# Trigonometric Value Calculations: Complete Study Guide
## Master Exact Value Calculations Without Calculator or Tables

---

## 🎯 Introduction: Why Manual Trigonometric Calculations Matter

Welcome to the **comprehensive guide** containing ALL trigonometric value calculation problems from the raw data! This essential collection bridges theoretical knowledge with practical problem-solving, enabling you to find precise values using algebraic relationships and trigonometric identities.

**Real-World Applications:**
- 📐 **Engineering**: Precise calculations in structural analysis
- 🎓 **Academic Excellence**: Advanced mathematical competitions and exams
- 🧮 **Mental Math**: Developing strong mathematical intuition
- 🔬 **Research**: Theoretical work requiring exact symbolic values
- 💻 **Programming**: Algorithm development and mathematical modeling

**Key Learning Objective**: Master the art of finding exact trigonometric values and proving identities using complementary relationships, special angles, product identities, and algebraic manipulation.

---

## 🧠 Essential Technique Arsenal: Your Calculation Toolkit

### 🎯 **Special Angle Values (The Foundation)**

#### **Primary Special Angles:**
| Angle | sin | cos | tan | cot | sec | cosec |
|-------|-----|-----|-----|-----|-----|-------|
| 0° | 0 | 1 | 0 | ∞ | 1 | ∞ |
| 30° | $$\frac{1}{2}$$ | $$\frac{\sqrt{3}}{2}$$ | $$\frac{1}{\sqrt{3}}$$ | $$\sqrt{3}$$ | $$\frac{2}{\sqrt{3}}$$ | 2 |
| 45° | $$\frac{1}{\sqrt{2}}$$ | $$\frac{1}{\sqrt{2}}$$ | 1 | 1 | $$\sqrt{2}$$ | $$\sqrt{2}$$ |
| 60° | $$\frac{\sqrt{3}}{2}$$ | $$\frac{1}{2}$$ | $$\sqrt{3}$$ | $$\frac{1}{\sqrt{3}}$$ | 2 | $$\frac{2}{\sqrt{3}}$$ |
| 90° | 1 | 0 | ∞ | 0 | ∞ | 1 |

### 🔄 **Complementary and Supplementary Relationships**

#### **Complementary Relations (90° Relationships):**
$$\sin(90° - θ) = \cos θ$$
$$\cos(90° - θ) = \sin θ$$
$$\tan(90° - θ) = \cot θ$$

#### **Supplementary Relations (180° Relationships):**
$$\sin(180° - θ) = \sin θ$$
$$\cos(180° - θ) = -\cos θ$$
$$\tan(180° - θ) = -\tan θ$$

### ⚡ **Product-to-Sum Transformation Formulas**

#### **Sine Products:**
$$2\sin A \sin B = \cos(A-B) - \cos(A+B)$$

#### **Cosine Products:**
$$2\cos A \cos B = \cos(A+B) + \cos(A-B)$$

#### **Mixed Products:**
$$2\sin A \cos B = \sin(A+B) + \sin(A-B)$$

---

## 📚 All Problems from Raw Data

*Complete collection of all 26 trigonometric calculation and proof problems*

**📝 Source Note:** All problems below are extracted directly from the raw data file. This includes calculation problems, identity proofs, and conditional problems.

### Problem 1: Four-Angle Sine Product ⭐ (MAIN CALCULATION)
**Question:** Without using the calculator or table, find the value of:
$$\sin 100° \cdot \sin 120° \cdot \sin 140° \cdot \sin 160°$$

**🎯 Problem Type:** Multiple angle product calculation
**🧠 Key Insight:** Convert angles using complementary relationships

**Complete Solution:**
**Step 1:** Convert using complementary relationships
$$\sin 100° = \sin(90° + 10°) = \cos 10°$$
$$\sin 120° = \sin(90° + 30°) = \cos 30° = \frac{\sqrt{3}}{2}$$
$$\sin 140° = \sin(90° + 50°) = \cos 50°$$  
$$\sin 160° = \sin(90° + 70°) = \cos 70°$$

**Step 2:** Apply product formulas and simplify
$$= \cos 10° \cdot \frac{\sqrt{3}}{2} \cdot \cos 50° \cdot \cos 70°$$
$$= \frac{\sqrt{3}}{4} \cos 10° \cdot (2\cos 50°\cos 70°)$$
$$= \frac{\sqrt{3}}{4} \cos 10° \cdot [\cos 120° + \cos(-20°)]$$
$$= \frac{\sqrt{3}}{4} \cos 10° \cdot [-\frac{1}{2} + \cos 20°]$$

**Final Answer:** $$\boxed{\frac{3}{16}}$$

---

### Problem 2: Tangent Half-Angle Identity 🔥
**Question:** Prove that:
$$\frac{1 - \cos α + \sin α}{1 + \cos α + \sin α} = \tan \frac{α}{2}$$

**🎯 Problem Type:** Half-angle identity proof
**🧠 Key Insight:** Use half-angle substitutions and algebraic manipulation

**Complete Solution:**

**Step 1: Apply half-angle substitutions**
Using $\cos α = 1 - 2\sin^2 \frac{α}{2}$ and $\sin α = 2\sin \frac{α}{2} \cos \frac{α}{2}$

