Skip to main content

Find the values: $\sin{\left(2\arccos{\frac{2}{5}}\right)}=?$, $\arcsin{a}+\arccos{a}=?$, $\arctan{a}+\arctan{\frac{1}{a}}=?$

The following exercises involve the simplification of numerical expressions involving inverse trigonometric functions. We use the co-function relationships and reciprocal relationships among trigonometric functions.

  1. $\sin{\left(2\arccos{\frac{2}{5}}\right)}=?$ $$\begin{align} &=2\sin{\left(\arccos{\frac{2}{5}}\right)}\cos{\left(\arccos{\frac{2}{5}}\right)}% =2\cdot\frac{2}{5}\sin{\left(\arccos{\frac{2}{5}}\right)}\\% &=\frac{4}{5}\sin{\left(\arcsin{\frac{\sqrt{5^2-2^2}}{5}}\right)}\\ &=\frac{4}{5}\cdot\frac{\sqrt{21}}{5}\\ &=\frac{4}{25}\sqrt{21}.□ \end{align}$$
  2. $\arcsin{a}+\arccos{a}=?$

    We observe that because sine and cosine functions are co-functions of each other, for any $a$ in the domain of inverse cosine and inverse sine functions, $$ y \equiv \sin(\theta) = \cos{\left(\frac{\pi}{2} - \theta\right)} \Rightarrow \arccos{y} = \frac{\pi}{2} -\arcsin{y}. $$

    Thus, the given equals $$ \arcsin{a} + \left(\frac{\pi}{2}-\arcsin{a}\right) = \frac{\pi}{2}.□ $$

  3. $\arctan{a}+\arctan{\frac{1}{a}}=?$

    We observe that because tangent and cotangent functions are co-functions of each other, for any $a$ in the domain of inverse tangent sine functions, $$ \DeclareMathOperator{\arccot}{arccot} y \equiv \tan(\theta) = \arccot{\left(\frac{\pi}{2} - \theta\right)} \Rightarrow \arccot{y} = \frac{\pi}{2} -\arctan{y}. $$

    Thus, the given equals $$ \arctan{a} + \left(\frac{\pi}{2}-\arccot{\frac{1}{a}}\right) = \frac{\pi}{2} + \arctan{a} -\arccot{\frac{1}{a}}. $$ Because tangent and cotangent are reciprocals of each other, $$ \arccot{\frac{1}{a}} = \arctan{a}, $$ and so, $$ \frac{\pi}{2} + \arctan{a} -\arccot{\frac{1}{a}} = \frac{\pi}{2} + \arctan{a} -\arctan{a} = \frac{\pi}{2}.□ $$

References

Labels:

Comments

Post a Comment

Popular posts from this blog

2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(4)

This problem appears at the Qualifying Examinations for Applicants for Japanese Government (MEXT) Scholarships 2020 . There are two mathematics exams: one for biology-related natural sciences (Mathematics A), and another for physics- and engineering-related natural sciences (Mathematics B). This problem is from the 2020 Mathematics (B) questionnaire . The official answer key is here . Problem 1(4) The division of a polynomial function $f(x)$ by $(x-1)^2$ gives the remainder of $x+1$, and that by $x^2$ gives the remainder $2x+3$. Thus, the remainder of the division of $f(x)$ by $x^2(x-1)$ is $$ \fbox{ A }x^2 + \fbox{ B }x + \fbox{ C } . $$ Solution We need to find the remainder when $f(x)$ is divided by $x^2(x-1)$. Because $x^2(x-1)$ is of order $n=3$, the remainder will be of at most the order $n=2$, which means that it is of the form $Ax^2 + Bx + C$. The problem is now to find the coefficients $A,B$ and $C$ such that $$ f(x) = Q(x)\
Post a Comment
Read more

Reasonable Majority Rule: Wise Decision-making in a Democracy

Let the people decide what is good for them. On the one hand, this principle is at the core of democracy and manifests in the phrase "majority rule." On the other hand, the protection of the voice of the minority is another principle a functioning democracy must enforce. Democracy requires both. The fundamental assumption is that, after free and informed debate, the majority will be reasonable enough to judge ideas based on their merits (and not on their emotional relationship with the proponents). Because in a democracy, the Government's wisdom is an extension of the people's wisdom, wise decision-making is the duty of every citizen. The "goodness" of a decision or an idea does not necessarily depend on the number of its proponents but on the independent practical assessment of available information. The difficulty in balancing the interests of opposing sides is highlighted on occasions when the minority is unwilling to accept the decision of the
Post a Comment
Read more

2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(1)

This problem appears at the Qualifying Examinations for Applicants for Japanese Government (MEXT) Scholarships 2020 . There are two mathematics exams: one for biology-related natural sciences (Mathematics A), and another for physics- and engineering-related natural sciences (Mathematics B). This problem is from the 2020 Mathematics (B) questionnaire . The official answer key is here . Problem 1(1) The largest one among natural numbers that are less than $$ \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020} $$ is $ \fbox { A } $. Solution The key idea is the following relationship that holds for positive numbers $p, q,$ and $b$. $$ \log_p{q} = \frac{\log_b{q}}{\log_b{p}}, $$ where $p\neq1$ and $b\neq1$. For convenience, let $$ L \equiv \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020}. $$ Selecting $b=e$, we obtain $$\begin{align} L &=\frac{\cancel{\log{3}}}{\log{2}}\cdot \frac{\cancel{\
Post a Comment
Read more