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:

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

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

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

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

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

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

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

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

Therefore,

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

My Professor's Solution:

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

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

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

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

\(\cos \theta = 0\) 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?

No vote yet

3 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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.

Log in to reply

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: \[ \mbox{If } \,a,b \ge 0 \quad \mbox{then} \quad a = b \qquad \Leftrightarrow \qquad a^2 = b^2 \]

Log in to reply

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.

Log in to reply

Log in to reply

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

Log in to reply

very brilliant solution for mark

Log in to reply

I agree.

Log in to reply