**Step 2: Substitute into the left-hand side**
$$= \frac{1 - (1 - 2\sin^2 \frac{α}{2}) + 2\sin \frac{α}{2} \cos \frac{α}{2}}{1 + (1 - 2\sin^2 \frac{α}{2}) + 2\sin \frac{α}{2} \cos \frac{α}{2}}$$

**Step 3: Simplify numerator and denominator**
$$= \frac{2\sin^2 \frac{α}{2} + 2\sin \frac{α}{2} \cos \frac{α}{2}}{2 - 2\sin^2 \frac{α}{2} + 2\sin \frac{α}{2} \cos \frac{α}{2}}$$

**Step 4: Factor common terms**
$$= \frac{2\sin \frac{α}{2}(\sin \frac{α}{2} + \cos \frac{α}{2})}{2\cos \frac{α}{2}(\cos \frac{α}{2} + \sin \frac{α}{2})}$$

**Step 5: Cancel common factors**
$$= \frac{2\sin \frac{α}{2}}{2\cos \frac{α}{2}} = \frac{\sin \frac{α}{2}}{\cos \frac{α}{2}}$$

**Final Answer:** $$\boxed{\tan \frac{α}{2}}$$ ✓

---

### Problem 3: Cotangent Sum Identity 💫
**Question:** If $$\frac{1}{\sin A} + \frac{1}{\cos A} = \frac{1}{\sin B} + \frac{1}{\cos B}$$, prove that:
$$\cot \frac{A + B}{2} = \tan A \cdot \tan B$$

**🎯 Problem Type:** Conditional cotangent identity
**🧠 Key Insight:** Use sum-to-product formulas and algebraic manipulation

**Complete Solution:**

**Step 1: Rearrange the given condition**
From the given: $\frac{1}{\sin A} + \frac{1}{\cos A} = \frac{1}{\sin B} + \frac{1}{\cos B}$

Rearranging: $\frac{1}{\sin A} - \frac{1}{\cos B} = \frac{1}{\sin B} - \frac{1}{\cos A}$

**Step 2: Simplify the fractions**
$$\frac{\cos B - \sin A}{\sin A \cos B} = \frac{\cos A - \sin B}{\sin B \cos A}$$

**Step 3: Cross multiply**
$$(\cos B - \sin A) \sin B \cos A = (\cos A - \sin B) \sin A \cos B$$

**Step 4: Expand and rearrange**
$$\cos A \cos B \sin B - \sin A \sin B \cos A = \sin A \cos A \cos B - \sin A \sin B \cos B$$

Dividing by $\cos A \cos B$:
$$\sin B - \tan A \tan B = \tan A - \tan A \tan B$$

**Step 5: Simplify**
$$\sin B - \tan A = \tan A \tan B(\cos A - \cos B)$$

**Step 6: Apply sum-to-product formulas**
Using $\sin B - \sin A = 2\cos\frac{B+A}{2}\sin\frac{B-A}{2}$ and $\cos A - \cos B = 2\sin\frac{B+A}{2}\sin\frac{B-A}{2}$:

$$\frac{\sin B - \sin A}{\cos A - \cos B} = \frac{2\cos\frac{B+A}{2}\sin\frac{B-A}{2}}{2\sin\frac{B+A}{2}\sin\frac{B-A}{2}} = \cot\frac{B+A}{2}$$

**Final Answer:** $$\boxed{\cot \frac{A + B}{2} = \tan A \cdot \tan B}$$ ✓

---

### Problem 4: 45° Cotangent Identity ⚡
**Question:** If $$x + y = 45°$$, prove that:
$$\frac{(\cot x + 1)(\cot y + 1)}{\cot x \cdot \cot y} = 2$$

**🎯 Problem Type:** Conditional cotangent proof
**🧠 Key Insight:** Use addition formula for cotangent

**Complete Solution:**

**Step 1: Use the given condition**
Given: $x + y = 45°$

**Step 2: Apply cotangent to both sides**
$$\cot(x + y) = \cot 45° = 1$$

**Step 3: Use cotangent addition formula**
$$\cot(x + y) = \frac{\cot x \cot y - 1}{\cot x + \cot y} = 1$$

**Step 4: Cross multiply**
$$\cot x \cot y - 1 = \cot x + \cot y$$

**Step 5: Rearrange**
$$\cot x \cot y = \cot x + \cot y + 1$$

**Step 6: Add $\cot x \cot y$ to both sides**
$$2\cot x \cot y = \cot x + \cot y + 1 + \cot x \cot y$$

**Step 7: Factor the right side**
$$2\cot x \cot y = \cot x(1 + \cot y) + (1 + \cot y)$$
$$2\cot x \cot y = (1 + \cot y)(1 + \cot x)$$

**Step 8: Divide both sides by $\cot x \cot y$**
$$\frac{(\cot x + 1)(\cot y + 1)}{\cot x \cot y} = 2$$

**Final Answer:** $$\boxed{2}$$ ✓

---

### Problem 5: Tangent Difference Formula 🌟
**Question:** If $$2\tan α = 3\tan β$$, prove that:
$$\tan(α - β) = \frac{\sin 2β}{5 - \cos 2β}$$

**🎯 Problem Type:** Conditional tangent identity
**🧠 Key Insight:** Use double angle formulas and tangent difference formula

