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A Trigo Solution

At our Math 17 class, we were asked to solve a trigo problem. The question is,

\(\cos \theta+\sin \theta =1\) ,such that \(0 \leq \theta < 2\pi \)

solve for \(\theta\)

My Solution:

\(\cos \theta+\sin \theta =1\)

From the eq. \(\cos^{2} \theta + \sin^{2} \theta = 1\)

\(\cos \theta \pm \sqrt{1-\cos^{2} \theta}=1 \)

\(\pm \sqrt{1-\cos^{2} \theta}=1-\cos \theta \)

Squaring Both Sides,

\(1-\cos^{2} \theta =1-2\cos \theta+\cos^{2} \theta \)

\(0=2\cos^{2} \theta-2\cos \theta\)

\(\cos^{2} \theta-\cos \theta=0\)

\(\cos \theta (\cos \theta-1)=0\)

\(\cos \theta=0\) and \(\cos \theta-1=0\)

Therefore,

\(\theta=90^\circ, 270^\circ\) and \(0^\circ \)

My Professor's Solution:

\(\cos \theta+\sin \theta =1\)

Squaring Both Sides

\(\cos^{2} \theta +2\cos \theta \sin \theta + \sin^{2} \theta = 1\)

\(2\cos \theta \sin \theta +1=1\)

\(2\cos \theta \sin \theta =0\)

\(\cos \theta = 0\) and \(\sin \theta =0\)

Thus,

\(\theta= 90^\circ, 270^\circ, 180^\circ\) and \(0^\circ \)

She wonders why my answer and her's are different. According to her, my solution is correct but came up with different answers. Could it be that there are some misconceptions? What would be the real answer?

Note by Jeffer Dave Cagubcob
4 years ago

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Both of your answers are wrong. Both errors come on from "squaring both sides". In your case, the effect of squaring both sides means that when you find \(\cos\theta=0\) as a solution, you take both \(90^\circ\) and \(270^\circ\) as possible answers (since you are allowing \(\sin\theta = \pm1\). If you look at the original equation, if \(\cos\theta=0\), we must have \(\sin\theta=1\), so \(270^\circ\) is not an answer.

Your professor is also finding the solutions to the equation \(\cos\theta+\sin\theta=-1\), so is getting twice as many answers as there are.

The best solution is: \[ \begin{array}{rcl} \sin\theta + \cos\theta & = & 1 \\ \tfrac{1}{\sqrt{2}}\sin\theta + \tfrac{1}{\sqrt{2}}\cos\theta & = & \tfrac{1}{\sqrt{2}} \\ \sin(\theta+45^\circ) & = & \sin45^\circ \\ \theta + 45^\circ & = & 45^\circ\,,\,135^\circ \\ \theta & = & 0^\circ\,,\,90^\circ \end{array} \]

Mark Hennings - 4 years ago

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Nice :O I was thinking of making use of the R-formula to approach this question

Ho Wei Haw - 4 years ago

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Thanks Mark, I agree with what you said about my Professor's mistake but I don't really get what is my mistake when I square both sides.

Jeffer Dave Cagubcob - 4 years ago

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Your arguement goes: \[ \begin{array}{rcl} \sin\theta + \cos\theta & = & 1 \\ \sin\theta & = & 1 - \cos\theta \\ \sin^2\theta & = & (1 - \cos\theta)^2 \end{array} \] The third line includes the option that \(-\sin\theta = 1 - \cos\theta\). In other words, in addition to solving \(\cos\theta+\sin\theta=1\) (\(\theta=0^\circ,90^\circ\)), you are also solving \(\cos\theta-\sin\theta = 1\) (\(\theta=0^\circ,270^\circ\)).

The problem with squaring both sides is that the argument does not go both ways. While it is true that \(a^2=b^2\) whenever \(a=b\), it is not always the case that \(a=b\) whenever \(a^2=b^2\). The method of squaring both sides is only totally safe when both sides of the equation are known to be positive (or both negative). In other words: \[ \mbox{If } \,a,b \ge 0 \quad \mbox{then} \quad a = b \qquad \Leftrightarrow \qquad a^2 = b^2 \]

Mark Hennings - 4 years ago

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@Mark Hennings Good explanation.

Always remember that when you manipulate equations (squaring, cubing, multiplying, adding, dividing, etc) you risk introducing additional solutions into the system. As such, you have to check that your solutions are indeed solutions, otherwise you will run into errors like claiming \( 1 = - 1 \).

Part of the reason why we don't realize this, is that we mostly deal with simultaneous linear equations, in which the operations that we do (addition of 2 equations, multiplication by a constant, substitution of variables) is well behaved. Outside of that, things could go wrong.

Calvin Lin Staff - 4 years ago

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@Mark Hennings Ahh.. Now i get it. So, lesson learned. XD . Thank you so much Mark. Also, I like your strategy in solving that problem.

Jeffer Dave Cagubcob - 4 years ago

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very brilliant solution for mark

Erick Sumargo - 4 years ago

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I agree.

Jeffer Dave Cagubcob - 4 years ago

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