12.3 Trigonometry and Optional Calculus Enrichment

12.3 Trigonometry and Optional Calculus Enrichment

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Unit Circle Signs & Quadrant Reduction (Bölgelerde İşaretler ve İndirgeme)

On the unit circle (birim çember), $\cos\theta$ is the $x$-coordinate and $\sin\theta$ is the $y$-coordinate. The signs of the four functions across Quadrants I, II, III, and IV follow the classic All – Sin – Tan – Cos rule (Turkish mnemonic: "Bütün Sınıf Kara Tahtada Coşar"):

• Quadrant I ($0 < \theta < \frac{\pi}{2}$): All positive ($+, +, +, +$)
• Quadrant II ($\frac{\pi}{2} < \theta < \pi$): Only $\sin\theta > 0$ (and $\csc\theta > 0$)
• Quadrant III ($\pi < \theta < \frac{3\pi}{2}$): Only $\tan\theta > 0$ and $\cot\theta > 0$
• Quadrant IV ($\frac{3\pi}{2} < \theta < 2\pi$): Only $\cos\theta > 0$ (and $\sec\theta > 0$)

To reduce any large angle to an acute angle $\alpha \in (0, \frac{\pi}{2})$:

1. Determine the sign first: Check the quadrant of the original angle and apply the sign of the original trigonometric function in that quadrant.
2. Check the reference axis:
   • Horizontal Axis ($\pi \pm \alpha$ or $2\pi \pm \alpha$, i.e., $180^\circ, 360^\circ$): The function name stays the same ($\sin \to \sin,\; \cos \to \cos,\; \tan \to \tan,\; \cot \to \cot$).
   • Vertical Axis ($\frac{\pi}{2} \pm \alpha$ or $\frac{3\pi}{2} \pm \alpha$, i.e., $90^\circ, 270^\circ$): The function switches to its co-function ($\sin \leftrightarrow \cos$ and $\tan \leftrightarrow \cot$).
   • Complementary Angles (Tümler Açılar): Whenever $\alpha + \beta = 90^\circ = \frac{\pi}{2}$, $\sin\alpha = \cos\beta$ and $\tan\alpha = \cot\beta$ (e.g., $\sin 20^\circ = \cos 70^\circ$ and $\tan 15^\circ \cdot \tan 75^\circ = \tan 15^\circ \cdot \cot 15^\circ = 1$).

Pythagorean, Sum-Product & Double-Angle Identities

• Pythagorean & Reciprocal Identities:

$$\sin^2 x + \cos^2 x = 1, \qquad \tan x = \frac{\sin x}{\cos x}, \qquad \cot x = \frac{\cos x}{\sin x}, \qquad \tan x \cdot \cot x = 1$$ $$1 + \tan^2 x = \sec^2 x = \frac{1}{\cos^2 x}, \qquad 1 + \cot^2 x = \csc^2 x = \frac{1}{\sin^2 x}$$

• Square of Sine Plus/Minus Cosine:

$$(\sin x \pm \cos x)^2 = \sin^2 x + \cos^2 x \pm 2\sin x \cos x = 1 \pm \sin(2x)$$

Whenever a question gives $\sin x - \cos x = k$ and asks for $\sin x \cos x$ or $\sin(2x)$, square both sides immediately!

• Sum & Difference Formulas (Toplam-Fark Formülleri):

$$\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b, \qquad \cos(a \pm b) = \cos a \cos b \mp \sin a \sin b$$ $$\tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$$

• Double-Angle Formulas (İki Kat Açı Formülleri):

$$\sin(2x) = 2\sin x \cos x, \qquad \tan(2x) = \frac{2\tan x}{1 - \tan^2 x}$$ $$\cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$$

(Pro-Tip: To eliminate $+1$ or $-1$ in expressions like $1 + \cos(2x)$ or $1 - \cos(2x)$, substitute $1 + \cos(2x) = 2\cos^2 x$ and $1 - \cos(2x) = 2\sin^2 x$!)