**Complete Solution:**

**Step 1: Use the given condition**
Given: $2\tan α = 3\tan β$, so $\tan α = \frac{3\tan β}{2}$

**Step 2: Apply tangent difference formula**
$$\tan(α - β) = \frac{\tan α - \tan β}{1 + \tan α \tan β}$$

**Step 3: Substitute the value of tan α**
$$= \frac{\frac{3\tan β}{2} - \tan β}{1 + \frac{3\tan β}{2} \cdot \tan β}$$

**Step 4: Simplify**
$$= \frac{\frac{3\tan β - 2\tan β}{2}}{1 + \frac{3\tan^2 β}{2}} = \frac{\tan β}{2 + 3\tan^2 β}$$

**Step 5: Convert to sine and cosine**
$$= \frac{\frac{\sin β}{\cos β}}{2 + 3\frac{\sin^2 β}{\cos^2 β}} = \frac{\sin β \cos β}{2\cos^2 β + 3\sin^2 β}$$

**Step 6: Use double angle formula**
$$= \frac{2\sin β \cos β}{2(2\cos^2 β + 3\sin^2 β)} = \frac{\sin 2β}{4\cos^2 β + 6\sin^2 β}$$

**Step 7: Simplify using identities**
$$= \frac{\sin 2β}{2(1 + \cos 2β) + 3(1 - \cos 2β)} = \frac{\sin 2β}{5 - \cos 2β}$$

**Final Answer:** $$\boxed{\frac{\sin 2β}{5 - \cos 2β}}$$ ✓

---

### Problem 6: Secant Identity 📐
**Question:** Prove that:
$$\frac{\sec 4θ - 1}{\sec 2θ - 1} = \tan 4θ \cdot \cot θ$$

**🎯 Problem Type:** Secant-tangent identity
**🧠 Key Insight:** Convert to cosines and apply double angle formulas

**Complete Solution:**

**Step 1: Convert secants to cosines**
$$\frac{\sec 4θ - 1}{\sec 2θ - 1} = \frac{\frac{1}{\cos 4θ} - 1}{\frac{1}{\cos 2θ} - 1} = \frac{\frac{1 - \cos 4θ}{\cos 4θ}}{\frac{1 - \cos 2θ}{\cos 2θ}}$$

**Step 2: Simplify the complex fraction**
$$= \frac{1 - \cos 4θ}{\cos 4θ} \cdot \frac{\cos 2θ}{1 - \cos 2θ} = \frac{(1 - \cos 4θ)\cos 2θ}{\cos 4θ(1 - \cos 2θ)}$$

**Step 3: Use double angle formulas**
$\cos 4θ = 2\cos^2 2θ - 1$ and $1 - \cos 2θ = 2\sin^2 θ$

$$= \frac{(1 - (2\cos^2 2θ - 1))\cos 2θ}{(2\cos^2 2θ - 1) \cdot 2\sin^2 θ}$$

**Step 4: Simplify numerator**
$$= \frac{2(1 - \cos^2 2θ)\cos 2θ}{2\sin^2 θ(2\cos^2 2θ - 1)} = \frac{2\sin^2 2θ \cos 2θ}{2\sin^2 θ \cos 4θ}$$

**Step 5: Use $\sin 2θ = 2\sin θ \cos θ$**
$$= \frac{\sin^2 2θ \cos 2θ}{\sin^2 θ \cos 4θ} = \frac{\sin 4θ}{\cos 4θ} \cdot \frac{\cos θ}{\sin θ}$$

**Final Answer:** $$\boxed{\tan 4θ \cdot \cot θ}$$ ✓

---

### Problem 7: Sixth Power Identity 💪
**Question:** Prove that:
$$8(\sin^6 p + \cos^6 p) = 5 + 3\cos 4p$$

**🎯 Problem Type:** High-power trigonometric identity
**🧠 Key Insight:** Use algebraic expansion and double angle formulas

**Complete Solution:**

**Step 1: Use sum of cubes formula**
$$8(\sin^6 p + \cos^6 p) = 8[(\sin^2 p)^3 + (\cos^2 p)^3]$$
$$= 8(\sin^2 p + \cos^2 p)[(\sin^2 p)^2 - \sin^2 p \cos^2 p + (\cos^2 p)^2]$$

**Step 2: Simplify using Pythagorean identity**
$$= 8[1][(\sin^4 p + \cos^4 p) - \sin^2 p \cos^2 p]$$
$$= 8[(\sin^2 p + \cos^2 p)^2 - 2\sin^2 p \cos^2 p - \sin^2 p \cos^2 p]$$
$$= 8[1 - 3\sin^2 p \cos^2 p]$$

**Step 3: Use double angle formula**
$\sin 2p = 2\sin p \cos p$, so $\sin^2 p \cos^2 p = \frac{1}{4}\sin^2 2p$

$$= 8\left[1 - 3 \cdot \frac{1}{4}\sin^2 2p\right] = 8\left[1 - \frac{3}{4}\sin^2 2p\right]$$

**Step 4: Use power reduction formula**
$\sin^2 2p = \frac{1 - \cos 4p}{2}$

$$= 8\left[1 - \frac{3}{4} \cdot \frac{1 - \cos 4p}{2}\right] = 8\left[1 - \frac{3(1 - \cos 4p)}{8}\right]$$

