Skip to main content

Solve: $\arcsin{\frac{3}{5}}=\arctan{x}$, $\arcsin{\frac{3}{5}}=2\arctan{x}$

Equations involving inverse trigonometric functions are solved for the unknown. The following relation is assumed proven: $$ \theta = \arcsin{\frac{a}{c}} = \arccos{\frac{\sqrt{c^2-a^2}}{c}}, $$ which is clear from the figure below.

  1. $\arcsin{\frac{3}{5}}=\arctan{x}$ $$\begin{align} \arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ &\Rightarrow x = \frac{3}{4}.□ \end{align}$$
  2. $\arcsin{\frac{3}{5}}=2\arctan{x}$ $$\begin{align} 2\arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ \frac{3}{4} &= \tan{\left(2\arctan{x}\right)} = \frac{2\tan{\left(\arctan{x}\right)}}{1-\tan^2{\left(\arctan{x}\right)}} \\ &= \frac{2x}{1-x^2}\\ &\Rightarrow 0 = 3x^2 +8x -3 = (3x-1)(3x+3) \\ &\Rightarrow x = \frac{1}{3} ( x = -1 \hbox{ is extraneous}).□ \end{align}$$

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(2)

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(2) Let $f(x) = 1 + \frac{1}{x-1}(x\neq 1)$. The solution of the equation $f\left(f\left(x\right)\right) = f\left(x\right)$ is $x = \fbox{A }, \fbox{B }$. Solution The right-hand side of the equation is $f(x)$, which is already given in the problem to be $$ f\left(x\right) = 1 + \frac{1}{x-1}. $$ The left-hand side of the equation is $$\begin{align} f\left(f\left(x\right)\right) &= 1 + \frac{1}{f(x) - 1}\\ &= 1 + \frac{1}{\left(1+\frac{1}{x-1}\right) - 1}\\ &= 1 + \frac{1}{\frac{1}{x-1}}\\ &= 1 + {x-1}\\ &= x. \end{align}...
Post a Comment
Read more

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 ...
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