How do I determine the sign of $\sin\theta$ in this question.

Question

$\alpha$ and $\beta$ are two different values of $\theta$ lying between $0$ and $2\pi$ which satisfy the equation $\begin{equation}6\cos\theta + 8\sin\theta = 9\end{equation}$. Find $\sin(\alpha+\beta).$

I solved and got $\cos(\alpha+\beta) = \frac{-28}{100}$.
So $\sin(\alpha+\beta) = \pm\frac{24}{25}$.
I think the sign of $\sin(\alpha+\beta)$ could be either of '+' and '-', however my book says it must be '+'. How do I determine the sign?

How did you solve for $\cos(\alpha + \beta)$? I bet there's a pretty good chance you could modify the derivation to solve for $\sin(\alpha + \beta)$ directly, without going through $\cos$ first. — , Feb 16 '14 at 23:07

Américo Tavares · Accepted Answer · 2014-02-16T22:48:31.407

Your book is correct. We don't need to find either $\alpha$ or $\beta$, i.e. to solve the given linear equation in $\cos\theta$ and $\sin\theta$, which could be transformed into a quadratic equation in $\tan\frac{\theta}{2}$.

[EDIT. Since you've commented that your course was a beginner's one, I edited this answer to add the derivation of all trigonomatric identities I used, from the addition and subtraction formulas.]

Equating the equations satisfied by $\alpha$ and $\beta$

\begin{eqnarray*} 6\cos \alpha +8\sin \alpha &=&9\tag{1} \\ 6\cos \beta +8\sin \beta &=&9,\tag{2} \end{eqnarray*}

and then simplyfying and rearranging, we get

\begin{eqnarray*} 3\left( \cos \alpha -\cos \beta \right) &=&4\left( \sin \beta -\sin \alpha \right) .\tag{3} \end{eqnarray*}

Applying the sum-to-product identities

\begin{eqnarray*} \cos \alpha -\cos \beta &=&-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \tag{4}\\ \sin \beta -\sin \alpha &=&-2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2},\tag{5} \end{eqnarray*}

assuming$^1$ that $\sin \frac{\alpha-\beta }{2}\ne 0 $ we have that

\begin{eqnarray*} -3\times 2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} &=&-4\times 2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}\tag{6} \\ 3\sin \frac{\alpha +\beta }{2} &=&4\cos \frac{\alpha +\beta }{2}\tag{7} \\ \tan \frac{\alpha +\beta }{2} &=&\frac{4}{3}.\tag{8} \end{eqnarray*}

Finally expressing $\sin \left( \alpha +\beta \right) $ in terms of $\tan \frac{\alpha +\beta }{2}$, using the double-angle identity we find

\begin{equation*} \sin \left( \alpha +\beta \right) =\frac{2\tan \frac{\alpha +\beta }{2}}{ 1+\tan ^{2}\frac{\alpha +\beta }{2}}=\frac{2\left( \frac{4}{3}\right) }{ 1+\left( \frac{4}{3}\right) ^{2}}=\frac{24}{25},\tag{9} \end{equation*}

which agrees with the $\cos \left( \alpha +\beta \right) $ value you found:

\begin{equation*} \cos \left( \alpha +\beta \right) =\frac{1-\tan ^{2}\frac{\alpha +\beta }{2} }{1+\tan ^{2}\frac{\alpha +\beta }{2}}=\frac{1-\left( \frac{4}{3}\right) ^{2} }{1+\left( \frac{4}{3}\right) ^{2}}=-\frac{7}{25}.\tag{10} \end{equation*}

EDIT. Justification of:

Identity $(4)$ \begin{eqnarray*} \cos \left( a\pm b\right) &=&\cos a\cos b\mp \sin a\sin b \\ &&\text{If }\left\{ \begin{array}{c} a+b=\alpha \\ a-b=\beta , \end{array} \right. \text{ then }\left\{ \begin{array}{c} a=\frac{\alpha +\beta }{2} \\ b=\frac{\alpha -\beta }{2} \end{array} \right. \text{ and} \\ \cos \alpha &=&\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} -\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \\ \cos \beta &=&\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} +\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \\ \cos \alpha -\cos \beta &=&-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}. \end{eqnarray*}
Identity $(5)$ \begin{eqnarray*} \sin \left( a\pm b\right) &=&\sin a\cos b\pm \sin b\cos a \\ &&\text{If }\left\{ \begin{array}{c} a+b=\alpha \\ a-b=\beta , \end{array} \right. \text{ then }\left\{ \begin{array}{c} a=\frac{\alpha +\beta }{2} \\ b=\frac{\alpha -\beta }{2} \end{array} \right. \text{ and} \\ \sin \alpha &=&\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} +\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\ \sin \beta &=&\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} -\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\ \sin \beta -\sin \alpha &=&-2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}. \end{eqnarray*}
identities $(9)$ and $(10)$ \begin{eqnarray*} \sin \left( \alpha +\beta \right) &=&2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}=\frac{2\sin \frac{\alpha +\beta }{2}\cos \frac{ \alpha +\beta }{2}}{\cos ^{2}\frac{\alpha +\beta }{2}+\sin ^{2}\frac{\alpha +\beta }{2}} \\ &=&\frac{2\tan \frac{\alpha +\beta }{2}}{1+\tan ^{2}\frac{\alpha +\beta }{2}} \\ && \\ \cos \left( \alpha +\beta \right) &=&\cos ^{2}\frac{\alpha +\beta }{2}-\sin ^{2}\frac{\alpha +\beta }{2}=\frac{\cos ^{2}\frac{\alpha +\beta }{2}-\sin ^{2}\frac{\alpha +\beta }{2}}{\cos ^{2}\frac{\alpha +\beta }{2}+\sin ^{2} \frac{\alpha +\beta }{2}} \\ &=&\frac{1-\tan ^{2}\frac{\alpha +\beta }{2}}{1+\tan ^{2}\frac{\alpha +\beta }{2}} \end{eqnarray*}