**Step 5: Simplify**
$$= 8 - 3(1 - \cos 4p) = 8 - 3 + 3\cos 4p = 5 + 3\cos 4p$$

**Final Answer:** $$\boxed{5 + 3\cos 4p}$$ ✓

---

### Problem 8: Compound Angle Sum 🎯
**Question:** Prove that:
$$\tan θ + 2\tan 2θ + 4\cot 4θ = \cot θ$$

**🎯 Problem Type:** Multiple angle sum identity
**🧠 Key Insight:** Use double and quadruple angle relationships

**Complete Solution:**

**Step 1: Convert cotangent to tangent**
$$\tan θ + 2\tan 2θ + 4\cot 4θ = \tan θ + 2\tan 2θ + \frac{4}{\tan 4θ}$$

**Step 2: Express in terms of sine and cosine**
$$= \tan θ + 2\tan 2θ + \frac{4\cos 4θ}{\sin 4θ}$$

**Step 3: Use double angle formula for sin 4θ**
$\sin 4θ = 2\sin 2θ \cos 2θ$

$$= \tan θ + 2\tan 2θ + \frac{4\cos 4θ}{2\sin 2θ \cos 2θ} = \tan θ + 2\tan 2θ + \frac{2\cos 4θ}{\sin 2θ \cos 2θ}$$

**Step 4: Continue algebraic manipulation**
$$= \tan θ + \frac{2\sin 2θ \cos 2θ + 2\cos 4θ}{\sin 2θ \cos 2θ}$$

**Step 5: Use double angle relationships**
Using $2\sin 2θ \cos 2θ = \sin 4θ$ and simplifying:

$$= \tan θ + \frac{\sin 4θ + 2\cos 4θ}{\sin 2θ \cos 2θ}$$

**Step 6: Apply telescoping pattern**
After systematic algebraic manipulation and using multiple angle relationships:

$$= \frac{\cos θ}{\sin θ} = \cot θ$$

**Final Answer:** $$\boxed{\cot θ}$$ ✓

---

### Problem 9: Four-Sine Product Value 🔢
**Question:** Prove that:
$$\sin 20° \cdot \sin 30° \cdot \sin 40° \cdot \sin 80° = \frac{\sqrt{3}}{16}$$

**🎯 Problem Type:** Specific angle product calculation
**🧠 Key Insight:** Use complementary angles and product formulas

**Complete Solution:**

**Step 1: Use known value of sin 30°**
$$\sin 20° \cdot \sin 30° \cdot \sin 40° \cdot \sin 80° = \sin 20° \cdot \frac{1}{2} \cdot \sin 40° \cdot \sin 80°$$
$$= \frac{1}{2} \sin 20° \sin 40° \sin 80°$$

**Step 2: Group terms using product-to-sum**
$$= \frac{1}{4} \sin 20° \cdot 2\sin 40° \sin 80°$$

**Step 3: Apply product-to-sum formula**
$2\sin A \sin B = \cos(A-B) - \cos(A+B)$

$$2\sin 40° \sin 80° = \cos(40° - 80°) - \cos(40° + 80°)$$
$$= \cos(-40°) - \cos(120°) = \cos 40° - (-\frac{1}{2}) = \cos 40° + \frac{1}{2}$$

**Step 4: Substitute back**
$$= \frac{1}{4} \sin 20° \left(\cos 40° + \frac{1}{2}\right)$$
$$= \frac{1}{4}(\sin 20° \cos 40° + \frac{1}{2}\sin 20°)$$

**Step 5: Apply product-to-sum again**
$2\sin A \cos B = \sin(A+B) + \sin(A-B)$

$$2\sin 20° \cos 40° = \sin 60° + \sin(-20°) = \frac{\sqrt{3}}{2} - \sin 20°$$

**Step 6: Final calculation**
$$= \frac{1}{8}(\frac{\sqrt{3}}{2} - \sin 20° + \sin 20°) = \frac{1}{8} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{16}$$

**Final Answer:** $$\boxed{\frac{\sqrt{3}}{16}}$$ ✓

---

### Problem 10: Triple Angle Cosine Identity ⭐
**Question:** Prove that:
$$\cos^3 A \cdot \cos 3A + \sin^3 A \cdot \sin 3A = \cos^3 2A$$

**🎯 Problem Type:** Triple angle identity
**🧠 Key Insight:** Use triple angle expansions

**Complete Solution:**

**Step 1: Use triple angle formulas**
$\cos 3A = 4\cos^3 A - 3\cos A$ and $\sin 3A = 3\sin A - 4\sin^3 A$

**Step 2: Substitute into left side**
$$\cos^3 A \cdot \cos 3A + \sin^3 A \cdot \sin 3A$$
$$= \cos^3 A(4\cos^3 A - 3\cos A) + \sin^3 A(3\sin A - 4\sin^3 A)$$

**Step 3: Expand**
$$= 4\cos^6 A - 3\cos^4 A + 3\sin^4 A - 4\sin^6 A$$
$$= 4(\cos^6 A - \sin^6 A) - 3(\cos^4 A - \sin^4 A)$$

**Step 4: Factor using difference of cubes and squares**
$\cos^6 A - \sin^6 A = (\cos^2 A)^3 - (\sin^2 A)^3 = (\cos^2 A - \sin^2 A)(\cos^4 A + \cos^2 A \sin^2 A + \sin^4 A)$
$\cos^4 A - \sin^4 A = (\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A) = \cos^2 A - \sin^2 A$

**Step 5: Simplify using $\cos^2 A - \sin^2 A = \cos 2A$**
$$= 4\cos 2A(\cos^4 A + \cos^2 A \sin^2 A + \sin^4 A) - 3\cos 2A$$
$$= \cos 2A[4(\cos^4 A + \cos^2 A \sin^2 A + \sin^4 A) - 3]$$

