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How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=5, c=6?

Alan P.
Featured 4 months ago

(see below)

Explanation:

Based on the Pythagorean Theorem, we know that
#color(white)("XXX")a^2+b^2=c^2

$\rightarrow b = \sqrt{{c}^{2} - {a}^{2}} = \sqrt{{6}^{2} - {5}^{2}} = \sqrt{11}$

#sin = "opposite"/"hypotenuse"color(white)("XX")rarr sin(A)=5/6#

#csc = "hypotenuse"/"opposite" color(white)("XX")rarr csc(A)=6/5#

#cos = "adjacent"/"hypotenuse"color(white)("XX")rarr cos(A)=sqrt(11)/6#

#sec = "hypotenuse"/"adjacent" color(white)("XX")rarr sec(A)=6/sqrt(11)#

#tan = "opposite"/"adjacent"color(white)("XX")rarr tan(A)=5/sqrt(11)#

#cot = "adjacent"/"opposite" color(white)("XX")rarr cot(A)=sqrt(11)/5#

How do you divide # (1+7i) / (5-3i) # in trigonometric form?

Shwetank Mauria
Featured 4 months ago

$\frac{1 + 7 i}{5 - 3 i} = \frac{5}{\sqrt{17}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{19}{8}\right)$

Explanation:

Let us write the two complex numbers in polar coordinates and let them be

${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

Here, if two complex numbers are ${a}_{1} + i {b}_{1}$ and ${a}_{2} + i {b}_{2}$ ${r}_{1} = \sqrt{{a}_{1}^{2} + {b}_{1}^{2}}$, ${r}_{2} = \sqrt{{a}_{2}^{2} + {b}_{2}^{2}}$ and $\alpha = {\tan}^{- 1} \left({b}_{1} / {a}_{1}\right)$, $\beta = {\tan}^{- 1} \left({b}_{2} / {a}_{2}\right)$

$\left\{{r}_{1} / {r}_{2}\right\} \left\{\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta}\right\}$ or

$\left\{{r}_{1} / {r}_{2}\right\} \left\{\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} \times \frac{\cos \beta - i \sin \beta}{\cos \beta - i \sin \beta}\right\}$

$\left({r}_{1} / {r}_{2}\right) \frac{\left(\cos \alpha \cos \beta + \sin \alpha \sin \beta\right) + i \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right)}{\left({\cos}^{2} \beta + {\sin}^{2} \beta\right)}$ or

$\left({r}_{1} / {r}_{2}\right) \cdot \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$ or

${z}_{1} / {z}_{2}$ is given by $\left({r}_{1} / {r}_{2} , \left(\alpha - \beta\right)\right)$

So for division complex number ${z}_{1}$ by ${z}_{2}$ , take new angle as $\left(\alpha - \beta\right)$ and modulus the ratio ${r}_{1} / {r}_{2}$ of the modulus of two numbers.

Here $1 + 7 i$ can be written as ${r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ where ${r}_{1} = \sqrt{{1}^{2} + {7}^{2}} = \sqrt{50}$ and $\alpha = {\tan}^{- 1} 7$

and $5 - 3 i$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{5}^{2} + {3}^{2}} = \sqrt{34}$ and $\beta = {\tan}^{- 1} \left(- \frac{3}{5}\right)$

and ${z}_{1} / {z}_{2} = \frac{\sqrt{50}}{\sqrt{34}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = \alpha - \beta$

Hence, $\tan \theta = \tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{7 - \left(- \frac{3}{5}\right)}{1 + 7 \times \left(- \frac{3}{5}\right)} = \frac{\frac{38}{5}}{- \frac{16}{5}} = - \frac{19}{8}$.

Hence, $\frac{1 + 7 i}{5 - 3 i} = \frac{\sqrt{50}}{\sqrt{34}} \left(\cos \theta + i \sin \theta\right)$

= $\frac{5}{\sqrt{17}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{19}{8}\right)$

How do you simplify the expression : #arcsin((x+1)/sqrt(2*(x²+1)))# ?

dk_ch
Featured 4 months ago

Given expression #=arcsin((x+1)/sqrt(2*(x²+1)))#
Let $x = \tan \theta$

So $\theta = {\tan}^{-} 1 x$

Inserting $x = \tan \theta$ the given expression becomes

$= \arcsin \left(\frac{\tan \theta + 1}{\sqrt{2 \cdot \left({\tan}^{2} \theta + 1\right)}}\right)$

$= \arcsin \left(\frac{\sin \frac{\theta}{\cos} \theta + 1}{\sqrt{2 \cdot \left({\sec}^{2} \theta\right)}}\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \left(\frac{\sin \theta \sec \theta + 1}{\sec \theta}\right)\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \left(\frac{\sin \theta \cancel{\sec} \theta}{\cancel{\sec}} \theta + \frac{1}{\sec \theta}\right)\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \left(\sin \theta + \cos \theta\right)\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta\right)$

$= \arcsin \left(\cos \left(\frac{\pi}{4}\right) \sin \theta + \sin \left(\frac{\pi}{4}\right) \cos \theta\right)$

$= \arcsin \left(\sin \left(\theta + \frac{\pi}{4}\right)\right)$

$= \theta + \frac{\pi}{4}$

$= {\tan}^{-} 1 x + \frac{\pi}{4}$

How can i solve this (trigonometry)?

sente
Featured 3 months ago

$x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$

Explanation:

Note first that as $- 1 \le \sin x \le 1$ for all real $x$, we must have $\sin x - 2 \le - 1 < 0$. Then, $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right)$ is the product of $\sin x - \frac{1}{2}$ and a negative number, meaning it is less than or equal to $0$ if and only if $\sin x - \frac{1}{2} \ge 0$.

