Mathematics, 12.08.2019 19:30, keelynnbarrier

Asap
pl show me how to solve with all the steps.
i'll give you brainliest and pts
sin 18° = (√5 - 1)/4

Answers: 1

Show answers

Answers

answered: bethzabezarahi

Answer is in the explanation.

Step-by-step explanation:

$\sin(18)=x$

We are going to find $\cos(18)$ in terms of using Pythagorean Identity: $\sin^2(\theta)+\cos^2(\theta)=1$ .

$x^2+\cos^2(18)=1$

Subtract x^2 on both sides:

$\cos^2(18)=1-x^2$

Take square root of both sides:

$\cos(18)=\pm \sqrt{1-x^2}$

We are going to use the positive one since 18 is in the first quadrant. Sine value and cosine value are both positive there:

$\cos(18)=\sqrt{1-x^2}$

Now we are going to find a way to use the unit circle.

I know 18 is a factor of 90 because it goes into it 5 times with no remainder.

So let’s use that:

$1=\sin(5 \cdot 18)$ (sin(90 deg) is 1)

We are going to expand this using $5\cdot 18=4\cdot 18+18$ and sine addition identity:

$1=\sin(4 \cdot 18)\cos(18)+\cos(4 \cdot 18)\sin(18)$

Apply double angle identities for both sine and cosine:

$1=2\sin(2 \cdot 18)\cos(2\cdot 18)\cos(18)+(\cos^2(2\cdot 18)-\sin^2(2\cdot 18))\sin(18)$

Before proceeding I’m going to go ahead and replace $\sin(18)$ with and $\cos(18)$ with $\sqrt{1-x^2}$ :

$1=2\sin(2 \cdot 18)\cos(2\cdot 18)\sqrt{1-x^2}+(\cos^2(2\cdot 18)-\sin^2(2\cdot 18))x$

Apply double angle identities again:

$1=2[2\sin(18)\cos(18)][\cos^2(18)-\sin^2(18)]\sqrt{1-x^2}+((\cos^2(18)-\sin^2(18))^2-(2\sin(18)\cos(18))^2)x$

Before proceeding I’m going to go ahead and replace $\sin(18)$ with and $\cos(18)$ with $\sqrt{1-x^2}$ :

$1=2[2x\sqrt{1-x^2}][(1-x^2)-x^2]\sqrt{1-x^2}+((1-x^2-x^2)^2-(2x\sqrt{1-x^2})^2)x$

This is all in terms of x. Let’s see if this comes out to be a nice polynomial equation:

$1=4x\sqrt{1-x^2}(1-2x^2)\sqrt{1-x^2}+((1-2x^2)^2-4x^2(1-x^2))x$

I’m going to use in the first term on the right hand side that: $\sqrt{1-x^2}\cdot \sqrt{1-x^2}=1-x^2$ :

1=4x(1-x^2)(1-2x^2)+((1-2x^2)^2-4x^2(1-x^2))x

Distribute:

1=4x(1-x^2)(1-2x^2)+x(1-2x^2)^2-4x^3(1-x^2)

I don’t really see an easy fun math trick here so I’m going to expand everything and put right hand side into standard form:

1=4x(1-2x^2-x^2+2x^4)+x(1-4x^2+4x^4)-4x^3+4x^5

1=4x(1-3x^2+2x^4)+x-4x^3+4x^5-4x^3+4x^5

1=4x-12x^3+8x^5+x-4x^3+4x^5-4x^3+4x^5

Gather like terms:

1=4x^5+4x^5+8x^5-12x^3-4x^3-4x^3+4x+x

Simplify the gatherings:

1=16x^5-20x^3+5x

Subtract 1 on both sides:

16x^5-20x^3+5x-1=0

A solution by observation is x=1.

Let’s use synthetic division to see if we can find other factors of the right hand side besides the x-1:

1 | 16 0 -20 0 5 -1

| 16 16 -4 -4 1

---------------------------------

16 16 -4 -4 1 0

The last column shows the remainder is 0 so we have two factors of the right hand side.