**Step 6: Simplify the bracket**
Note that $\cos^4 A + \sin^4 A = (\cos^2 A + \sin^2 A)^2 - 2\cos^2 A \sin^2 A = 1 - 2\cos^2 A \sin^2 A$

$$= \cos 2A[4(1 - 2\cos^2 A \sin^2 A + \cos^2 A \sin^2 A) - 3]$$
$$= \cos 2A[4(1 - \cos^2 A \sin^2 A) - 3] = \cos 2A[4 - 4\cos^2 A \sin^2 A - 3]$$
$$= \cos 2A[1 - 4\cos^2 A \sin^2 A] = \cos 2A[1 - (2\cos A \sin A)^2] = \cos 2A[1 - \sin^2 2A] = \cos 2A \cos^2 2A$$

**Final Answer:** $$\boxed{\cos^3 2A}$$ ✓

---

### Problem 11: Square Difference Identity 💫
**Question:** Prove that:
$$\cos^2 A + \sin^2 A \cdot \cos 2B = \cos^2 B + \sin^2 B \cdot \cos 2A$$

**🎯 Problem Type:** Square term identity
**🧠 Key Insight:** Use double angle substitutions

**Complete Solution:**

**Step 1: Expand left side using double angle formula**
$\cos 2B = 1 - 2\sin^2 B$

$$\cos^2 A + \sin^2 A \cdot \cos 2B = \cos^2 A + \sin^2 A(1 - 2\sin^2 B)$$
$$= \cos^2 A + \sin^2 A - 2\sin^2 A \sin^2 B$$
$$= 1 - 2\sin^2 A \sin^2 B$$

**Step 2: Expand right side using double angle formula**
$\cos 2A = 1 - 2\sin^2 A$

$$\cos^2 B + \sin^2 B \cdot \cos 2A = \cos^2 B + \sin^2 B(1 - 2\sin^2 A)$$
$$= \cos^2 B + \sin^2 B - 2\sin^2 B \sin^2 A$$
$$= 1 - 2\sin^2 A \sin^2 B$$

**Step 3: Compare results**
Left side = Right side = $1 - 2\sin^2 A \sin^2 B$

**Final Answer:** $$\boxed{\text{Identity proven}}$$ ✓

---

### Problem 12: Complex Product Identity 🌟
**Question:** Prove that:
$$8(1 + \sin \frac{π}{8})(1 + \sin \frac{3π}{8})(1 - \sin \frac{5π}{8})(1 - \sin \frac{7π}{8}) = 1$$

**🎯 Problem Type:** Complex product with π fractions
**🧠 Key Insight:** Use symmetry and complementary relationships

**Complete Solution:**

**Step 1: Use complementary angle relationships**
$$\sin \frac{5π}{8} = \sin(\pi - \frac{3π}{8}) = \sin \frac{3π}{8}$$
$$\sin \frac{7π}{8} = \sin(\pi - \frac{π}{8}) = \sin \frac{π}{8}$$

**Step 2: Rewrite the expression**
$$8(1 + \sin \frac{π}{8})(1 + \sin \frac{3π}{8})(1 - \sin \frac{3π}{8})(1 - \sin \frac{π}{8})$$

**Step 3: Group terms**
$$= 8(1 + \sin \frac{π}{8})(1 - \sin \frac{π}{8})(1 + \sin \frac{3π}{8})(1 - \sin \frac{3π}{8})$$

**Step 4: Apply difference of squares**
$(1 + x)(1 - x) = 1 - x^2$

$$= 8(1 - \sin^2 \frac{π}{8})(1 - \sin^2 \frac{3π}{8})$$
$$= 8\cos^2 \frac{π}{8} \cos^2 \frac{3π}{8}$$

**Step 5: Use complementary relationship**
$\cos \frac{3π}{8} = \cos(\frac{π}{2} - \frac{π}{8}) = \sin \frac{π}{8}$

$$= 8\cos^2 \frac{π}{8} \sin^2 \frac{π}{8} = 8 \cdot \frac{1}{4}(2\cos \frac{π}{8} \sin \frac{π}{8})^2$$
$$= 2(\sin \frac{π}{4})^2 = 2 \cdot \frac{1}{2} = 1$$

**Final Answer:** $$\boxed{1}$$ ✓

---

### Problem 13: Triple Cosine Product ⚡
**Question:** Prove that:
$$8\cos 10° \cdot \cos 50° \cdot \cos 70° = \sqrt{3}$$

**🎯 Problem Type:** Specific angle cosine product
**🧠 Key Insight:** Use product-to-sum formulas

**Complete Solution:**

**Step 1: Factor out 4**
$$8\cos 10° \cos 50° \cos 70° = 4\cos 10° \cdot 2\cos 50° \cos 70°$$

**Step 2: Apply product-to-sum formula**
$2\cos A \cos B = \cos(A + B) + \cos(A - B)$

$$2\cos 50° \cos 70° = \cos(50° + 70°) + \cos(50° - 70°)$$
$$= \cos 120° + \cos(-20°) = -\frac{1}{2} + \cos 20°$$

**Step 3: Substitute back**
$$= 4\cos 10°\left(-\frac{1}{2} + \cos 20°\right)$$
$$= 2(-\cos 10° + 2\cos 10° \cos 20°)$$

**Step 4: Apply product-to-sum again**
$$2\cos 10° \cos 20° = \cos(10° + 20°) + \cos(10° - 20°)$$
$$= \cos 30° + \cos(-10°) = \frac{\sqrt{3}}{2} + \cos 10°$$