Law of Cosines (Kosinüs Teoremi) & Law of Sines (Sinüs Teoremi)

In any triangle $ABC$ with sides $a, b, c$ opposite angles $A, B, C$ and circumradius $R$:

$$a^2 = b^2 + c^2 - 2bc\cos A, \qquad \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R, \qquad \operatorname{Area}(ABC) = \frac{1}{2}bc\sin A$$

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2. Optional Calculus Enrichment (Limit, Türev ve İntegral)

Limits & Continuity (Limit ve Süreklilik)

• Piecewise Continuity (Süreklilik): A function $f(x)$ is continuous at a transition point $x = c$ if and only if its left-hand limit, right-hand limit, and function value all agree:

$$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$

• Trigonometric Zero Limit:

$$\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}, \qquad \lim_{x \to 0} \frac{\tan(ax)}{bx} = \frac{a}{b}$$

• Indeterminate $\frac{0}{0}$ Forms & L'Hôpital's Rule: When direct substitution gives $0/0$, first try factoring or rationalization. If the differentiability, nonzero denominator derivative and derivative-ratio limit conditions explained below hold, L’Hôpital’s rule allows differentiating the numerator and denominator separately:

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Derivatives (Türev)

• Differentiation Rules:

$$(x^n)' = nx^{n-1}, \qquad (u \cdot v)' = u'v + uv', \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}, \qquad [f(g(x))]' = f'(g(x)) \cdot g'(x)$$ $$(\sqrt{u})' = \frac{u'}{2\sqrt{u}}, \qquad (\sin x)' = \cos x, \qquad (\cos x)' = -\sin x, \qquad (e^u)' = u'e^u, \qquad (\ln u)' = \frac{u'}{u}$$

• Geometric Interpretation (Teğetin Eğimi) & Local Extrema (Ekstremum):
  • The slope of the tangent line to $y = f(x)$ at $x = x_0$ is $m_T = f'(x_0)$, and the tangent line equation is $y - f(x_0) = f'(x_0)(x - x_0)$.
  • At a differentiable interior local maximum or minimum, $f'(x)=0$ is necessary. It does not prove an extremum: $f(x)=x^3$ has $f'(0)=0$ without one. If the function is continuous at the candidate and the derivative changes from positive to negative across it, a local maximum follows; a negative-to-positive change proves a local minimum. This sign test is sufficient, not necessary for every possible function.

Definite Integrals & Fast Area Shortcuts (İntegral ve Alan)

• Power Rule & Fundamental Theorem of Calculus:

$$\int x^n\,dx = \frac{x^{n+1}}{n + 1} + C \quad (n \ne -1), \qquad \int_a^b f'(x)\,dx = f(b) - f(a)$$

• Symmetric Interval $[-a, a]$ Shortcut:
  • If $f(-x) = -f(x)$ (odd function / tek fonksiyon, such as $x^3, x^5, \sin x, x\cos x$), then $\int_{-a}^a f(x)\,dx = 0$.
  • If $f(-x) = f(x)$ (even function / çift fonksiyon, such as $x^2, \cos x, \lvert x \rvert$), then $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$.
• Parabolic Area Shortcuts (No Integration Required!):
  1. Vertex-Rectangle Split ($S$ vs. $2S$): A parabola $y = ax^2$ with vertex at $(0, 0)$ passing through $(x_0, y_0)$ splits the $x_0 \times y_0$ bounding rectangle into $\frac{1}{3}x_0 y_0$ (under the curve against the $x$-axis) and $\frac{2}{3}x_0 y_0$ (above the curve against the $y$-axis).
  2. Parabola-Line Enclosed Area (1/6 Kuralı): The area enclosed between a parabola $y = ax^2 + bx + c$ and a line $y = mx + n$ that intersect at $x_1$ and $x_2$ is:

$$\text{Area} = \frac{\lvert a \rvert}{6} \lvert x_2 - x_1 \rvert^3$$

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Summary Reference Table

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Worked Examples

Worked Example 1: Trigonometric Reduction & Law of Cosines

Problem:

1. Given that $\tan x = \frac{3}{4}$ for $x \in \left(0, \frac{\pi}{2}\right)$, evaluate $E = \frac{\sin(\pi - x) + \cos\!\left(\frac{3\pi}{2} + x\right)}{\tan\!\left(\frac{\pi}{2} + x\right)}$.
2. In $\triangle ABC$, $AB = 5$, $AC = 8$, and $m(\angle A) = 60^\circ$. Find the length of side $BC = a$.

