Right Triangles And Trigonometry Construction Of The Unit: Fill & Download for Free

The truth is even with a calculator the best you can do is an approximation.Our trig works pretty well when the Pythagorean Theorem says [math]1+1=2[/math] (a 45,45,90 triangle) or [math]1+3=4[/math] (a 30,60,90 triangle). Here we have [math].64 + .36 = 1,[/math] way too complicated for trig as usually taught. Clearing the fractions that’s [math]16+9=25,[/math] a 3, 4, 5 right triangle.Of course the answer is [math]\theta = \arccos .8[/math] and as we know the sides of our right triangle [math]\theta = \arccos \frac 4 5 = \arctan \frac 3 4 = \arcsin \frac 3 5. [/math] We have various ways of producing an approximation that I’m sure the other answers will tell you. [math]\theta[/math] is irrational, probably transcendental.My point is why do we need [math]\theta[/math]? We have a perfectly good 3,4,5 right triangle that we’d have no problem constructing, drawing on graph paper, even doing trigonometry with because we know all its trig functions. The only thing we don’t know, really can’t know, is the exact angle corresponding to a slope of [math]\frac 3 4.[/math]It’s only our way of doing trigonometry that conditions us to think we even need to know the angle. We have everything we need without knowing [math]\theta.[/math]

The three fundamental Pythagorean Identities, which I’m sure one can find in any Algebra-Trigonometry textbook, are as follows:sin² θ + cos² θ = 1tan² θ + 1 = sec² θ1 + cot² θ = csc² θwhere angle θ is any angle in standard position in the xy-plane.Consistent with the definition of an identity, the above identities are true for all values of the variable, in this case angle θ, for which the functions involved are defined.The Pythagorean Identities are so named because they are ultimately derived from a utilization of the Pythagorean Theorem, i.e., c² = a² + b², where c is the length of the hypotenuse of a right triangle and a and b are the lengths of the other two sides.This derivation can be easily seen when considering the special case of the unit circle (r = 1). For any angle θ in standard position in the xy-plane and whose terminal side intersects the unit circle at the point (x, y), that is a distance r = 1 from the origin, we can construct a right triangle with hypotenuse c = r, with height a = y and with base b = x so that:c² = a² + b² becomes:r² = y² + x² = 1²y² + x² = 1We also know from our study of the unit circle that x = r(cos θ) = (1)(cos θ) = cos θ and y = r(sin θ) = (1)(sin θ) = sin θ; therefore, substituting, we get:(sin θ)² + (cos θ)² = 11.) sin² θ + cos² θ = 1 which is the first Pythagorean Identity.Now, if we divide through equation 1 by cos² θ, we get the second Pythagorean Identity as follows:(sin² θ + cos² θ)/cos² θ = 1/cos² θ(sin² θ/cos² θ) + (cos² θ/cos² θ) = 1/cos² θ(sin θ/cos θ)² + 1 = (1/cos θ)²(tan θ)² + 1 = (sec θ)²2.) tan² θ + 1 = sec² θNow, if we divide through equation 1 by sin² θ, we get the third Pythagorean Identity as follows:(sin² θ + cos² θ)/sin² θ = 1/sin² θ(sin² θ/sin² θ) + (cos² θ/sin² θ) = 1/sin² θ1 + (cos θ/sin θ)² = (1/sin θ)²1 + (cot θ)² = (csc θ)²3.) 1 + cot² θ = csc² θ

