Solve for A: (0° ≤ A ≤ 90°) cos(A − 20) = sin(A + 30)

here are the solutions to the questions.

QUESTION 1

1.1 Solve for A ($0^\circ \le A \le 90^\circ$)
$\cos(A - 20^\circ) = \sin(A + 30^\circ)$

Step 1: Convert $\sin(A + 30^\circ)$ to a cosine function using the co-function identity $\sin \theta = \cos(90^\circ - \theta)$.

$$\sin(A + 30^\circ) = \cos(90^\circ - (A + 30^\circ))$$ $$\sin(A + 30^\circ) = \cos(90^\circ - A - 30^\circ)$$ $$\sin(A + 30^\circ) = \cos(60^\circ - A)$$

Step 2: Equate the arguments of the cosine functions.

$$\cos(A - 20^\circ) = \cos(60^\circ - A)$$

For $\cos X = \cos Y$, the general solution is $X = \pm Y + k \cdot 360^\circ$, where $k$ is an integer.
Considering the positive case:

$$A - 20^\circ = 60^\circ - A + k \cdot 360^\circ$$ $$2A = 80^\circ + k \cdot 360^\circ$$ $$A = 40^\circ + k \cdot 180^\circ$$

Step 3: Find the value of A within the given range ($0^\circ \le A \le 90^\circ$).
For $k=0$:

$$A = 40^\circ + 0 \cdot 180^\circ = 40^\circ$$

This value is within the specified range.
(If we consider the negative case $A - 20^\circ = -(60^\circ - A) + k \cdot 360^\circ$, it leads to $-20^\circ = -60^\circ + k \cdot 360^\circ$, which simplifies to $40^\circ = k \cdot 360^\circ$, meaning $k = \frac{1}{9}$, which is not an integer, so no solutions from this case.)

The solution is \boxed{\text{A = 40^\circ}}.

1.2 Prove the following identity

$$\frac{\sin 2x}{1 + \cos 2x} = \tan x$$

Step 1: Start with the Left Hand Side (LHS) and apply double angle identities.
Recall the identities: $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = 2 \cos^2 x - 1$.

$$\text{LHS} = \frac{\sin 2x}{1 + \cos 2x}$$

Substitute the identities into the expression:

$$\text{LHS} = \frac{2 \sin x \cos x}{1 + (2 \cos^2 x - 1)}$$

Step 2: Simplify the expression.

$$\text{LHS} = \frac{2 \sin x \cos x}{2 \cos^2 x}$$

Cancel out $2 \cos x$ from the numerator and denominator:

$$\text{LHS} = \frac{\sin x}{\cos x}$$

Step 3: Recognize the tangent identity.

$$\text{LHS} = \tan x$$

Since LHS = RHS, the identity is proven.

1.3 Simplify without using a calculator, $\frac{1}{\sec 75^\circ}$

Step 1: Use the reciprocal identity for secant.
Recall that $\sec \theta = \frac{1}{\cos \theta}$.

$$\frac{1}{\sec 75^\circ} = \cos 75^\circ$$

Step 2: Use the angle addition formula for cosine.
Recall $\cos(A+B) = \cos A \cos B - \sin A \sin B$.
We can write $75^\circ$ as $45^\circ + 30^\circ$.

$$\cos 75^\circ = \cos(45^\circ + 30^\circ)$$ $$\cos 75^\circ = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ$$

Step 3: Substitute the known exact values of trigonometric functions.

$$\cos 45^\circ = \frac{\sqrt{2}}{2}$$ $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$ $$\sin 45^\circ = \frac{\sqrt{2}}{2}$$ $$\sin 30^\circ = \frac{1}{2}$$ $$\cos 75^\circ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$$ $$\cos 75^\circ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$

Step 4: Combine the terms.

$$\cos 75^\circ = \boxed{\frac{\sqrt{6} - \sqrt{2}}{4}}$$

1.4 Simplify: $\frac{\sin(-\theta) \cos(90^\circ - \theta) \tan^2(90^\circ - \theta)}{\sin(180^\circ - \theta) \cos(360^\circ - \theta)}$

Step 1: Apply reduction formulas and co-function identities to each term.

• $\sin(-\theta) = -\sin \theta$
• $\cos(90^\circ - \theta) = \sin \theta$
• $\tan(90^\circ - \theta) = \cot \theta \implies \tan^2(90^\circ - \theta) = \cot^2 \theta$
• $\sin(180^\circ - \theta) = \sin \theta$
• $\cos(360^\circ - \theta) = \cos \theta$

Step 2: Substitute these simplified terms into the expression.

$$\text{Expression} = \frac{(-\sin \theta)(\sin \theta)(\cot^2 \theta)}{(\sin \theta)(\cos \theta)}$$

Step 3: Simplify the expression by canceling common terms and using $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

$$\text{Expression} = \frac{-\sin^2 \theta \cot^2 \theta}{\sin \theta \cos \theta}$$

Cancel one $\sin \theta$ from the numerator and denominator:

$$\text{Expression} = \frac{-\sin \theta \cot^2 \theta}{\cos \theta}$$

Substitute $\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}$:

$$\text{Expression} = \frac{-\sin \theta \left(\frac{\cos^2 \theta}{\sin^2 \theta}\right)}{\cos \theta}$$ $$\text{Expression} = \frac{-\frac{\cos^2 \theta}{\sin \theta}}{\cos \theta}$$ $$\text{Expression} = \frac{-\cos^2 \theta}{\sin \theta \cos \theta}$$

Cancel one $\cos \theta$ from the numerator and denominator:

$$\text{Expression} = \frac{-\cos \theta}{\sin \theta}$$

Step 4: Recognize the cotangent identity.

$$\text{Expression} = \boxed{-\cot \theta}$$

QUESTION 2

2.1 Differentiate using the first principle: $y = 3x^2 - 2x$

The first principle of differentiation is given by $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Given $f(x) = 3x^2 - 2x$.

Step 1: Find $f(x+h)$.

$$f(x+h) = 3(x+h)^2 - 2(x+h)$$ $$f(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h$$ $$f(x+h) = 3x^2 + 6xh + 3h^2 - 2x - 2h$$

Step 2: Find $f(x+h) - f(x)$.

$$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h) - (3x^2 - 2x)$$ $$f(x+h) - f(x) = 3x^2 + 6xh + 3h^2 - 2x - 2h - 3x^2 + 2x$$ $$f(x+h) - f(x) = 6xh + 3h^2 - 2h$$

Step 3: Divide by $h$.

$$\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 2h}{h}$$ $$\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h - 2)}{h}$$ $$\frac{f(x+h) - f(x)}{h} = 6x + 3h - 2$$

Step 4: Take the limit as $h \to 0$.

$$f'(x) = \lim_{h \to 0} (6x + 3h - 2)$$ $$f'(x) = 6x + 3(0) - 2$$ $$f'(x) = \boxed{6x - 2}$$

2.2 Expand to four terms by using the binomial theorem. $(2x - 4)^{\frac{1}{2}}$

Step 1: Rewrite the expression to fit the form $(a+b)^n$ or $(1+u)^n$.
Factor out $2x$ from the expression:

$$(2x - 4)^{\frac{1}{2}} = [2x(1 - \frac{4}{2x})]^{\frac{1}{2}} = (2x)^{\frac{1}{2}} (1 - \frac{2}{x})^{\frac{1}{2}}$$

Let $n = \frac{1}{2}$ and $u = -\frac{2}{x}$. The generalized binomial theorem is $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \dots$

Step 2: Calculate the first four terms of the expansion for $(1 - \frac{2}{x})^{\frac{1}{2}}$.

• Term 1: 1
• Term 2: $nu = \frac{1}{2}\left(-\frac{2}{x}\right) = -\frac{1}{x}$
• Term 3: $\frac{n(n-1)}{2!} u^2 = \frac{\frac{1}{2}(\frac{1}{2}-1)}{2} \left(-\frac{2}{x}\right)^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2} \left(\frac{4}{x^2}\right) = \frac{-\frac{1}{4}}{2} \left(\frac{4}{x^2}\right) = -\frac{1}{8} \cdot \frac{4}{x^2} = -\frac{1}{2x^2}$
• Term 4: $\frac{n(n-1)(n-2)}{3!} u^3 = \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{6} \left(-\frac{2}{x}\right)^3 = \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{6} \left(-\frac{8}{x^3}\right) = \frac{\frac{3}{8}}{6} \left(-\frac{8}{x^3}\right) = \frac{3}{48} \left(-\frac{8}{x^3}\right) = \frac{1}{16} \left(-\frac{8}{x^3}\right) = -\frac{1}{2x^3}$

So, $(1 - \frac{2}{x})^{\frac{1}{2}} = 1 - \frac{1}{x} - \frac{1}{2x^2} - \frac{1}{2x^3} + \dots$

Step 3: Multiply the expansion by $(2x)^{\frac{1}{2}} = \sqrt{2x}$.

$$(2x - 4)^{\frac{1}{2}} = \sqrt{2x} \left(1 - \frac{1}{x} - \frac{1}{2x^2} - \frac{1}{2x^3} + \dots \right)$$ $$(2x - 4)^{\frac{1}{2}} = \sqrt{2x} - \frac{\sqrt{2x}}{x} - \frac{\sqrt{2x}}{2x^2} - \frac{\sqrt{2x}}{2x^3} + \dots$$

To simplify the terms, we can write $\frac{\sqrt{2x}}{x} = \frac{\sqrt{2x}\sqrt{x}}{x\sqrt{x}} = \frac{\sqrt{2}x}{x\sqrt{x}} = \frac{\sqrt{2}}{\sqrt{x}} = \sqrt{\frac{2}{x}}$.
Alternatively, $\frac{\sqrt{2x}}{x} = \frac{\sqrt{2}\sqrt{x}}{x} = \frac{\sqrt{2}}{\sqrt{x}}$.
Let's keep the form $\frac{\sqrt{2x}}{x}$ or rationalize the denominator for each term.
$\frac{\sqrt{2x}}{x} = \frac{\sqrt{2x} \cdot \sqrt{2x}}{x \cdot \sqrt{2x}} = \frac{2x}{x\sqrt{2x}} = \frac{2}{\sqrt{2x}}$.
$\frac{\sqrt{2x}}{2x^2} = \frac{\sqrt{2x} \cdot \sqrt{2x}}{2x^2 \cdot \sqrt{2x}} = \frac{2x}{2x^2\sqrt{2x}} = \frac{1}{x\sqrt{2x}}$.
$\frac{\sqrt{2x}}{2x^3} = \frac{\sqrt{2x} \cdot \sqrt{2x}}{2x^3 \cdot \sqrt{2x}} = \frac{2x}{2x^3\sqrt{2x}} = \frac{1}{x^2\sqrt{2x}}$.

So the expansion is:

$$\boxed{\sqrt{2x} - \frac{2}{\sqrt{2x}} - \frac{1}{x\sqrt{2x}} - \frac{1}{x^2\sqrt{2x}} + \dots}$$

2.3 Differentiate by using quotient rule $y = \frac{3}{2x+1}$

The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$.
Here, $u = 3$ and $v = 2x+1$.

Step 1: Find the derivatives of $u$ and $v$.

$$u = 3 \implies u' = 0$$ $$v = 2x+1 \implies v' = 2$$

Step 2: Apply the quotient rule formula.

$$\frac{dy}{dx} = \frac{(0)(2x+1) - (3)(2)}{(2x+1)^2}$$ $$\frac{dy}{dx} = \frac{0 - 6}{(2x+1)^2}$$ $$\frac{dy}{dx} = \boxed{\frac{-6}{(2x+1)^2}}$$

2.4 Given the function $y = x^3 - 2x^2 - 5x + 6$. Determine, with the aid of differentiation, the co-ordinates of maximum and minimum turning points an distinguish between the maximum and minimum turning points by using the second derivative.

Step 1: Find the first derivative ($\frac{dy}{dx}$) and set it to zero to find the x-coordinates of the turning points.

$$y = x^3 - 2x^2 - 5x + 6$$ $$\frac{dy}{dx} = 3x^2 - 4x - 5$$

Set $\frac{dy}{dx} = 0$:

$$3x^2 - 4x - 5 = 0$$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-5)}}{2(3)}$$ $$x = \frac{4 \pm \sqrt{16 + 60}}{6}$$ $$x = \frac{4 \pm \sqrt{76}}{6}$$ $$x = \frac{4 \pm 2\sqrt{19}}{6}$$ $$x = \frac{2 \pm \sqrt{19}}{3}$$

The x-coordinates of the turning points are $x_1 = \frac{2 + \sqrt{19}}{3}$ and $x_2 = \frac{2 - \sqrt{19}}{3}$.

Step 2: Find the second derivative ($\frac{d^2y}{dx^2}$).

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 4x - 5)$$ $$\frac{d^2y}{dx^2} = 6x - 4$$

Step 3: Use the second derivative to distinguish between maximum and minimum turning points.

• For $x_1 = \frac{2 + \sqrt{19}}{3}$:

  $$\frac{d^2y}{dx^2} \Big|_{x_1} = 6\left(\frac{2 + \sqrt{19}}{3}\right) - 4$$ $$= 2(2 + \sqrt{19}) - 4$$ $$= 4 + 2\sqrt{19} - 4 = 2\sqrt{19}$$

  Since $2\sqrt{19} > 0$, this is a minimum turning point.

• For $x_2 = \frac{2 - \sqrt{19}}{3}$:

  $$\frac{d^2y}{dx^2} \Big|_{x_2} = 6\left(\frac{2 - \sqrt{19}}{3}\right) - 4$$ $$= 2(2 - \sqrt{19}) - 4$$ $$= 4 - 2\sqrt{19} - 4 = -2\sqrt{19}$$

  Since $-2\sqrt{19} < 0$, this is a maximum turning point.

Step 4: Calculate the y-coordinates for each turning point.
Substitute the x-values back into the original function $y = x^3 - 2x^2 - 5x + 6$.
A simplified form for $y$ can be derived from $3x^2 - 4x - 5 = 0 \implies x^2 = \frac{4x+5}{3}$:

$$y = x(x^2) - 2x^2 - 5x + 6$$ $$y = x\left(\frac{4x+5}{3}\right) - 2\left(\frac{4x+5}{3}\right) - 5x + 6$$ $$y = \frac{4x^2+5x - 8x-10 - 15x + 18}{3}$$ $$y = \frac{4x^2 - 18x + 8}{3}$$

Substitute $x^2 = \frac{4x+5}{3}$ again:

$$y = \frac{4\left(\frac{4x+5}{3}\right) - 18x + 8}{3}$$ $$y = \frac{\frac{16x+20}{3} - 18x + 8}{3}$$ $$y = \frac{16x+20 - 54x + 24}{9}$$ $$y = \frac{-38x + 44}{9}$$

• For the minimum turning point ($x_1 = \frac{2 + \sqrt{19}}{3}$): $$y_1 = \frac{-38\left(\frac{2 + \sqrt{19}}{3}\right) + 44}{9}$$ $$y_1 = \frac{-76 - 38\sqrt{19} + 132}{27}$$ $$$$