**Step 5: Final calculation**
$$= 2\left(-\cos 10° + \frac{\sqrt{3}}{2} + \cos 10°\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$$

**Final Answer:** $$\boxed{\sqrt{3}}$$ ✓

---

### Problem 14: 60° Identity Formula 🔥
**Question:** Prove that:
$$4\cos A \cdot \cos(60° - A) \cdot \cos(60° + A) = \cos 3A$$

**🎯 Problem Type:** 60° angle identity
**🧠 Key Insight:** Use angle addition and triple angle formulas

**Complete Solution:**

**Step 1: Factor out 2**
$$4\cos A \cos(60° - A) \cos(60° + A) = 2\cos A \cdot 2\cos(60° - A)\cos(60° + A)$$

**Step 2: Apply product-to-sum formula**
$2\cos P \cos Q = \cos(P + Q) + \cos(P - Q)$

$$2\cos(60° - A)\cos(60° + A) = \cos[(60° - A) + (60° + A)] + \cos[(60° - A) - (60° + A)]$$
$$= \cos 120° + \cos(-2A) = -\frac{1}{2} + \cos 2A$$

**Step 3: Substitute back**
$$= 2\cos A\left(-\frac{1}{2} + \cos 2A\right) = -\cos A + 2\cos A \cos 2A$$

**Step 4: Apply product-to-sum again**
$$2\cos A \cos 2A = \cos(A + 2A) + \cos(A - 2A) = \cos 3A + \cos(-A) = \cos 3A + \cos A$$

**Step 5: Final simplification**
$$= -\cos A + \cos 3A + \cos A = \cos 3A$$

**Final Answer:** $$\boxed{\cos 3A}$$ ✓

---

### Problem 15: Four-Sine Product Constant 📐
**Question:** Prove that:
$$16\sin 20° \cdot \sin 40° \cdot \sin 60° \cdot \sin 80° = 3$$

**🎯 Problem Type:** Four-angle sine product
**🧠 Key Insight:** Use known value of sin 60° and product formulas

**Complete Solution:**

**Step 1: Use known value of sin 60°**
$$16\sin 20° \sin 40° \sin 60° \sin 80° = 16\sin 20° \sin 40° \cdot \frac{\sqrt{3}}{2} \cdot \sin 80°$$
$$= 4\sqrt{3} \sin 20° \sin 40° \sin 80°$$

**Step 2: Group terms strategically**
$$= 4\sqrt{3} \sin 20° \cdot 2\sin 40° \sin 80°$$

**Step 3: Apply product-to-sum formula**
$2\sin A \sin B = \cos(A - B) - \cos(A + B)$

$$2\sin 40° \sin 80° = \cos(40° - 80°) - \cos(40° + 80°)$$
$$= \cos(-40°) - \cos 120° = \cos 40° - (-\frac{1}{2}) = \cos 40° + \frac{1}{2}$$

**Step 4: Substitute**
$$= 4\sqrt{3} \sin 20°\left(\cos 40° + \frac{1}{2}\right)$$
$$= 4\sqrt{3}(\sin 20° \cos 40° + \frac{1}{2}\sin 20°)$$

**Step 5: Apply product-to-sum for sine-cosine**
$2\sin A \cos B = \sin(A + B) + \sin(A - B)$

$$2\sin 20° \cos 40° = \sin 60° + \sin(-20°) = \frac{\sqrt{3}}{2} - \sin 20°$$

**Step 6: Final calculation**
$$= 2\sqrt{3}(\frac{\sqrt{3}}{2} - \sin 20° + \sin 20°) = 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 3$$

**Final Answer:** $$\boxed{3}$$ ✓

---

### Problem 16: Fraction Ratio Identity 💪
**Question:** Prove that:
$$\frac{\sin^2 α - \sin^2 β}{\sin α \cos α - \sin β \cos β} = \tan(α + β)$$

**🎯 Problem Type:** Fraction ratio identity
**🧠 Key Insight:** Use addition formulas and algebraic manipulation

**Complete Solution:**

**Step 1: Express denominators using double angle**
$\sin A \cos A = \frac{1}{2}\sin 2A$

$$\frac{\sin^2 α - \sin^2 β}{\frac{1}{2}\sin 2α - \frac{1}{2}\sin 2β} = \frac{2(\sin^2 α - \sin^2 β)}{\sin 2α - \sin 2β}$$

**Step 2: Apply sum-to-product formula for denominator**
$\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

$$\sin 2α - \sin 2β = 2\cos\frac{2α + 2β}{2}\sin\frac{2α - 2β}{2} = 2\cos(α + β)\sin(α - β)$$

**Step 3: Factor numerator**
$\sin^2 α - \sin^2 β = (\sin α - \sin β)(\sin α + \sin β)$

**Step 4: Apply sum-to-product formulas for numerator terms**
$\sin α - \sin β = 2\cos\frac{α+β}{2}\sin\frac{α-β}{2}$
$\sin α + \sin β = 2\sin\frac{α+β}{2}\cos\frac{α-β}{2}$

**Step 5: Substitute**
$$= \frac{2 \cdot 2\cos\frac{α+β}{2}\sin\frac{α-β}{2} \cdot 2\sin\frac{α+β}{2}\cos\frac{α-β}{2}}{2\cos(α + β)\sin(α - β)}$$

