Post Reply 
proof left as an exercise
06-06-2022, 11:41 PM
Post: #1
proof left as an exercise
Just stumbled upon this and thought some of you might enjoy solving this as well.
Quote:Prove that the following expression can be expressed as being identical to a single standard trigonometric function of an integer input with all inputs being in degrees.
You may use a calculator or trig table as a guide, but the results from the calculator or trig table will not be accepted as part of the proof.

\(
\begin{align}
\frac{2\cos(30^{\circ})}{1+4\sin(70^{\circ})} = \,?
\end{align}
\)
Find all posts by this user
Quote this message in a reply
06-07-2022, 05:05 AM
Post: #2
RE: proof left as an exercise
not a proof but it's equal to: tan(20), right?
Find all posts by this user
Quote this message in a reply
06-07-2022, 05:32 AM
Post: #3
RE: proof left as an exercise
Correct, that was the calculator or trig table part.
Find all posts by this user
Quote this message in a reply
06-07-2022, 05:36 PM
Post: #4
RE: proof left as an exercise
Proof: 2*cos(30°) / (1 + 4*sin(70°)) = tan(20°)

2*cos(30°) / (1 + 4*cos(20°)) ?=? tan(20°)             // sin(70°) = cos(20°)
2*cos(30°) ?=? tan(20°) + 4*sin(20°)                     // cross multiply
cos(20°)*cos(30°) ?=? sin(20°)/2 + sin(40°)            // multiply by cos(20°)/2
sin(40°) ?=? cos(20°)*cos(30°) - sin(20°)*sin(30°)

cos(50°) = cos(20°+30°)                                       // trig angle sum identity

Proof is complete, (read it bottom-up)
Find all posts by this user
Quote this message in a reply
06-07-2022, 06:17 PM
Post: #5
RE: proof left as an exercise
Top-down proof (equivalent to previous post)

2*cos(30°) / (1+4*sin(70°))
= cos(30°) / (sin(30°) + 2*cos(20°))
= sin(20°)*cos(30°) / (sin(20°)*sin(30°) + sin(40°))
= sin(20°)*cos(30°) / (sin(20°)*sin(30°) + cos(50°))
= sin(20°)*cos(30°) / (cos(20°)*cos(30°))
= tan(20°)
Find all posts by this user
Quote this message in a reply
06-08-2022, 01:50 AM
Post: #6
RE: proof left as an exercise
Another proof, letting t = tan(10°)

tan(30°) = 1/√3 = (3t-t³) / (1-3t²)                  // triple-angle formula
⇒ t³ = √3*t² + 3t - 1/√3
⇒ t4 = 6t² + 8/√3*t - 1

2*cos(30°) / (1 + 4*cos(20°)) ?=? tan(20°)

√3 / (1 + 4*(1-t²)/(1+t²)) ?=? 2t / (1-t²)        // tangent half-angle formula
√3*(1+t²) / (5-3t²) ?=? 2t / (1-t²)                  // cross multiply
√3*(1-t4) ?=? 10t - 6t³

√3*(1-t4) + 6t³ - 10t
= √3 - √3*(6t² + 8/√3*t - 1) + 6*(√3*t² + 3t - 1/√3) - 10t
= 0      QED
Find all posts by this user
Quote this message in a reply
06-08-2022, 11:12 AM
Post: #7
RE: proof left as an exercise
(06-08-2022 01:50 AM)Albert Chan Wrote:  Another proof, letting t = tan(10°)

tan(30°) = 1/√3 = (3t-t³) / (1-3t²)                  // triple-angle formula
⇒ t³ = √3*t² + 3t - 1/√3

We could simplify previous proof, by getting t² in terms of t

t²*t = √3*t² + 3t - 1/√3
t² = (3t - 1/√3) / (t - √3)

2*cos(30°) / (1 + 4*cos(20°))
= √3 / (1 + 4*(1-t²)/(1+t²))
= (1/√3 - t) / (1 + (1/√3)*t)
= (tan(30°) - tan(10°)) / (1 + tan(30°)*tan(10°))
= tan(20°)
Find all posts by this user
Quote this message in a reply
06-08-2022, 11:18 PM
Post: #8
RE: proof left as an exercise
We can use the triple angle formulae:

