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sammy [17]
3 years ago
14

Trig help please Find the exact value of each trigonometric equation

Mathematics
2 answers:
gavmur [86]3 years ago
7 0

The exact value for the equation is true but I don't really think that's the question so anyways...

- 15.) The exact form for this equation is -13pi/3 and the decimal form -13.613...

- 16.) The exact form for this equation is 23pi/4 and the decimal form 18/064...

- 17.) The exact form is -7pi/2 as the decimal is -10.995...

- 18.) The exact is -29pi/6 and the decimal is -15.184...

GaryK [48]3 years ago
6 0

We know any trig problem that asks for exact values probably has something to do with 30° or 45° and their multiples.  That's \pi/6 and \pi/4; we're apparently doing radians in this one.

General rules off the top of my head: Coterminal angles (gotten by adding or subtracting multiples of 2π) have the same values for their trig functions , cosine is even, sine is odd, cosine negate supplementary angles, sine of supplementary angles is unchanged, and the cosine of an angle is the sine of the complementary angle.

15

\cos (- \frac{13\pi}{3}) = \cos( 13\pi/3-6(2\pi)) =\cos(\pi/3) = \frac 1 2

16

\csc(\frac{23 \pi}{4}) = \dfrac{1}{\sin (23\pi/4 - 3(8\pi/4))} = \dfrac{1}{\sin(-\pi/4)}= \dfrac{1}{- 1 /\sqrt{2}} = - \sqrt{2}

17

\sec(-\frac {7 \pi}{2}) = \dfrac{1}{\cos(-7\pi/2+ (4/2)(2\pi) )}= \dfrac{1}{\cos(\pi/2)} = \dfrac 1 0

That one is undefined

18

\cot(-\frac{29\pi}{6}) = \cot(-29\pi/6 + (18/6) (2 \pi)) = \cot(7\pi/6) \\= \tan(\pi/2 - 7\pi/6) = \tan(-4\pi/6)= \tan(-2\pi/3 + \pi) = \tan(\pi/3)= \sqrt{3}

Whoever created this math homework problem needs a lesson in writing and typesetting math.  Let's list the errors:

Exact -- capitalized

each equation -- there are no equations

0 to 2 pi for theta -- do they want us to find the values or find the thetas but not evaluate the trig function?

theta is spelled out, not typeset

trig functions shouldn't be typeset in italics

sec -(7 pi/2)  is a typo

Sometimes there's a space after the problem number sometimes there isn't

This is awful.  Demand more of your teachers and online exercises!

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35.2792 cm from one end (The square)

And 27.7208 cm from the other end (The circle)

b. See (b) explanation below

Step-by-step explanation:

Given

Length of Wire ,= 63cm

Let L be the length of one side of the square

Circumference of a circle = 2πr

Perimeter of a square = 4L

a. To minimise

4L + 2πr = 63 ----- make r the subject of formula

2πr = 63 - 4L

r = (63 - 4L)/2π

r = (31.5 - 2L)/π

Let X = Area of the Square. + Area of the circle

X = L² + πr²

Substitute (31.5 - 2L)/π for r

So,

X² = L² + π((31.5 - 2L)/π)²

X² = L² + π(31.5 - 2L)²/π²

X² = L² + (31.5 - 2L)²/π

X² = L² + (992.25 - 126L + 4L²)/π

X² = L² + 992.25/π - 126L/π +4L²/π ------ Collect Like Terms

X² = 992.25/π - 126L/π + 4L²/π + L²

X² = 992.25/π - 126L/π (4/π + 1)L² ---- Arrange in descending order of power

X² = (4/π + 1)L² - 126L/π + 992.25/π

The coefficient of L² is positive so this represents a parabola that opens upward, so its vertex will be at a minimum

To find the x-cordinate of the vertex, we use the vertex formula

i.e

L = -b/2a

L = - (-126/π) / (2 * (4/π + 1)

L = (126/π) / ( 2 * (4 + π)/π)

L = (126/π) /( (8 + 2π)/π)

L = 126/π * π/(8 + 2π)

L = (126)/(8 + 2π)

L = 63/(4 + π)

So, for the minimum area, the side of a square will be 63/(4 + π)

= 8.8198 cm ---- Approximated

We will need to cut the wire at 4 times the side of the square. (i.e. the four sides of the square)

I.e.

4 * (63/(4 + π)) cm

Or

35.2792 cm from one end.

Subtract this result from 63, we'll get the other end.

i.e. 63 - 35.2792

= 27.7208 cm from the other end

b. To maximize

Now for the maximum area.

The problem is only defined for 0 ≤ L ≤ 63/4 which gives

0 ≤ L ≤ 15.75

When L=0, the square shrinks to 0 and the whole 63 cm wire is made into a circle.

Similarly, when L =15.75 cm, the whole 63 cm wire is made into a square, the circle shrinks to 0.

Since the parabola opens upward, the maximum value is at one endpoint of the interval, either when

L=0 or when L = 15.75.

It is well known that if a piece of wire is bent into a circle or a square, the circle will have more area, so we will assume that the maximum area would be when we "cut" the wire 0, or no, centimeters from the

end, and bend the whole wire into a circle. That is we don't cut the wire at

all.

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