Respuesta :
Answer:
Answer is in the explanation.
Step-by-step explanation:
[tex]\sin(18)=x[/tex]
We are going to find [tex]\cos(18)[/tex] in terms of [tex]x[/tex] using Pythagorean Identity: [tex]\sin^2(\theta)+\cos^2(\theta)=1[/tex].
[tex]x^2+\cos^2(18)=1[/tex]
Subtract [tex]x^2[/tex] on both sides:
[tex]\cos^2(18)=1-x^2[/tex]
Take square root of both sides:
[tex]\cos(18)=\pm \sqrt{1-x^2}[/tex]
We are going to use the positive one since 18 is in the first quadrant. Sine value and cosine value are both positive there:
[tex]\cos(18)=\sqrt{1-x^2}[/tex]
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:
[tex]1=\sin(5 \cdot 18)[/tex] (sin(90 deg) is 1)
We are going to expand this using [tex]5\cdot 18=4\cdot 18+18[/tex] and sine addition identity:
[tex]1=\sin(4 \cdot 18)\cos(18)+\cos(4 \cdot 18)\sin(18)[/tex]
Apply double angle identities for both sine and cosine:
[tex]1=2\sin(2 \cdot 18)\cos(2\cdot 18)\cos(18)+(\cos^2(2\cdot 18)-\sin^2(2\cdot 18))\sin(18)[/tex]
Before proceeding I’m going to go ahead and replace [tex]\sin(18)[/tex] with [tex]x[/tex] and [tex]\cos(18)[/tex] with [tex]\sqrt{1-x^2}[/tex]:
[tex]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[/tex]
Apply double angle identities again:
[tex]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[/tex]
Before proceeding I’m going to go ahead and replace [tex]\sin(18)[/tex] with [tex]x[/tex] and [tex]\cos(18)[/tex] with [tex]\sqrt{1-x^2}[/tex]:
[tex]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[/tex]
This is all in terms of x. Let’s see if this comes out to be a nice polynomial equation:
[tex]1=4x\sqrt{1-x^2}(1-2x^2)\sqrt{1-x^2}+((1-2x^2)^2-4x^2(1-x^2))x[/tex]
I’m going to use in the first term on the right hand side that: [tex]\sqrt{1-x^2}\cdot \sqrt{1-x^2}=1-x^2[/tex]:
[tex]1=4x(1-x^2)(1-2x^2)+((1-2x^2)^2-4x^2(1-x^2))x[/tex]
Distribute:
[tex]1=4x(1-x^2)(1-2x^2)+x(1-2x^2)^2-4x^3(1-x^2)[/tex]
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:
[tex]1=4x(1-2x^2-x^2+2x^4)+x(1-4x^2+4x^4)-4x^3+4x^5[/tex]
[tex]1=4x(1-3x^2+2x^4)+x-4x^3+4x^5-4x^3+4x^5[/tex]
[tex]1=4x-12x^3+8x^5+x-4x^3+4x^5-4x^3+4x^5[/tex]
Gather like terms:
[tex]1=4x^5+4x^5+8x^5-12x^3-4x^3-4x^3+4x+x[/tex]
Simplify the gatherings:
[tex]1=16x^5-20x^3+5x[/tex]
Subtract 1 on both sides:
[tex]16x^5-20x^3+5x-1=0[/tex]
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:
[tex](x-1)(16x^4+16x^3-4x^2-4x+1)[/tex]
We already know that [tex]\sin(18) neq 1[/tex] so the solution must occur in that second factor.
So we are still looking to solve:
[tex]16x^4+16x^3-4x^2-4x+1=0[/tex].
[tex]4x^2(4x^2+4x+1)-8x^2-4x+1=0[/tex]
[tex]4x^2(2x+1)^2-4x(2x+1)+1=0[/tex]
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.)
[tex]2x+1=\frac{4x\pm \sqrt{(4x)^2-4(4x^2)(1)}}{2(4x^2)}[/tex]
[tex]2x+1=\frac{4x\pm \sqrt{16x^2-16x^2}}{8x^2}[/tex]
[tex]2x+1=\frac{4x}{8x^2}[/tex]
[tex]2x+1=\frac{1}{2x}[/tex]
Multiply both sides by 2x:
[tex]2x(2x+1)=1[/tex]
Distribute:
[tex]4x^2+2x=1[/tex]
Subtract 1 on both sids:
[tex]4x^2+2x-1=0[/tex]
Use quadratic formula to solve for x. This time the answer will not be in terms of x. :)
[tex]x=\frac{-2 \pm \sqrt{(2)^2-4(4)(-1)}}{2(4)}[/tex]
[tex]x=\frac{-2 \pm \sqrt{4+16}}{8}[/tex]
[tex]x=\frac{-2 \pm \sqrt{20}}{8}[/tex]
20 has a factor that is a perfect square:
[tex]x=\frac{-2 \pm \sqrt{4} \sqrt{5}}{8}[/tex]
[tex]x=\frac{-2 \pm 2 \sqrt{5}}{8}[/tex]
Divide top and bottom by 2 since all three terms have a common factor 2:
[tex]x=\frac{-1 \pm 1 \sqrt{5}}{4}[/tex]
[tex]x=\frac{-1 \pm \sqrt{5}}{4}[/tex]
So we have that x is either:
[tex]\frac{-1+\sqrt{5}}{4} \text{ or } \frac{-1-\sqrt{5}}{4}[/tex].
The last solution is negative and recall 18 degrees in the first quadrant so:
[tex]x=\frac{-1+\sqrt{5}}{4}[/tex]
That is:
[tex]\sin(18)=\frac{-1+\sqrt{5}}{4}[/tex]
[tex]\sin(18)=\frac{\sqrt{5}-1}{4}[/tex] by commutative property.