The good thing about investing a bit of time and effort into learning how to use a new tool is that that tool, the reverse order or reverse time line problem-solving approach, is versatile and can be applied with equal success across various branches of mathematics.(See this approach in action in this Quora post, in a strange problem where eggs are thrown off of a building, and in this Quora answer on trigonometry, scroll down, where a problem is first created and then solved - both times in reverse order)The essence of the reverse order method is to first make a bold assumption that a problem has been solved already - how it was solved we don’t know yet and we don’t care - this is just a temporary assumption (snatched, admittedly, out of thin air).That assumption, however, may (but is not guaranteed to) pave the way to a forward solution. Armed with the knowledge of the forward solution we may (but are not guaranteed to) be able to reverse its steps to produce the solution sought-after splicing into it what is given.In this problem:Assume, temporarily, that the equation of a tangent [math]\tau[/math] to the graph [math]g[/math] of a horns up parabola[math]y(x) = ax^2, \; a > 0 \; (a \in \mathbb{R}) \tag{1}[/math]has been found somehow.What does this assumption buy us?We know that two distinct points in a (Euclidean) plane define a unique straight line. But we are given one point already - name that point [math]T[/math]. Therefore, one point is missing but is needed.How do we synthesise that missing point?Well, it is natural to think that a tangent to [math]g[/math] must pass through a point [math]P[/math] on [math]g[/math], right? Or so it seems.Therefore, we temporarily take it that [math]P \in g[/math] is given:Hence, if [math]P_x[/math] is the (orthogonal) projection of [math]P[/math] on [math]Ox[/math] then by construction [math]\triangle TP_xP[/math] is right. But by the geometric interpretation of a derivative of a function at a point the trigonometric tangent (as a function) of the angle [math]\theta[/math] must be that derivative. Symbolically:[math]\tan\theta = y_x'(P) = 2a|OP_x| \tag{2}[/math]On the other hand, in isolation, our tangent is a ratio of two sides of a right triangle:[math]\tan\theta = \dfrac{|PP_x|}{|TP_x|} = \dfrac{a|OP_x|^2}{|TP_x|} \tag{3}[/math]Equate (2) and (3):[math]2a|OP_x| = \dfrac{a|OP_x|^2}{|TP_x|} \tag*{}[/math][math]|TP_x| = \dfrac{a|OP_x|^2}{2a|OP_x|} = \dfrac{|OP_x|}{2} \tag{4}[/math]and it means that our forward solution is over: from (4) it follows that the given point [math]T[/math] cuts [math]OP_x[/math] exactly in half.We now run through the forward solution (in our minds): [math]P \in g[/math], project [math]P[/math] on [math]Ox[/math], cut that projection in half to locate [math]T[/math] (the given) - and, by reversing these steps, generate the solution sought-after:double the given distance [math]OT[/math] to obtain [math]P_x[/math]:construct a perpendicular [math]v[/math] to [math]Ox[/math] through [math]P_x[/math] until [math]v[/math] intersects [math]g[/math] at [math]P[/math]:construct the tangent [math]\tau[/math], read off the coordinates of the points [math]T[/math] and [math]P[/math], generate the equation sought-after:In your case:[math]a = 1, \; T(2, 0), \; P_x(4, 0), \; P(4, 16) \tag*{}[/math]Therefore, for the gradient [math]m[/math] we have:[math]m = \dfrac{16 - 0}{4 - 2} = \dfrac{16}{2} = 8 \tag*{}[/math]Using [math]T[/math] as the point, the equation requested (in a point-slope form) is:[math]y - 0 = 8(x - 2) \tag*{}[/math][math]y(x) = 8x - 16 \tag*{}[/math]Or so it seems - recall the weird comment above.All this time we quietly assumed that [math]g[/math] is given.What if it’s not? Let’s challenge ourselves.What if we are given the coordinate axes and the [math]x[/math]-intercept [math]T[/math] only - no [math]g[/math] is available to locate [math]P[/math]? Can we construct a solution then?This is insane!Not really. We actually have an ocean of useful information that we can squeeze for clues: if two distinct straight lines intersect (in a Euclidean plane) then they do so at a unique point - the origin; we thus have two distinct points (with [math]T[/math]); the coordinate axes are perpendicular - we feel a right triangle coming up. But where?Try solving this problem yourself before reading on. Hint: try reverser order, again.Assuming that [math]g[/math] and [math]P[/math] are where they are supposed to be - extend the tangent [math]\tau[/math] all the way until it intersects the [math]y[/math]-axis at [math]A[/math]. But then the (highlighted) right triangles [math]AOT[/math] and [math]TP_xP[/math] are congruent:and the problem is solved: double the distance [math]OT[/math] along the [math]y[/math]-axis starting from [math]O[/math], square the result (in place) and shift it upward to line up one of its extremities with origin to locate [math]A[/math] (to avoid the clutter I’ve done the squaring in GeoGebra by the book but omitted the scaffolding):We see that [math]A[/math] is a trace evidence of [math]P[/math]: as [math]P[/math] runs along [math]g[/math], [math]A[/math] runs along the negative portion of the [math]y[/math]-axis (and conversely).Lastly, since all the numbers involved so far are Euclidean, it is possible to construct a solution with Euclidean tools only - a detailed description of these steps is given in this Quora post (to be explicit - we will also require that a line segment of unit length is given to us).In your case:[math]T(2, 0), \; A(0, -16) \tag*{}[/math]For [math]m[/math]:[math]m = \dfrac{-16 - 0}{0 - 2} = 8 \tag*{}[/math]and using [math]A[/math] as the point, the requested equation is:[math]y - (-16) = 8(x - 0) \tag*{}[/math][math]y(x) = 8x - 16 \tag*{}[/math]Extra for experts: as long as [math]n[/math] remains a natural number, the above represents a straight-edge-and-compass-only solution for a family of functions:[math]y_n(x) = ax^n, \; a \in \mathbb{R}, \; n \in \mathbb{N} \tag*{}[/math]because (from (3)):[math]|TP_x| = \dfrac{a|OP_x|^n}{na|OP_x|^{n-1}} = \dfrac{|OP_x|}{n} \tag*{}[/math]