Question 1(a): Identity Proof
Prove the identity:
$$ \frac{\csc\alpha + \cot\alpha}{\sec\alpha + \tan\alpha} \equiv \frac{\sec\alpha - \tan\alpha}{\csc\alpha - \cot\alpha} $$
Key Identities:
Difference of Squares: \( a^2 - b^2 = (a-b)(a+b) \)
Pythagorean Identity 1: \( \csc^2\alpha - \cot^2\alpha = 1 \)
Pythagorean Identity 2: \( \sec^2\alpha - \tan^2\alpha = 1 \)
Step 1: Start with the Left Hand Side (LHS)
$$ \text{LHS} = \frac{\csc\alpha + \cot\alpha}{\sec\alpha + \tan\alpha} $$
Step 2: Multiply by Conjugates
To create Pythagorean identities, we multiply the numerator by \((\csc\alpha - \cot\alpha)\) and the denominator by \((\sec\alpha - \tan\alpha)\). To keep the fraction equivalent, we must apply these to both the top and bottom.
$$ = \frac{(\csc\alpha + \cot\alpha)}{(\sec\alpha + \tan\alpha)} \cdot \frac{(\csc\alpha - \cot\alpha)}{(\csc\alpha - \cot\alpha)} \cdot \frac{(\sec\alpha - \tan\alpha)}{(\sec\alpha - \tan\alpha)} $$
Step 3: Group terms to form Difference of Squares
$$ = \frac{[(\csc\alpha + \cot\alpha)(\csc\alpha - \cot\alpha)] \cdot (\sec\alpha - \tan\alpha)}{[(\sec\alpha + \tan\alpha)(\sec\alpha - \tan\alpha)] \cdot (\csc\alpha - \cot\alpha)} $$
Step 4: Apply Identities
Applying difference of squares:
$$ = \frac{(\csc^2\alpha - \cot^2\alpha) \cdot (\sec\alpha - \tan\alpha)}{(\sec^2\alpha - \tan^2\alpha) \cdot (\csc\alpha - \cot\alpha)} $$
Since \(\csc^2\alpha - \cot^2\alpha = 1\) and \(\sec^2\alpha - \tan^2\alpha = 1\):
$$ = \frac{(1) \cdot (\sec\alpha - \tan\alpha)}{(1) \cdot (\csc\alpha - \cot\alpha)} $$
$$ = \frac{\sec\alpha - \tan\alpha}{\csc\alpha - \cot\alpha} = \text{RHS} \quad \text{(Q.E.D.)} $$
Question 1(b): Identity Proof
Prove the identity:
$$ \frac{\sin 2\theta + \sin 5\theta}{\cos 2\theta - \cos 5\theta} \equiv \cot\frac{3\theta}{2} $$
Sum-to-Product Formulas:
\(\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})\)
\(\cos A - \cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})\)
Step 1: Rearrange for clarity (optional)
It is often easier to put the larger angle first to track negative signs clearly.
$$ \text{LHS} = \frac{\sin 5\theta + \sin 2\theta}{-(\cos 5\theta - \cos 2\theta)} \quad \text{or apply formula directly to original.} $$
Let's apply formulas directly to the original with \(A=2\theta\) and \(B=5\theta\).
Step 2: Apply Sum-to-Product to Numerator
$$ \sin 2\theta + \sin 5\theta = 2\sin\left(\frac{2\theta+5\theta}{2}\right)\cos\left(\frac{2\theta-5\theta}{2}\right) $$
$$ = 2\sin\left(\frac{7\theta}{2}\right)\cos\left(\frac{-3\theta}{2}\right) $$
Since \(\cos(-x) = \cos(x)\):
$$ = 2\sin\left(\frac{7\theta}{2}\right)\cos\left(\frac{3\theta}{2}\right) $$
Step 3: Apply Sum-to-Product to Denominator
$$ \cos 2\theta - \cos 5\theta = -2\sin\left(\frac{2\theta+5\theta}{2}\right)\sin\left(\frac{2\theta-5\theta}{2}\right) $$
$$ = -2\sin\left(\frac{7\theta}{2}\right)\sin\left(\frac{-3\theta}{2}\right) $$
Since \(\sin(-x) = -\sin(x)\), the negative signs cancel:
$$ = -2\sin\left(\frac{7\theta}{2}\right)\left[-\sin\left(\frac{3\theta}{2}\right)\right] = 2\sin\left(\frac{7\theta}{2}\right)\sin\left(\frac{3\theta}{2}\right) $$
Step 4: Simplify the Fraction
$$ \text{LHS} = \frac{2\sin(\frac{7\theta}{2})\cos(\frac{3\theta}{2})}{2\sin(\frac{7\theta}{2})\sin(\frac{3\theta}{2})} $$
Cancel common terms \(2\) and \(\sin(\frac{7\theta}{2})\):
$$ = \frac{\cos(\frac{3\theta}{2})}{\sin(\frac{3\theta}{2})} $$
$$ = \cot\left(\frac{3\theta}{2}\right) = \text{RHS} \quad \text{(Q.E.D.)} $$
Question 1(c): Identity Proof
Prove:
$$ \frac{\sin A - \sin B}{\sin A + \sin B} = \cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right) $$
Sum-to-Product Formulas:
\(\sin A - \sin B = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})\)
\(\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})\)
Step 1: Apply formulas to Numerator and Denominator
$$ \text{Numerator} = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) $$
$$ \text{Denominator} = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) $$
Step 2: Substitute and Group Terms
$$ \text{LHS} = \frac{2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})}{2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})} $$
Cancel the 2s and group by angle argument:
$$ = \left[ \frac{\cos(\frac{A+B}{2})}{\sin(\frac{A+B}{2})} \right] \cdot \left[ \frac{\sin(\frac{A-B}{2})}{\cos(\frac{A-B}{2})} \right] $$
Since \(\frac{\cos x}{\sin x} = \cot x\) and \(\frac{\sin x}{\cos x} = \tan x\):
$$ = \cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right) = \text{RHS} \quad \text{(Q.E.D.)} $$
Question 2: Exact Value
Show that \( \cos\frac{\pi}{8} \) can be written exactly as \( \frac{\sqrt{2+\sqrt{2}}}{2} \).
Method 1: Half-Angle Formula
$$ \cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}} $$
Step 1: Identify \(\theta\)
We know that \( \frac{\pi}{8} = \frac{1}{2} \times \frac{\pi}{4} \). Let \( \theta = \frac{\pi}{4} \).
Step 2: Apply the Formula
$$ \cos\frac{\pi}{8} = \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} $$
(Positive root because \(\frac{\pi}{8}\) is acute).
Step 3: Substitute Known Value
We know \( \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
$$ \cos\frac{\pi}{8} = \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} $$
Step 4: Simplify
$$ = \sqrt{\frac{\frac{2+\sqrt{2}}{2}}{2}} = \sqrt{\frac{2+\sqrt{2}}{4}} = \frac{\sqrt{2+\sqrt{2}}}{2} $$
Method 2: Double Angle Formula
Formula: \( \cos 2A = 2\cos^2 A - 1 \)
Step 1: Set up the equation
Let \( A = \frac{\pi}{8} \). Then \( 2A = \frac{\pi}{4} \).
Substituting into the formula:
$$ \cos\left(\frac{\pi}{4}\right) = 2\cos^2\left(\frac{\pi}{8}\right) - 1 $$
Step 2: Substitute value and rearrange
$$ \frac{\sqrt{2}}{2} = 2\cos^2\left(\frac{\pi}{8}\right) - 1 $$
Add 1 to both sides:
$$ 2\cos^2\left(\frac{\pi}{8}\right) = 1 + \frac{\sqrt{2}}{2} = \frac{2+\sqrt{2}}{2} $$
Step 3: Solve for cos(pi/8)
Divide by 2:
$$ \cos^2\left(\frac{\pi}{8}\right) = \frac{2+\sqrt{2}}{4} $$
Take the square root (positive, as angle is in Q1):
$$ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{2+\sqrt{2}}{4}} = \frac{\sqrt{2+\sqrt{2}}}{2} $$
$$ = \frac{\sqrt{2+\sqrt{2}}}{2} $$
Question 3(i): Exact Value of Tangent
It can be shown that \( \tan(\frac{\pi}{8}) = \sqrt{2}-1 \). Find a similar expression for \( \tan(\frac{3\pi}{8}) \).
Method 1: Co-function Identity
\( \tan(\frac{\pi}{2} - x) = \cot x = \frac{1}{\tan x} \)
Step 1: Relate the Angles
Notice that \( \frac{\pi}{2} = \frac{4\pi}{8} \). So, \( \frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8} \).
Step 2: Apply Identity
$$ \tan\left(\frac{3\pi}{8}\right) = \tan\left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \cot\left(\frac{\pi}{8}\right) = \frac{1}{\tan(\frac{\pi}{8})} $$
Step 3: Substitute and Rationalize
$$ = \frac{1}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1 $$
Method 2: Sum of Angles Formula
Formula: \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Step 1: Break down the angle
We can write \( \frac{3\pi}{8} \) as the sum of known angles:
$$ \frac{3\pi}{8} = \frac{\pi}{4} + \frac{\pi}{8} $$
Let \( A = \frac{\pi}{4} \) and \( B = \frac{\pi}{8} \).
Step 2: Substitute into formula
We know \( \tan\frac{\pi}{4} = 1 \) and \( \tan\frac{\pi}{8} = \sqrt{2}-1 \).
$$ \tan\left(\frac{3\pi}{8}\right) = \frac{1 + (\sqrt{2}-1)}{1 - (1)(\sqrt{2}-1)} $$
Step 3: Simplify and Rationalize
$$ = \frac{\sqrt{2}}{1 - \sqrt{2} + 1} = \frac{\sqrt{2}}{2 - \sqrt{2}} $$
Rationalize the denominator by multiplying by \( (2+\sqrt{2}) \):
$$ = \frac{\sqrt{2}(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} $$
$$ = \frac{2\sqrt{2} + 2}{4 - 2} = \frac{2(\sqrt{2}+1)}{2} $$
$$ = \sqrt{2} + 1 $$
$$ \tan\left(\frac{3\pi}{8}\right) = \sqrt{2} + 1 $$
Question 3(ii): Solving System of Equations
Find all real solutions for the system in exact form:
$$ (1) \quad \tan\left(\frac{x}{2}\right)\tan y = \sqrt{2}-1 $$
$$ (2) \quad \tan\left(\frac{x}{2} + y\right) = 1 + \sqrt{2} $$
Step 1: Substitute Known Values
Let \( A = \frac{x}{2} \) and \( B = y \).
From Q3(i), we know:
\(\sqrt{2}-1 = \tan\frac{\pi}{8}\)
\(1+\sqrt{2} = \tan\frac{3\pi}{8}\)
The system becomes:
$$ (1) \quad \tan A \tan B = \tan\frac{\pi}{8} $$
$$ (2) \quad \tan(A+B) = \tan\frac{3\pi}{8} $$
Step 2: Solve for (A+B)
From equation (2):
$$ \tan(A+B) = \tan\frac{3\pi}{8} $$
The general solution for \( \tan \theta = \tan \alpha \) is \( \theta = n\pi + \alpha \).
$$ A + B = n\pi + \frac{3\pi}{8} \quad \text{where } n \in \mathbb{Z} $$
Step 3: Expand tan(A+B) to find (tan A + tan B)
Using the expansion formula: \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Substitute the values from the system equations:
$$ 1+\sqrt{2} = \frac{\tan A + \tan B}{1 - (\sqrt{2}-1)} $$
$$ 1+\sqrt{2} = \frac{\tan A + \tan B}{1 - \sqrt{2} + 1} $$
$$ 1+\sqrt{2} = \frac{\tan A + \tan B}{2 - \sqrt{2}} $$
Solve for \( (\tan A + \tan B) \):
$$ \tan A + \tan B = (1+\sqrt{2})(2-\sqrt{2}) $$
$$ = 2 - \sqrt{2} + 2\sqrt{2} - 2 $$
$$ \tan A + \tan B = \sqrt{2} $$
Step 4: Form a Quadratic Equation
We now have the Sum and Product of the terms \(u = \tan A\) and \(v = \tan B\):
Sum \(S = u + v = \sqrt{2} \)
Product \(P = u \cdot v = \sqrt{2}-1 \)
These are roots of the quadratic equation \( t^2 - St + P = 0 \):
$$ t^2 - \sqrt{2}t + (\sqrt{2}-1) = 0 $$
Step 5: Solve the Quadratic
Using the quadratic formula:
$$ t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
$$ t = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(\sqrt{2}-1)}}{2} $$
$$ t = \frac{\sqrt{2} \pm \sqrt{2 - 4\sqrt{2} + 4}}{2} $$
$$ t = \frac{\sqrt{2} \pm \sqrt{6 - 4\sqrt{2}}}{2} $$
Simplify the square root: Note that \( (2-\sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2} \).
So, \( \sqrt{6-4\sqrt{2}} = 2-\sqrt{2} \).
$$ t = \frac{\sqrt{2} \pm (2-\sqrt{2})}{2} $$
Case 1 (Plus): \( t_1 = \frac{\sqrt{2} + 2 - \sqrt{2}}{2} = \frac{2}{2} = 1 \)
Case 2 (Minus): \( t_2 = \frac{\sqrt{2} - (2 - \sqrt{2})}{2} = \frac{2\sqrt{2}-2}{2} = \sqrt{2}-1 \)
Step 6: Determine x and y
We have two possibilities for the pair \( (\tan\frac{x}{2}, \tan y) \):
Possibility 1: \( \tan\frac{x}{2} = 1 \) and \( \tan y = \sqrt{2}-1 \)
$$ \frac{x}{2} = m\pi + \frac{\pi}{4} \implies x = 2m\pi + \frac{\pi}{2}, \quad m \in \mathbb{Z} $$
$$ y = k\pi + \frac{\pi}{8}, \quad k \in \mathbb{Z} $$
Possibility 2: \( \tan\frac{x}{2} = \sqrt{2}-1 \) and \( \tan y = 1 \)
$$ \frac{x}{2} = m\pi + \frac{\pi}{8} \implies x = 2m\pi + \frac{\pi}{4}, \quad m \in \mathbb{Z} $$
$$ y = k\pi + \frac{\pi}{4}, \quad k \in \mathbb{Z} $$
Set 1: \( x = 2m\pi + \frac{\pi}{2}, \quad y = k\pi + \frac{\pi}{8} \)
Set 2: \( x = 2m\pi + \frac{\pi}{4}, \quad y = k\pi + \frac{\pi}{4} \)
(where \(m, k \in \mathbb{Z}\))