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If #A = <7 ,-5 ,6 >#, #B = <4 ,-8 ,9 ># and #C=A-B#, what is the angle between A and C?

Douglas K.
Featured 2 months ago

#theta ~~ 102.7°#

Explanation:

Compute vector C:

$\overline{C} = \overline{A} - \overline{B} = \left(7 - 4\right) \hat{i} + \left(- 5 - - 8\right) \hat{j} + \left(6 - 9\right) \hat{k}$

$\overline{C} = 3 \hat{i} + 3 \hat{j} - 3 \hat{k}$

Compute the dot-product of vectors A and C:

$\overline{A} \cdot \overline{C} = \left(7\right) \left(3\right) + \left(- 5\right) \left(3\right) + \left(6\right) \left(- 3\right)$

$\overline{A} \cdot \overline{C} = \left(7\right) \left(3\right) + \left(- 5\right) \left(3\right) + \left(6\right) \left(- 3\right)$

$\overline{A} \cdot \overline{C} = - 12$

There is another form for the equation of the dot-product that contains the angle between the two vectors:

$\overline{A} \cdot \overline{C} = | \overline{A} | | \overline{C} | \cos \left(\theta\right) = - 12$

Compute the magnitude of vector A:

$| \overline{A} | = \sqrt{{7}^{2} + {\left(- 5\right)}^{2} + {6}^{2}}$

$| \overline{A} | = \sqrt{110}$

Compute the magnitude of vector C:

$| \overline{C} | = \sqrt{{3}^{2} + {3}^{2} + {\left(- 3\right)}^{2}}$

$| \overline{C} | = 3 \sqrt{3}$

$\sqrt{110} \left(3 \sqrt{3}\right) \cos \left(\theta\right) = - 12$

#cos(theta) = -12/(sqrt(110)(3sqrt(3))#

$\theta = {\cos}^{-} 1 \left(- \frac{4}{\sqrt{330}}\right)$

#theta ~~ 102.7°#

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 2 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 evaluate #sec(cos^-1(1/2))# without a calculator?

sente
Featured 2 months ago

$\sec \left({\cos}^{- 1} \left(\frac{1}{2}\right)\right) = 2$

Explanation:

Another way, without calculating ${\cos}^{- 1} \left(\frac{1}{2}\right)$:

Consider a right triangle with an angle $\theta = {\cos}^{- 1} \left(\frac{1}{2}\right)$. Then $\cos \left(\theta\right) = \frac{1}{2}$, meaning the ratio of its adjacent side to the hypotenuse is $\frac{1}{2}$. Thus the ratio of the hypotenuse to its adjacent side, that is, $\sec \left(\theta\right)$, is $\frac{2}{1} = 2$.

Thus $\sec \left({\cos}^{- 1} \left(\frac{1}{2}\right)\right) = \sec \left(\theta\right) = 2$

Note that this same reasoning shows that in general, $\sec \left({\cos}^{- 1} \left(x\right)\right) = \frac{1}{x}$

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

dk_ch
Featured 2 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 do you graph #r = 12cos(theta)#?

Douglas K.
Featured 2 months ago

Set your compass to a radius of 6, put the center point at $\left(6 , 0\right)$, and draw a circle.

Explanation:

Multiply both sides of the equation by r:

${r}^{2} = 12 r \cos \left(\theta\right)$

Substitute ${x}^{2} + {y}^{2}$ for ${r}^{2}$ and $x$ for $r \cos \left(\theta\right)$

${x}^{2} + {y}^{2} = 12 x$

The standard form of this type of equation (a circle) is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

To put the equation in this form, we need to complete the square for both the x and y terms. The y term is easy, we merely subtract $0$ in the square:

${x}^{2} + {\left(y - 0\right)}^{2} = 12 x$

To complete the square for the x terms, we add $- 12 x + {h}^{2}$ both sides of the equation:

${x}^{2} - 12 x + {h}^{2} + {\left(y - 0\right)}^{2} = {h}^{2}$

We can use the pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ to find the value of h:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} - 12 x + {h}^{2}$

$- 2 h x = - 12 x$

$h = 6$

Substitute 6 for h into the equation of the circle:

${x}^{2} - 12 x + {6}^{2} + {\left(y - 0\right)}^{2} = {6}^{2}$

We know that the first three terms are a perfect square with h = 6:

${\left(x - 6\right)}^{2} + {\left(y - 0\right)}^{2} = {6}^{2}$

This is a circle with a radius of 6 and a center point $\left(6 , 0\right)$

The inverse of the function sin(x) + cos(x) is...?

Douglas K.
Featured 1 month ago

Explanation:

Here is the graph:

This looks like a cosine function that has been multiplied by an amplitude and phase shift to the right:

$f \left(x\right) = A \cos \left(x - \phi\right) = \sin \left(x\right) + \cos \left(x\right)$

My graphing tool allows me to obtain values of points and I can tell you that $A = \sqrt{2}$

$\cos \left(x - \phi\right) = \frac{1}{\sqrt{2}} \sin \left(x\right) + \frac{1}{\sqrt{2}} \cos \left(x\right)$

This fits the trigonometric identity:

$\cos \left(x - \phi\right) = \sin \left(x\right) \sin \left(\phi\right) + \cos \left(x\right) \cos \left(\phi\right)$

where $\cos \left(\phi\right) = \sin \left(\phi\right) = \frac{1}{\sqrt{2}}$

This happens at $\phi = \frac{\pi}{4}$

The graphing tool confirms that the x coordinates are shift by $\frac{\pi}{4}$.

$f \left(x\right) = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$

Substitute ${f}^{-} 1 \left(x\right)$ for every x:

$f \left({f}^{-} 1 \left(x\right)\right) = \sqrt{2} \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

The left side becomes x by definition:

$x = \sqrt{2} \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

Make the $\sqrt{2}$ on the right disappear by multiplying both sides by $\frac{\sqrt{2}}{2}$:

$\frac{\sqrt{2}}{2} x = \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

Use the inverse cosine on both sides:

${\cos}^{-} 1 \left(\frac{\sqrt{2}}{2} x\right) = {f}^{-} 1 \left(x\right) - \frac{\pi}{4}$

Solve for ${f}^{-} 1 \left(x\right)$:

${f}^{-} 1 \left(x\right) = {\cos}^{-} 1 \left(\frac{\sqrt{2}}{2} x\right) + \frac{\pi}{4}$

To confirm that this is truly an inverse, verify that $f \left({f}^{-} 1 \left(x\right)\right) = {f}^{-} 1 \left(f \left(x\right)\right) = x$

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