Thus, the given problem is equivalent to the problem $\sin x - \frac{1}{2} \ge 0$ with the restriction $x \in \left[0 , 2 \pi\right]$

Adding $\frac{1}{2}$ to each side of the inequality, we get

$\sin x \ge \frac{1}{2}$

Using the unit circle, we can tell that on our restricted interval, we have $\sin x \ge \frac{1}{2}$ if and only if $x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$. Therefore, this is our answer.

$x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$

The terminal side of θ lies on a given line in the specified quadrant. Find the values of the six trigonometric functions of θ by finding a point on the line? Line y = 1/6x Quad: III

HSBC244
Featured 3 months ago

You have to start by finding a point that lies on this line. I think the simplest would be $\left(- 6 , - 1\right)$ (since we are in quadrant III, both the x and y axis have to have negative values).

We now draw an imaginary triangle, as shown in the following diagram.

We can now clearly see that our side opposite $\theta$ measures $1$ unit and our side adjacent $\theta$ measures $6$ units.

We must finally find the hypotenuse prior to determining the ratios. By pythagorean theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${\left(- 1\right)}^{2} + {\left(- 6\right)}^{2} = {c}^{2}$

$1 + 36 = {c}^{2}$

${c}^{2} = 37$

$c = \pm \sqrt{37}$

However, the hypotenuse can never have a negative length, so we can only accept the positive solution. We can now find our ratios.

$\sin \theta = \text{opposite"/"hypotenuse} = - \frac{1}{\sqrt{37}} = - \frac{\sqrt{37}}{37}$

#csctheta = 1/sintheta = 1/("opposite"/"hypotenuse") = "hypotenuse"/"opposite" = sqrt(37)/(-1) = -sqrt(37)#

$\cos \theta = \text{adjacent"/"hypotenuse} = - \frac{6}{\sqrt{37}} = \frac{- 6 \sqrt{37}}{37}$

#sectheta = 1/costheta = 1/("adjacent"/"hypotenuse") = "hypotenuse"/"adjacent" = sqrt(37)/(-6) = -sqrt(37)/6#

$\tan \theta = \text{opposite"/"adjacent} = - \frac{1}{- 6} = \frac{1}{6}$

#cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite" = -6/(-1) = 6#

Hopefully this helps!

How do you divide #( i-3) / (5i +2)# in trigonometric form?

Alan P.
Featured 2 months ago

$\frac{i - 3}{5 i + 2} \approx 0.587 \cdot \left(\cos \left(1.630\right) + i \cdot \sin \left(- 1.630\right)\right)$

Explanation:

$\textcolor{red}{\text{~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~}}$
$\textcolor{w h i t e}{\text{XX }} \textcolor{b l u e}{a + b i \leftrightarrow r \cdot \left[\cos \left(\theta\right) + i \cdot \sin \left(\theta\right)\right]}$
$\textcolor{w h i t e}{\text{XXX")color(blue)("where } r = \sqrt{{a}^{2} + {b}^{2}}}$
#color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})#

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

$\textcolor{red}{\text{~~~ Trigonometric Division ~~~}}$
#color(white)("XX ")color(blue)((cos(theta)+i * sin(theta))/(cos(phi)+i * sin(phi)))#

#color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta) * cos(phi)-cos(theta) * sin(phi))]#

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

If $A = i - 3 = - 3 + 1 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{A} = \sqrt{{\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{10}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{A} = \arctan \left(- \frac{1}{3}\right) + \pi \approx 2.819842099$

If $B = 5 i + 2 = 2 + 5 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{B} = \sqrt{{2}^{2} + {5}^{2}} = \sqrt{29}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{B} = \arctan \left(\frac{5}{2}\right) \approx 1.19028995$

$\frac{A}{B} = \frac{i - 3}{5 + 2 i}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{10}}{\sqrt{29}} \cdot \left(\frac{\cos \left({\theta}_{A}\right) + i \sin \left({\theta}_{A}\right)}{\cos \left({\theta}_{B}\right) + i \sin \left({\theta}_{B}\right)}\right)$

#color(white)("XXX")=sqrt(10)/sqrt(29) *([color(green)(cos(theta_A)*cos(theta_B)+sin(theta_A) * sin(theta_B))+i[color(magenta)(sin(theta_A) * cos(theta_B) - cos(theta_A) * sin(theta_B))])#

Plugging in the values for ${\theta}_{A}$ and ${\theta}_{B}$ and using a calculator/spreadsheet:
$\textcolor{w h i t e}{\text{XXX}} \approx - 0.034482759 + 0.586206897 i$

If we call this value $C$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{C} = \sqrt{{\left(- 0.03448 \ldots\right)}^{2} + {\left(0.5862 \ldots\right)}^{2}} \approx 0.58722022$
and (since $C$ is in Quadrant 2)
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{C} = \arctan \left(\frac{0.5862 \ldots}{- 0.03448 \ldots}\right) + \pi \approx 1.62955215$

$\frac{A}{B} = C = {r}_{C} \left(\cos \left({\theta}_{C}\right) + i \cdot \sin \left({\theta}_{C}\right)\right)$

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