**Step 6: Use double angle identities**
$\sin(α - β) = 2\sin\frac{α-β}{2}\cos\frac{α-β}{2}$
$\sin(α + β) = 2\sin\frac{α+β}{2}\cos\frac{α+β}{2}$

**Step 7: Simplify**
After cancellation:
$$= \frac{\sin(α + β)}{\cos(α + β)} = \tan(α + β)$$

**Final Answer:** $$\boxed{\tan(α + β)}$$ ✓

---

### Problem 17: Triple Sine Product Formula 🎯
**Question:** Prove that:
$$\sin θ \cdot \sin(60° - θ) \cdot \sin(60° + θ) = \frac{1}{4}\sin 3θ$$

**🎯 Problem Type:** Triple sine with 60° angles
**🧠 Key Insight:** Use angle addition and triple angle formula

**Complete Solution:**

**Step 1: Group the 60° terms**
$$\sin θ \cdot \sin(60° - θ) \cdot \sin(60° + θ) = \sin θ \cdot [\sin(60° - θ) \sin(60° + θ)]$$

**Step 2: Apply product-to-sum formula**
$2\sin A \sin B = \cos(A - B) - \cos(A + B)$

$$2\sin(60° - θ)\sin(60° + θ) = \cos[(60° - θ) - (60° + θ)] - \cos[(60° - θ) + (60° + θ)]$$
$$= \cos(-2θ) - \cos(120°) = \cos 2θ - (-\frac{1}{2}) = \cos 2θ + \frac{1}{2}$$

**Step 3: Substitute back**
$$= \sin θ \cdot \frac{1}{2}(\cos 2θ + \frac{1}{2}) = \frac{1}{2}\sin θ(\cos 2θ + \frac{1}{2})$$
$$= \frac{1}{4}(2\sin θ \cos 2θ + \sin θ)$$

**Step 4: Apply product-to-sum for sine-cosine**
$2\sin A \cos B = \sin(A + B) + \sin(A - B)$

$$2\sin θ \cos 2θ = \sin(θ + 2θ) + \sin(θ - 2θ) = \sin 3θ + \sin(-θ) = \sin 3θ - \sin θ$$

**Step 5: Final calculation**
$$= \frac{1}{4}(\sin 3θ - \sin θ + \sin θ) = \frac{1}{4}\sin 3θ$$

**Final Answer:** $$\boxed{\frac{1}{4}\sin 3θ}$$ ✓

---

### Problem 18: Cosecant-Secant Identity ⭐
**Question:** Prove that:
$$\sqrt{3} \cosec 20° - \sec 20° = 4$$

**🎯 Problem Type:** Reciprocal function identity
**🧠 Key Insight:** Convert to sines and cosines

**Complete Solution:**

**Step 1: Convert to sine and cosine**
$$\sqrt{3} \cosec 20° - \sec 20° = \frac{\sqrt{3}}{\sin 20°} - \frac{1}{\cos 20°}$$

**Step 2: Find common denominator**
$$= \frac{\sqrt{3} \cos 20° - \sin 20°}{\sin 20° \cos 20°}$$

**Step 3: Recognize the numerator pattern**
Note that $\sqrt{3} = \cot 30° = \frac{\cos 30°}{\sin 30°}$

$$\sqrt{3} \cos 20° - \sin 20° = \frac{\cos 30°}{\sin 30°} \cos 20° - \sin 20°$$
$$= \frac{\cos 30° \cos 20° - \sin 30° \sin 20°}{\sin 30°}$$

**Step 4: Apply cosine addition formula**
$\cos(A + B) = \cos A \cos B - \sin A \sin B$

$$= \frac{\cos(30° + 20°)}{\sin 30°} = \frac{\cos 50°}{\frac{1}{2}} = 2\cos 50°$$

**Step 5: Substitute back**
$$= \frac{2\cos 50°}{\sin 20° \cos 20°} = \frac{2\cos 50°}{\frac{1}{2}\sin 40°} = \frac{4\cos 50°}{\sin 40°}$$

**Step 6: Use complementary angle**
$\sin 40° = \sin(90° - 50°) = \cos 50°$

$$= \frac{4\cos 50°}{\cos 50°} = 4$$

**Final Answer:** $$\boxed{4}$$ ✓

---

### Problem 19: Fourth Power Sine Identity 💫
**Question:** Prove that:
$$\sin^4 x = \frac{1}{8}(3 - 4\cos 2x + \cos 4x)$$

**🎯 Problem Type:** Fourth power reduction
**🧠 Key Insight:** Use power reduction formulas repeatedly

**Complete Solution:**

**Step 1: Use power reduction formula**
$\sin^2 x = \frac{1 - \cos 2x}{2}$

$$\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2$$

**Step 2: Expand the square**
$$= \frac{(1 - \cos 2x)^2}{4} = \frac{1 - 2\cos 2x + \cos^2 2x}{4}$$

**Step 3: Apply power reduction to $\cos^2 2x$**
$\cos^2 2x = \frac{1 + \cos 4x}{2}$

$$= \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4}$$

**Step 4: Simplify**
$$= \frac{1 - 2\cos 2x + \frac{1}{2} + \frac{\cos 4x}{2}}{4}$$
$$= \frac{\frac{3}{2} - 2\cos 2x + \frac{\cos 4x}{2}}{4}$$

**Step 5: Final simplification**
$$= \frac{1}{4} \cdot \frac{3 - 4\cos 2x + \cos 4x}{2} = \frac{1}{8}(3 - 4\cos 2x + \cos 4x)$$

**Final Answer:** $$\boxed{\frac{1}{8}(3 - 4\cos 2x + \cos 4x)}$$ ✓

---

### Problem 20: Sum-to-Product Application 🌟
**Question:** Prove that:
$$2\cos 70° \cdot \cos 20° = \cos 50°$$

**🎯 Problem Type:** Sum-to-product identity
**🧠 Key Insight:** Direct application of product-to-sum formula

**Complete Solution:**

**Step 1: Apply product-to-sum formula**
$2\cos A \cos B = \cos(A + B) + \cos(A - B)$

$$2\cos 70° \cos 20° = \cos(70° + 20°) + \cos(70° - 20°)$$

**Step 2: Simplify angles**
$$= \cos 90° + \cos 50°$$

**Step 3: Use known value**
$\cos 90° = 0$

$$= 0 + \cos 50° = \cos 50°$$

**Final Answer:** $$\boxed{\cos 50°}$$ ✓

---

### Problem 21: Multiple Angle Chain ⚡
**Question:** Find relationships between:
$$\sin^2 α - \sin^2 β$$ and $$\sin α \cos α - \sin β \cos β$$

**🎯 Problem Type:** Relationship derivation
**🧠 Key Insight:** Use double angle identities

**Complete Solution:**

**Step 1: Express using double angle formulas**
Numerator: $\sin^2 α - \sin^2 β$
Denominator: $\sin α \cos α - \sin β \cos β = \frac{1}{2}\sin 2α - \frac{1}{2}\sin 2β = \frac{1}{2}(\sin 2α - \sin 2β)$

**Step 2: Use power reduction**
$\sin^2 θ = \frac{1 - \cos 2θ}{2}$

$$\sin^2 α - \sin^2 β = \frac{1 - \cos 2α}{2} - \frac{1 - \cos 2β}{2} = \frac{\cos 2β - \cos 2α}{2}$$

**Step 3: Form the ratio**
$$\frac{\sin^2 α - \sin^2 β}{\sin α \cos α - \sin β \cos β} = \frac{\frac{\cos 2β - \cos 2α}{2}}{\frac{\sin 2α - \sin 2β}{2}} = \frac{\cos 2β - \cos 2α}{\sin 2α - \sin 2β}$$

**Step 4: Apply sum-to-product formulas**
$\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$
$\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

$$= \frac{-2\sin\frac{2β+2α}{2}\sin\frac{2β-2α}{2}}{2\cos\frac{2α+2β}{2}\sin\frac{2α-2β}{2}}$$

**Step 5: Simplify**
$$= \frac{-2\sin(α+β)\sin(β-α)}{2\cos(α+β)\sin(α-β)} = \frac{-\sin(α+β)\sin(β-α)}{\cos(α+β)\sin(α-β)}$$

Since $\sin(β-α) = -\sin(α-β)$:
$$= \frac{\sin(α+β)}{\cos(α+β)} = \tan(α + β)$$

**Final Answer:** $$\boxed{\tan(α + β)}$$ ✓

---

---

## ❌ Common Student Mistakes (Avoid These!)

### 🚫 **Mistake #1: Forgetting Angle Quadrants**
**The Error:** "sin 120° = sin 60°" (ignoring quadrant)
**The Fix:** Remember quadrant rules and reference angle relationships

### 🚫 **Mistake #2: Misapplying Product Formulas**
**The Error:** Confusing product-to-sum with sum-to-product formulas
**The Fix:** Create a reference sheet and practice systematically

### 🚫 **Mistake #3: Algebraic Manipulation Errors**
**The Error:** Incorrect expansion or factoring in complex expressions
**The Fix:** Work step-by-step and verify each algebraic step

### 🚫 **Mistake #4: Special Angle Memory Errors**
**The Error:** Confusing sin 30° with cos 30° values
**The Fix:** Use consistent memorization techniques and regular practice

---

## 🎓 Examination Success Strategies

### ⏱️ **Time Management for Complex Problems**
1. **Quick Scan** (30 seconds): Identify problem type and required techniques
2. **Strategy Selection** (1 minute): Choose the most efficient approach
3. **Systematic Calculation** (5-8 minutes): Apply chosen method step-by-step
4. **Verification** (1 minute): Check result using alternative method

### 📝 **Problem Type Distribution**
- **Direct Calculations**: 20% (like Problem 1, 9, 13, 15)
- **Identity Proofs**: 60% (majority of problems)
- **Conditional Problems**: 20% (Problems 3, 4, 5)

### 🎯 **Master Strategy Checklist**
- ✅ All special angle values memorized
- ✅ Product-to-sum formulas mastered
- ✅ Double and triple angle formulas ready
- ✅ Complementary angle relationships automatic
- ✅ Algebraic manipulation skills sharp

---

**Remember:** This collection represents authentic examination-level problems that test deep understanding of trigonometric relationships. Master these 26+ problems, and you'll have comprehensive coverage of trigonometric value calculations and identity proofs.

**Final Encouragement:** Every complex identity breaks down into fundamental relationships. With systematic practice of these authentic problems, you'll develop the pattern recognition and algebraic skills needed for advanced trigonometry success!
