Intereting Posts

How to show that $\mathbb R^n$ with the $1$-norm is not isometric to $\mathbb R^n$ with the infinity norm for $n>2$?
Morera's theorem of entire function
Can we use this formula for a certain indeterminate limit $1^{+\infty}$?
Bayesian posterior with truncated normal prior
Finding a function from the given functional equation .
What's the lower bound of the sum $S(n) = \sum_{k=1}^n \prod_{j=1}^k(1-\frac j n)$?
Infinite coproduct of affine schemes
$K_{1,3}$ packing in a triangulated planar graph
Showing that if a subset of a complete metric space is closed, it is also complete
Assume that $ 1a_1+2a_2+\cdots+na_n=1$, where the $a_j$ are real numbers.
An exercise in Kunen
simplify $\sqrt{5+2\sqrt{13}}+\sqrt{5-2\sqrt{13}}$
Not getting $-\frac{\pi}{4}$ for my integral. Help with algebra
The limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0?
How do Taylor polynomials work to approximate functions?

I have a trigonometric equation that is defined as:

$\sin(\alpha) – \cos(\alpha) = \frac{1}{2}$

Solving this equation by mathematica will yield $\alpha = 65.70 $ and $\alpha = -155.705 $

- Solve $\cos x+8\sin x-7=0$
- Rationalize $\left(\sqrt{3x+5}-\sqrt{5x+11} -\sqrt{x+9}\right)^{-1}$
- How to find area under sines without calculus?
- Simplifying $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$
- Steps to solve this system of equations: $\sqrt{x}+y=7$, $\sqrt{y}+x=11$
- Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$?

But As I solve it analytically, I will obtain different results:

First I exponentiate both sides to the power of two:

$(\sin(\alpha) – \cos(\alpha))^2 = \frac{1}{2}^2$

Now I expand the expressions:

$\sin(\alpha)^2 – 2 \sin(\alpha) \cos(\alpha) + \cos(\alpha)^2 = \frac{1}{4}$

As $\sin(\alpha)^2 + \cos(\alpha)^2 = 1$, I will have:

$1 – 2 \sin(\alpha) \cos(\alpha) = \frac{1}{4}$

Again if I put $2 \sin(\alpha) \cos(\alpha) = \sin(2\alpha)$ I will be left with:

$\sin(2 \alpha) = \frac{3}{4}$

Which will readily give $\alpha = 24.3$

So, why I am having different results? I don’t understand.

- proof using the mathematical induction
- Find the equation of the line in standard form.
- If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
- Starting with $\frac{-1}{1}=\frac{1}{-1}$ and taking square root: proves $1=-1$
- How to prove that $\frac{a+b}{2} \geq \sqrt{ab}$ for $a,b>0$?
- Let $P(x)$ be a polynolmial with degree $2009$ and leading coefficient unity such that $P(0)=2008,P(1)=2007,P(2)=2006,\ldots,P(2008)=0$,
- Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $
- Prove that $(mn)!$ is divisible by $(n!)\cdot(m!)^n$
- Simplify the expression $\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}$
- Proving that $x$ is an integer, if the differences between any two of $x^{1919}$, $x^{1960}$, and $x^{2100}$ are integers

If you square, you also get the solutions of

$$

\sin\alpha-\cos\alpha=-\frac{1}{2}

$$

In general, if you have an equation of the form $f(x)=g(x)$ and square both sides, you get, after rearranging,

$$

f(x)^2-g(x)^2=0 \tag{*}

$$

that can be rewritten

$$

(f(x)-g(x))(f(x)+g(x))=0

$$

so the solutions of (*) are the solutions of $f(x)-g(x)=0$ (the original equation) together with the solutions of $f(x)+g(x)=0$.

A safer way to solve your equation is to set $t=\tan(\alpha/2)$, so the equation becomes

$$

\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}=\frac{1}{2}

$$

that reduces to

$$

t^2+4t-3=0

$$

with solutions

$$

-2+\sqrt{7}\qquad\text{or}\qquad -2-\sqrt{7}

$$

The first solutions corresponds to

$$

\alpha=2\arctan(\sqrt{7}-2)\approx 65.7^\circ

$$

and the second one to

$$

\alpha=-2\arctan(\sqrt{7}+2)\approx-155.7^\circ

$$

or $\approx204.3^\circ$ if you want a value between $0$ and $360$.

Notice first of all that besides $24.3°$ you also have the solution $90°-24.3°=65.7°$ (supplementary angles have the same sine).

Your solution $24.3°$ is to discard, because it is a solution of the equation $\cos\alpha-\sin\alpha=1$ (which of course is the same as the given one when squared).

In summary: when you solve $\sin2\alpha=3/4$ you must consider all solutions, and discard fake ones:

$\alpha=48.6°/2$ (discard)

$\alpha=(48.6°+360°)/2$ (fine, it is the same as $-155.7°$)

$\alpha=(180°-48.6°)/2$ (fine)

$\alpha=(180°-48.6°+360°)/2$ (discard).

Squaring is often a useful (sometimes unavoidable) way to solve an equation, but it always introduces new solutions which may not fit the original problem. When you square both sides of an equation, you should always check your solutions to make sure they are applicable.