\(
\begin{align}
\sin(3 \theta) &= 3 \sin \theta - 4 \sin^{3} \theta \\
\cos(3 \theta) & = 4 \cos^{3}\theta - 3 \cos \theta \\
\end{align}
\)

The first formula leads to:

\(
\begin{align}
\sin(3 \theta)
&= 3 \sin \theta - 4 \sin^{3} \theta \\
&= \sin \theta \, [3 - 4 \sin^{2} \theta] \\
&= \sin \theta \, [4 \cos^{2} \theta - 1] \\
\end{align}
\)

The second formula leads to:

\(
2 \cos(3 \theta) - 1 = 8 \cos^{3}\theta - 6 \cos \theta - 1 \\
\)

For \( \theta = 20^\circ \) we get:

\(
\begin{align}
2 \cos(3 \theta) - 1 &= \\
2 \cos(3 \cdot 20^\circ) - 1 &= \\
2 \cos(60^\circ) - 1 &= \\
2 \cdot \tfrac{1}{2} - 1 &= 0 \\
\end{align}
\)

And thus:

\(
8 \cos^{3}(20^\circ) - 6 \cos(20^\circ) - 1 = 0
\)

From this we conclude:

\(
\begin{align}
8 \cos^{3}(20^\circ) - 2 \cos(20^\circ) &= 1 + 4 \cos(20^\circ) \\
2 \cos(20^\circ) [4 \cos^{2}(20^\circ) - 1] &= 1 + 4 \cos(20^\circ) \\
\end{align}
\)

Time to plug all that into the formula:

\(
\begin{align}
\frac{2\cos(30^{\circ})}{1+4\sin(70^{\circ})}
&= \frac{2\sin(60^{\circ})}{1+4\cos(20^{\circ})} \\
\\
&= \frac{2\sin(3 \cdot 20^{\circ})}{1+4\cos(20^{\circ})} \\
\\
&= \frac{2 \sin(20^\circ) [4 \cos^{2}(20^\circ) - 1]}{2 \cos(20^\circ) [4 \cos^{2}(20^\circ) - 1]} \\
\\
&= \frac{\sin(20^\circ)}{\cos(20^\circ)} \\
\\
&= \tan(20^\circ)
\end{align}
\)



Or then we use the product to sum identity:

\(
2 \cos \theta \cos \varphi = \cos(\theta + \varphi )+\cos(\theta - \varphi)
\)

We use \(\cos(60^\circ) = \frac{1}{2}\) and start with:

\(
\begin{align}
\cos(10^\circ)
&= 2 \cos(60^\circ) \cos(10^\circ) \\
&= \cos(70^\circ) + \cos(50^\circ) \\
\\
\cos(50^\circ) + \cos(10^\circ) &= \cos(70^\circ) + 2 \cos(50^\circ) \\
2 \cos(30^\circ) \cos(20^\circ) &= \sin(20^\circ) + 2 \sin(40^\circ) \\
&= \sin(20^\circ) + 4 \sin(20^\circ) \cos(20^\circ) \\
&= \sin(20^\circ)[1 + 4 \cos(20^\circ)] \\
\end{align}
\)

This leads to:

\(
\begin{align}
\frac{2 \cos(30^\circ)}{1 + 4 \cos(20^\circ)} = \frac{\sin(20^\circ)}{\cos(20^\circ)} = \tan(20^\circ)
\end{align}
\)

Or then:

\(
\begin{align}
\frac{2 \cos(30^\circ)}{1 + 4 \sin(70^\circ)} = \tan(20^\circ)
\end{align}
\)
Find all posts by this user
Quote this message in a reply
06-09-2022, 12:35 AM
Post: #9
RE: proof left as an exercise
(06-08-2022 11:18 PM)Thomas Klemm Wrote:  We can use the triple angle formulae:

\(
\begin{align}
\sin(3 \theta) &= 3 \sin \theta - 4 \sin^{3} \theta \\
\cos(3 \theta) & = 4 \cos^{3}\theta - 3 \cos \theta \\
\end{align}
\)

I noticed an easier way

sin(3θ)/sin(θ) = 4*cos(θ)^2 - 1
cos(3θ)/cos(θ) = 4*cos(θ)^2 - 3

sin(3θ)/sin(θ) = cos(3θ)/cos(θ) + 2

This is all is need for the proof:

2*cos(30°) / (1+4*sin(70°))
= 2*sin(60°) / (1+4*cos(20°))
= 2*sin(20°) * (cos(60°)/cos(20°) + 2) / (1+4*cos(20°))
= tan(20°) * (1+4*cos(20°)) / (1+4*cos(20°))
= tan(20°)
Find all posts by this user
Quote this message in a reply
07-01-2022, 07:51 PM (This post was last modified: 07-06-2022 01:33 PM by Albert Chan.)
Post: #10
RE: proof left as an exercise
Code:
sin((2*k-1)*x) / sin(x) = 1 + 2*[cos(2x) + cos(4x) + ... + cos((2*k-2)*x)]
sin(( 2*k )*x) / sin(x) = 2*[cos(x) + cos(3x) + cos(5x) + ... + cos((2*k-1)*x)]

Another way, using above sin(n*x)/sin(x) identities

1 + 4*cos(20°)
= 2*cos(20°) + 2*(cos(60°) + cos(20°))
= 2*cos(20°) + 4*cos(20°)*cos(40°)
= 2*cos(20°) * (2*cos(40°)+1)
= 2*cos(20°) * sin(60°)/sin(20°)
= 2*sin(60°) / tan(20°)

--> 2*cos(30°) / (1 + 4*sin(70°)) = tan(20°)

---

Using complementary angle, we can get ratio of cos over cos (or, sin over cos)

XCAS> Q := makelist(k->f(k*x),1,4)
XCAS> normal(Q(f=sin, x=pi/2-y))      → [cos(y),sin(2*y),-cos(3*y),-sin(4*y)]
XCAS> normal(Q(f=cos, x=pi/2-y))      → [sin(y),-cos(2*y),-sin(3*y),cos(4*y)]

(x = pi/2-y) transform affected both side of identity; it is better to reverse sum order.
Code:
cos((2k-1)*y) / cos(y) = 2*[cos((2k-2)*y) - cos((2k-4)*y) + cos((2k-6)*y) - ... ] + (-1)^(k+1)
sin(( 2k )*y) / cos(y) = 2*[sin((2k-1)*y) - sin((2k-3)*y) + sin((2k-5)*y) - ...   + (-1)^(k+1)*sin(y)]

Example, redo the proof, using cos over cos identity:

cos(3x)/cos(x) = (2*cos(2x) - 1)

1 + 4*cos(20°)
= 2*cos(60°) + 4*cos(20°)
= 2*cos(20°)*(2*cos(40°)-1) + 4*cos(20°)
= 2*cos(20°) * (2*cos(40°)+1)
= 2*cos(20°) * sin(60°)/sin(20°)
= 2*sin(60°) / tan(20°)
Find all posts by this user
Quote this message in a reply
07-02-2022, 11:44 PM (This post was last modified: 07-04-2022 11:34 AM by Albert Chan.)
Post: #11
RE: proof left as an exercise
Using complex numbers, proof turns out very simple !

Let z = cis(20°), 1° = pi/180

cos(60°) = (z^3+1/z^3)/2 = 1/2      → (z^3+1/z^3) = 1

1 + 4*cos(20°)
= (z^3+1/z^3) + 2*(z+1/z)
= (z^2+1)*(z^4+z^2+1) / z^3
= (z^2+1)/(z^2-1) * (z^6-1)/z^3
= (2*cos(20°)) / (2i*sin(20°)) * (2i*sin(60°))
= 2*sin(60°) / tan(20°)

--> 2*cos(30°) / (1 + 4*sin(70°)) = tan(20°)
Find all posts by this user
Quote this message in a reply
Post Reply 




User(s) browsing this thread: