Find the smallest value of angle ##α + β ##

[chwala]

Homework Statement

See attached.

Relevant Equations

Trigonometry.

In my approach i have,

$α + β = \tan^{-1} \left[ \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1-\dfrac{a}{a+1} ⋅\dfrac{1}{2a+1}}\right]$

...

$α + β = \tan^{-1} \left[ \dfrac{2a^2+3a+1}{(a+1)(2a+1)}\right] \div \left[\dfrac{2a^2+2a+1}{(a+1)(2a+1)}\right]$

$α + β = \tan^{-1} \left[\dfrac{(a+1)(2a+1)}{2a^2+2a+1}\right]$

Let

$\left[\dfrac{(a+1)(2a+1)}{2a^2+2a+1}\right]=0$

$⇒a=-1, a=-\dfrac{1}{2}$

checking my working and latex... a minute...

 

[SammyS]

Homework Statement: See attached.
Relevant Equations: Trigonometry.

View attachment 331693

In my approach i have,

$α + β = \tan^{-1} \left[ \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1-\dfrac{a}{a+1} ⋅\dfrac{1}{2a+1}}\right]$

...

$α + β = \tan^{-1} \left[ \dfrac{2a^2+3a+1}{(a+1)(2a+1)}\right] \div \left[\dfrac{2a^2+2a+1}{(a+1)(2a+1)}\right]$checking my working and latex... a minute...

Check your algebra.

 

[Mark44]

In addition to the error that @SammyS points out in his quote of your work, there is also this:

Let $\left[\dfrac{(a+1)(2a+1)}{2a^2+2a+1}\right]=0$

What justification is there for arbitrarily setting the quantity on the left side to zero?

 

[SammyS]

Homework Statement: See attached.
Relevant Equations: Trigonometry.

View attachment 331693

In my approach i have,

$α + β = \tan^{-1} \left[ \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1-\dfrac{a}{a+1} ⋅\dfrac{1}{2a+1}}\right]$

Rather than jumping to the above expression, it would be helpful to those reading your thread if you would start off with something like:

$\displaystyle \tan \alpha + \beta = \dfrac{\tan \alpha + \tan \beta }{1- \tan \alpha \tan \beta}$

$\displaystyle \quad \quad = \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1-\dfrac{a}{a+1} ⋅\dfrac{1}{2a+1}}$

You can simplify that expression quite a bit. Why not do that before applying the $\displaystyle \arctan$ function?
I suggest multiplying the numerator and denominator by $\displaystyle \ (a+1)(2a+1) \$, then simplify. In fact, you might not have to use the $\displaystyle \arctan$ function at all.

 

[chwala]

Homework Statement: See attached.
Relevant Equations: Trigonometry.

View attachment 331693

In my approach i have,

$α + β = \tan^{-1} \left[ \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1-\dfrac{a}{a+1} ⋅\dfrac{1}{2a+1}}\right]$

...

$α + β = \tan^{-1} \left[ \dfrac{2a^2+3a+1}{(a+1)(2a+1)}\right] \div\left[\dfrac{2a^2+2a+1}{(a+1)(2a+1)}\right]$

$α + β = \tan^{-1} \left[\dfrac{(a+1)(2a+1)}{2a^2+2a+1}\right]$

Let

$\left[\dfrac{(a+1)(2a+1)}{2a^2+2a+1}\right]=0$

$⇒a=-1, a=-\dfrac{1}{2}$

checking my working and latex... a minute...

$α + β = \tan^{-1} \left[ \dfrac{2a^2+2a+1}{(a+1)(2a+1)}\right] \div \left[\dfrac{2a^2+2a+1}{(a+1)(2a+1)}\right]$$α + β = \tan^{-1} \left[ \dfrac{2a^2+2a+1}{2a^2+3a+1}\right] × \left[\dfrac{2a^2+3a+1}{2a^2+2a+1}\right]$

$α + β = \tan^{-1} (1) = \dfrac{π}{4}$

 

[Mark44]

Again, you don't have to start each line with "$\alpha + \beta =$" when the goal is to simplify a complicated expression. Instead, you can use = to string together equal quantities.

 

[pasmith]

Homework Statement: See attached.
Relevant Equations: Trigonometry.

View attachment 331693

In my approach i have,

$α + β = \tan^{-1} \left[ \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1-\dfrac{a}{a+1} ⋅\dfrac{1}{2a+1}}\right]$

...