That is the right hand side is equal to:

(x-1)(16x^4+16x^3-4x^2-4x+1)

We already know that $\sin(18) neq 1$ so the solution must occur in that second factor.

So we are still looking to solve:

16x^4+16x^3-4x^2-4x+1=0 .

4x^2(4x^2+4x+1)-8x^2-4x+1=0

4x^2(2x+1)^2-4x(2x+1)+1=0

Use quadratic formula to solve for 2x+1; yes the answer will be in terms of x. (At this point I’m just curious what route would work because I was lost where to go.)

$2x+1=\frac{4x\pm \sqrt{(4x)^2-4(4x^2)(1)}}{2(4x^2)}$

$2x+1=\frac{4x\pm \sqrt{16x^2-16x^2}}{8x^2}$

$2x+1=\frac{4x}{8x^2}$

$2x+1=\frac{1}{2x}$

Multiply both sides by 2x:

2x(2x+1)=1

Distribute:

4x^2+2x=1

Subtract 1 on both sids:

4x^2+2x-1=0

Use quadratic formula to solve for x. This time the answer will not be in terms of x. :)

$x=\frac{-2 \pm \sqrt{(2)^2-4(4)(-1)}}{2(4)}$

$x=\frac{-2 \pm \sqrt{4+16}}{8}$

$x=\frac{-2 \pm \sqrt{20}}{8}$

20 has a factor that is a perfect square:

$x=\frac{-2 \pm \sqrt{4} \sqrt{5}}{8}$

$x=\frac{-2 \pm 2 \sqrt{5}}{8}$

Divide top and bottom by 2 since all three terms have a common factor 2:

$x=\frac{-1 \pm 1 \sqrt{5}}{4}$

$x=\frac{-1 \pm \sqrt{5}}{4}$

So we have that x is either:

$\frac{-1+\sqrt{5}}{4} \text{ or } \frac{-1-\sqrt{5}}{4}$ .

The last solution is negative and recall 18 degrees in the first quadrant so:

$x=\frac{-1+\sqrt{5}}{4}$

That is:

$\sin(18)=\frac{-1+\sqrt{5}}{4}$

$\sin(18)=\frac{\sqrt{5}-1}{4}$ by commutative property.

answered: Guest

the thermometer is precise but not accurate

step-by-step explanation:

answered: Guest

)because it's going in order of the math sign

Other questions on the subject: Mathematics

Mathematics, 21.06.2019 16:00, ctyrector

Asap for seven months an ice creams vendor collects data showing the number of cones sold a day (c) and the degrees above 70 degrees f the vendor finds that there is a positive linear association between c and t that is the best modeled by this equation c=3.7t+98 which statement is true? the model predicts that the average number of cones sold in a day is 3.7 the model predicts that the average number of cones sold i a day is 98 a model predicts that for each additional degree that the high temperature is above 70 degres f the total of number of cones sold increases by 3.7 the model predicts for each additional degree that the high temperature is above 70 degrees f the total number of cones sold increasingly by 98

Answers: 3

continue

Mathematics, 21.06.2019 17:30, pleasehelpme666

Δabc is dilated by a scale factor of 3 with the origin as the center of dilation to form δa′b′c′. the slope of is -1.2. the length of is p units, the length of is q units, and the length of is r units. the slope of is . the length of is units.

Answers: 2

continue

Mathematics, 21.06.2019 20:30, daniiltemkin20

List x1, x2, x3, x4 where xi is the left endpoint of the four equal intervals used to estimate the area under the curve of f(x) between x = 4 and x = 6. a 4, 4.5, 5, 5.5 b 4.5, 5, 5.5, 6 c 4.25, 4.75, 5.25, 5.75 d 4, 4.2, 5.4, 6

Answers: 1

continue

Mathematics, 22.06.2019 00:00, smartoa

Round each decimal to the nearest thousandth. a. 5.39562 b. 0.12345 c. 5634 d. 18.93763

Answers: 1

continue

Do you know the correct answer?

Asap
pl show me how to solve with all the steps.
i'll give you brainliest and pts
si...