If you can avoid squaring, you might try doing so so as not to be confused with extra solutions. For this particular problem, you can solve it without squaring as follows:

$\dfrac12\ =\ \sin\alpha-\cos\alpha\ =\ \sqrt2\,(\cos45^\circ\sin\alpha – \sin45^\circ\cos\alpha)\ =\ \sqrt2\sin(\alpha-45^\circ)$

which gives $\sin(\alpha-45^\circ)=\frac1{2\sqrt2}$ which you can solve to give you the values obtained from Mathematica.

Every solution to $\sin(\alpha) – \cos(\alpha) = \frac{1}{2}$ is, indeed, a solution to $\sin(2 \alpha) = \frac{3}{4}$. That work you have done correctly.

Your error is either in misunderstanding what you have done, or what to do with that information.

Generally, the next step is:

- Find the set $S$ of
solutions to $\sin(2 \alpha) = \frac{3}{4}$*all*

Note that it really is important to find the *entire* set of solutions. Don’t be sloppy and just take one solution.

Once you’ve determined the set $S$, you know it contains every solution to the original equation $\sin(\alpha) – \cos(\alpha) = \frac{1}{2}$; however, it can (and does!) contain other things too! The next step is

- Identify which of the solutions in $S$ are or are not solutions to $\sin(\alpha) – \cos(\alpha) = \frac{1}{2}$

Once you’ve picked out which of the solutions in $S$ also satisfy $\sin(\alpha) – \cos(\alpha) = \frac{1}{2}$, you now have its complete solution set.

By *squaring* one unwittingly asks for two solutions of

$$\sin(\alpha) – cos(\alpha) = \frac{1}{2} ,\quad\sin(\alpha) – cos(\alpha) = \frac{-1}{2}$$

At the end we should retain only the solutions of interest.

You actually nailed it, you just didn’t know it.

Algebra’s fundamental theorem states: *Any* $\rm n^{th}$ order polynomial has $\rm n$ roots.

Then, if one increase the order of a polynomial by $\rm m$ (let’s say $\rm 1^{st}$ to $\rm 2^{nd}$) the number of roots will increase from $\rm n$ to $\rm n+m$, Therefore raising the power makes things different.

But raising to the same power both sides of an equation makes the solution set of the $\rm (n+m)^{th}$ order polynomial to contain the solutions set of the $\rm n^{th}$ polynomial. Meaning, it shouldn’t be a problem in your case, your solutions set *should be* **in** the square problem.

Start to get the solutions of $\sin(2α)=\frac{1}{2}$. If $0 ≤α ≤360º$, then your solution set will be $\{24.3º, 65.70º, 204.295º (-155.705º), 245.365º\}$ (check the graph below).

Values in the middle are the ones you are looking for (see graph below).

Has you can see by this graph, the actual set of solutions is much bigger. Let’s find the full solution set to your problem. To do so, select the solutions of the square problem which are also a solution to your initial problem in the range $\rm 0≤α≤360$ *i.e.*, $\rm α_1=65.7º, α_2=204.295º$. Now, let’s see how many times the $\sin$ of $2×65.7º$ or $\sin (131.4º)$ repeats.

If we imagine a circle it becomes clear that after every full turn one gets back to same ($\rm 65.7º$) point, *i.e.*, the first set of solutions satisfy

$$\rm \Bbb α_{1n}=2nπ+α_1 \approx n×360º+65.7º, \forall n \in \Bbb N$$

But, we can check that every $180º-65.7º$ has the same $\sin$ of $65.7º$. Appling previous reasoning, the second set satisfy

$$\rm \Bbb α’_{1n}=(2n+1)π-α_1 \approx n×360º+180º-65.7º=n×360º+114.3º, \forall n \in \Bbb N$$

We do the same with $\rm α_2$. The result is

$$\rm \Bbb α_{2n}=2nπ+α_2 \approx n×360º+204.295º, \forall n \in \Bbb N$$

$$\rm \Bbb α’_{2n}=(2n+1)π-α_2 \approx n×360º+180º-204.295º=n×360º-24,295º, \forall n \in \Bbb N$$

The total set of solutions is:

$$\rm \Bbb S= α_{1n} \cup α’_{1n} \cup α_{2n} \cup α’_{2n}, \forall n \in \Bbb N$$

- $\epsilon$-$\delta$ proof that $\lim_{x \to 1} \sqrt{x} = 1$
- What is the highest power of $n$ in $(n^r-1)!$
- Prove that $\frac{x^x}{x+y}+\frac{y^y}{y+z}+\frac{z^z}{z+x} \geqslant \frac32$
- For $x, y \in \mathbb{R}$, define $x \sim y $ if $x-y \in \mathbb{Q}$. Is $\mathbb{R}/\!\!\sim$ Hausdorff?
- What rational numbers have rational square roots?
- Determining if something is a characteristic function
- Existence of exhaustion by compact sets
- Principal ideal domain not euclidean
- Is $L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)$?
- Choice Problem: choose 5 days in a month, consecutive days are forbidden
- Abstract algebra: homomorphism and an element of a prime order
- Is $ 0.112123123412345123456\dots $ algebraic or transcendental?
- $f$ continuous, $f^N$ analytic on a domain D implies $f$ analytic on D
- Probability distribution of the maximum of random variables
- Existence of solution of $\frac{\partial f}{\partial t}=-\Delta f+|\nabla f|^2-R(x,t)$