--

$^1$ The assumption $\sin \frac{\alpha -\beta }{2}\neq 0$ is valid because if $ 0\neq \frac{\alpha -\beta }{2}=\pi $, then without loss of generality we might assume that $0=\beta <\alpha =2\pi $, but we would get a contradition:

\begin{eqnarray*} 6\cos \left( 2\pi \right) +8\sin \left( 2\pi \right) &\neq &9 \\ 6\cos \left( 0\right) +8\sin \left( 0\right) &\neq &9. \end{eqnarray*}

Prosthaphaeresis Formulas (http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html) are fine. But, we need to explain why $$\sin\frac{\alpha-\beta}2\ne0$$ — lab bhattacharjee, Feb 12 '14 at 17:29
I was thinking in that line. What's more interesting is the Right Hand Side is $(9)$ is immaterial here. But, $$(6\cos\theta+8\sin\theta)^2\le(6^2+8^2)(\cos^2\theta+\sin^2\theta)\le100\iff-10\le 6\cos\theta+8\sin\theta\le10$$ Then what if the RHS $>10$? — lab bhattacharjee, Feb 12 '14 at 17:58

score 0 · Answer 2 · answered Feb 12 '14 at 16:25

0

HINT:

You won't face such confusion if you adapt the following method:

Use Weierstrass substitution to form a Quadratic Equation in $\tan \frac x2$

Then use Trigonometric Addition Formulas to find $\displaystyle\tan\left(\frac{\alpha+\beta}2\right)$

Finally use this

answered Feb 12 '14 at 16:25

lab bhattacharjee

274,582

While I was solving the question, I got $\cos\alpha + \cos\beta = 27/25$ and $\cos\alpha\cdot\cos\beta = 17/100$. That would imply that both $\cos\alpha$ and $\sin\beta$ are positive. – Shaurya Gupta Feb 12 '14 at 16:34
@shauryagupta, how have you solved ? – lab bhattacharjee Feb 12 '14 at 16:36
My course is a beginner's course, so I don't think that book writer would have gone about doing the question the way you are. – Shaurya Gupta Feb 12 '14 at 16:36
Squaring both sides of $6\cos\theta - 9 = -8\sin\theta$ and then replacing $sin^2\theta$ by $1-\cos^2\theta$ and using vieta's formula. – Shaurya Gupta Feb 12 '14 at 16:37
@shauryagupta, on squaring, you have immediately introduced an extraneous root, right? – lab bhattacharjee Feb 12 '14 at 16:44
Can you please tell me why? – Shaurya Gupta Feb 12 '14 at 16:47
@shauryagupta, one solution is the root of $$6\cos\theta-9=8\sin\theta$$ right? Are you allowed to use inverse trig functions? – lab bhattacharjee Feb 12 '14 at 16:48
Yes, I am allowed to use inverse trig functions. – Shaurya Gupta Feb 12 '14 at 16:51
I understand the extraneous root thing...So how should have I solved the question? – Shaurya Gupta Feb 12 '14 at 17:08

lsp · Answer 3 · 2014-02-12T16:33:07.830

0

$$\sin \theta=\frac{2\tan \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}}$$ $$\cos \theta=\frac{1-\tan^2 \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}}$$ Substitute the above two equations in the main equation and let $\tan \frac{\theta}{2} = x$. Now you get a quadratic equation $x$ from where you can find $x=\tan \frac{\theta}{2}$. Then find $\sin (\alpha+\beta)$ from here.

edited Feb 12 '14 at 16:33

answered Feb 12 '14 at 16:25

lsp

4,745

How do I determine the sign of $\sin\theta$ in this question.

3 Answers3