$α + β = \tan^{-1} \left[ \dfrac{2a^2+3a+1}{(a+1)(2a+1)}\right] \div \left[\dfrac{2a^2+2a+1}{(a+1)(2a+1)}\right]$

<br /> \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \frac{a}{a+1}\frac{1}{2a+1}} = <br /> \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \frac{a}{a+1}\frac{1}{2a+1}} \cdot \frac{(a+1)(2a+1)}{(a+1)(2a+1)} =<br /> \frac{a(2a+1) + (a+1)}{(2a+1)(a+1) - a} =<br /> \frac{ 2a^2 + 2a + 1}{2a^2 + 2a + 1}<br /> = 1.

 

[chwala]

<br /> \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \frac{a}{a+1}\frac{1}{2a+1}} =<br /> \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \frac{a}{a+1}\frac{1}{2a+1}} \cdot \frac{(a+1)(2a+1)}{(a+1)(2a+1)} =<br /> \frac{a(2a+1) + (a+1)}{(2a+1)(a+1) - a} =<br /> \frac{ 2a^2 + 2a + 1}{2a^2 + 2a + 1}<br /> = 1.

@pasmith they wanted the smallest angle or you were showing something else...

 

[Mark44]

@pasmith they wanted the smallest angle or you were showing something else...

The wording of the problem isn't clear, IMO. The smallest positive value for $\alpha + \beta$ is $\frac \pi 4$. There are smaller negative values that satisfy the two given equations.

 

[chwala]

The wording of the problem isn't clear, IMO. The smallest positive value for $\alpha + \beta$ is $\frac \pi 4$. There are smaller negative values that satisfy the two given equations.

I beg to disagree. Interpreting the problem as it is one can only deduce that the required smallest angle $=45^0$.

 

[Mark44]

I beg to disagree. Interpreting the problem as it is one can only deduce that the required smallest angle $=45^0$.

The given equations are $\tan(\alpha) = \frac a {a + 1}$ and $\tan(\beta) = \frac 1 {2a + 1}$. These lead to the equation $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.

As you've seen, $\alpha + \beta = \frac \pi 4$ is a solution, but there are an infinite number of solutions to this equation above, one of which is $-\frac{3\pi} 4$, which is smaller than $\frac \pi 4$. There are many others that are even more negative; i.e., smaller. This is why I said that the problem statement isn't clear, and would have been improved by asking for the smallest positive value of $\alpha + \beta$.

Of course, once you introduce the inverse tangent, whose principal domain is $(\frac{-\pi}2, \frac \pi 2)$, then there is only one solution.

 

[SammyS]

The given equations are $\tan(\alpha) = \frac a {a + 1}$ and $\tan(\beta) = \frac 1 {2a + 1}$. These lead to the equation $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.

As you've seen, $\alpha + \beta = \frac \pi 4$ is a solution, but there are an infinite number of solutions to this equation above, one of which is $-\frac{3\pi} 4$, which is smaller than $\frac \pi 4$. There are many others that are even more negative; i.e., smaller. This is why I said that the problem statement isn't clear, and would have been improved by asking for the smallest positive value of $\alpha + \beta$.

I beg to AGREE with Mark on this point.

In general, when using Inverse trig functions to solve trig equations, we must exercise care.

The solution to an equation such as $\displaystyle \ \tan(\theta) = u\$ is $\displaystyle \ \theta = \arctan(u) +k\pi \, ,\$ where $k$ is an integer.

Of course, once you introduce the inverse tangent, whose principal domain is $(\frac{-\pi}2, \frac \pi 2)$, then there is only one solution.

As a general method, the following is not particularly good at getting an overall solution in closed form, but it does use the principal domain of $\displaystyle \arctan \, .$

Alternate approach:
If $\displaystyle \tan(\alpha) = \dfrac a {a + 1} \, , \$ then using $\displaystyle \arctan \$

we have $\displaystyle \alpha = \arctan(\tan(\alpha)) = \arctan \, \dfrac a {a + 1} \, . \$

Similarly, $\displaystyle \beta = \arctan(\tan(\beta)) = \arctan \, \dfrac {1} {2a + 1} \, . \$

For $\displaystyle -1 < a < \frac {-1} 2 \, ,$ each of the above expressions gives a negative result for $\alpha$ and for $\beta$.

Using any particular value for $a$ in this interval gives a value of $\displaystyle \, \frac {-3\pi} 4 \,$ for $\displaystyle \alpha + \beta =\arctan \, \dfrac a {a + 1} + \arctan \, \dfrac {1} {2a + 1} \, .$

For example, try $\displaystyle a=\frac{-3}{4} \$ on a scientific calculator. .