proof left as an exercise - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: proof left as an exercise (/thread-18443.html) proof left as an exercise - Thomas Klemm - 06-06-2022 11:41 PM 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} RE: proof left as an exercise - Ángel Martin - 06-07-2022 05:05 AM not a proof but it's equal to: tan(20), right? RE: proof left as an exercise - Thomas Klemm - 06-07-2022 05:32 AM Correct, that was the calculator or trig table part. RE: proof left as an exercise - Albert Chan - 06-07-2022 05:36 PM 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) RE: proof left as an exercise - Albert Chan - 06-07-2022 06:17 PM 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°) RE: proof left as an exercise - Albert Chan - 06-08-2022 01:50 AM 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 RE: proof left as an exercise - Albert Chan - 06-08-2022 11:12 AM (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°) RE: proof left as an exercise - Thomas Klemm - 06-08-2022 11:18 PM 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} RE: proof left as an exercise - Albert Chan - 06-09-2022 12:35 AM (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°) RE: proof left as an exercise - Albert Chan - 07-01-2022 07:51 PM 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°) RE: proof left as an exercise - Albert Chan - 07-02-2022 11:44 PM 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°)