For example, $x=+(a-2)$ gives

$(a+2)\sin\alpha+(2a-1)\cos\alpha=(2a+1)$

$(2a-1)\sin\alpha-(a+2)\cos\alpha=a-2$

whose solution is $\sin\alpha=4/5,\cos\alpha=3/5$. Putting $-(a-2)$ for $x$ similarly gives a linear system with solution $\sin\alpha=2a/(a^2+1),\cos\alpha=(a^2-1)/(a^2+1)$.

The remainder of the solution, and the comments that follow, are identical to Method 1.

Oscar Lanzi
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2

Hint:

Use Weierstrass substitution to form a quadratic equation $$\tan\dfrac\alpha2=t$$

$$(a+2)\cdot\dfrac{2t}{1+t^2}+(2a-1)\cdot\dfrac{1-t^2}{1+t^2}=2a+1$$

$$\iff-t^2(2a+1+2a-1)+2t(a+2) +2a-1-(2a+1)=0$$

$$\iff2at^2-t(a+2)+1=0$$

Now use $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$

Oscar Lanzi
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lab bhattacharjee
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-1

The problem is that $a$ is a constant, not a variable. Therefore equating by parts does not work here: if we separate $2a - 1$ into $2(a-1) + 1$, $2(a-2) + 3$ and so on, we end up with a different set of solutions each time. If you try substituting $\sin \alpha$ and $\cos \alpha$ back into the original question, the LHS does not equal $2a-1$, so clearly something must have gone wrong.

Instead, notice that as this is an identity, so it must hold for all $a$. Substituting $a = -2, \frac{1}{2}$ to cancel one of the terms, we get:

$$-5 \cos a = -5 \Rightarrow \cos a = 1$$ $$\frac{5}{2} \sin a = 0 \Rightarrow \sin a = 0$$

Toby Mak
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  • I don't think the OP mentioned that this is an identity, if it is then his method would be fine assuming he didn't solve a different equation from the title. – kingW3 Oct 05 '20 at 11:21
  • Indeed, as I wrote in the body of the question, $a$ can only be assumed to be a constant from the context of the question, which is why I was unsure of the correctness of my method. –  Oct 05 '20 at 11:26
  • @Toby Mak There was a typo in the question title, which I have corrected? Could you please update your answer? Also, I don't understand why we will get a different answer each time "we separate $2a-1$ into $2(a-1)+1$ and so on" (in the context of your answer as it is now) –  Oct 05 '20 at 11:34