Step-by-Step Solution:

1. Draw a $3$-$4$-$5$ right triangle for $x \in (0, \pi/2)$: $\sin x = \frac{3}{5}$, $\cos x = \frac{4}{5}$, and $\cot x = \frac{4}{3}$. Now reduce each angle:
   • $\pi - x$ is in Quadrant II (horizontal axis $\pi$, $\sin$ is $+$): $\sin(\pi - x) = +\sin x = \frac{3}{5}$.
   • $\frac{3\pi}{2} + x$ is in Quadrant IV (vertical axis $\frac{3\pi}{2}$, $\cos$ is $+$, switches to $\sin$): $\cos\!\left(\frac{3\pi}{2} + x\right) = +\sin x = \frac{3}{5}$.
   • $\frac{\pi}{2} + x$ is in Quadrant II (vertical axis $\frac{\pi}{2}$, $\tan$ is $-$, switches to $\cot$): $\tan\!\left(\frac{\pi}{2} + x\right) = -\cot x = -\frac{4}{3}$.
   • Therefore, $E = \frac{3/5 + 3/5}{-4/3} = \frac{6}{5} \cdot \left(-\frac{3}{4}\right) = -\frac{9}{10}$.
2. Apply the Law of Cosines opposite $\angle A = 60^\circ$ (where $\cos 60^\circ = \frac{1}{2}$):

$$a^2 = 5^2 + 8^2 - 2(5)(8)\cos 60^\circ = 25 + 64 - 80\left(\frac{1}{2}\right) = 89 - 40 = 49 \implies a = 7$$

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Worked Example 2: L'Hôpital's Rule & Parabolic Area Shortcut

Problem:

1. Evaluate $L = \lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x^2 - 4}$.
2. Find the area of the finite region enclosed between the parabola $y = x^2 - 2x - 3$ and the line $y = x + 1$.

Step-by-Step Solution:

1. Substituting $x = 2$ yields $\frac{\sqrt{9} - 3}{4 - 4} = \frac{0}{0}$. Apply L'Hôpital's Rule by differentiating numerator and denominator with respect to $x$:

$$L = \lim_{x \to 2} \frac{\frac{1}{2\sqrt{x + 7}}}{2x} = \frac{\frac{1}{2\sqrt{9}}}{2(2)} = \frac{1/6}{4} = \frac{1}{24}$$

1. Set the two equations equal to find the $x$-coordinates of the intersection points:

$$x^2 - 2x - 3 = x + 1 \implies x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0$$

The roots are $x_1 = -1$ and $x_2 = 4$, so the root gap is $\lvert x_2 - x_1 \rvert = 4 - (-1) = 5$, and the leading coefficient is $a = 1$. By the $\frac{\lvert a \rvert}{6}\lvert x_2 - x_1 \rvert^3$ shortcut:

$$\text{Area} = \frac{1}{6} \cdot 5^3 = \frac{125}{6}$$

Conditions on the Calculus Shortcuts

Use radians in trigonometric limits and derivatives; in $\sin(ax)/(bx)$ require $b\ne0$. L’Hôpital’s rule needs differentiable numerator and denominator in a punctured neighbourhood, nonzero denominator derivative there, an indeterminate form, and an existing derivative-ratio limit. A 0/0 substitution alone does not authorize it. Factoring or conjugates often give a simpler proof. For the worked radical limit, rationalization gives $1/[(\sqrt{x+7}+3)(x+2)]$, whose limit is also $1/24$.

A derivative sign change is a sufficient test for a local extremum when the function is continuous at the point; stationary points are only candidates. Also inspect points where the derivative is undefined and interval endpoints. Integral symmetry assumes integrability, and the displayed fundamental-theorem identity holds, for example, when $f$ has a continuous derivative. Area is nonnegative; the rectangle shortcut assumes positive $x_0,y_0$, or uses absolute lengths after reflecting the configuration. This enrichment does not establish an official calculus requirement.

Review the Method

The derivative f'(x_0) gives the tangent line slope at x_0 and locates candidate stationary points, with a sign-change test for extrema, while the definite integral int_a^b f'(x) dx = f(b) - f(a) and the parabolic area shortcut A = |a|/6 * |x_2 - x_1|^3 solve area problems in seconds.