Trigonometric Functions Miscellaneous Exercise: NCERT Class 11 Maths

Understanding the Miscellaneous Exercise

The Miscellaneous Exercise in Chapter 3, "Trigonometric Functions," is unique because it doesn't limit itself to a single concept or type of problem. Instead, it acts as a comprehensive assessment, drawing upon all the identities, formulas, and principles covered throughout the chapter. This includes basic trigonometric ratios, compound angle formulas, multiple and submultiple angle formulas, sum and product formulas, and even the properties of inverse trigonometric functions (though predominantly focused on direct functions for Class 11).

Solving these problems requires a strong foundation in algebraic manipulation, a keen eye for pattern recognition, and the ability to choose the most efficient identity. Often, a problem can be solved in multiple ways, but some paths are significantly simpler than others. The goal is to develop a strategic approach rather than just memorising steps. This exercise prepares you not just for the chapter test, but also builds critical thinking for higher-level mathematics.

Essential Trigonometric Identities for Problem Solving

Compound Angle Formulas

These identities relate trigonometric functions of sums or differences of angles. Key formulas include $\sin(A \pm B)$, $\cos(A \pm B)$, and $\tan(A \pm B)$. For example, $\sin(A+B) = \sin A \cos B + \cos A \sin B$.

Multiple and Submultiple Angle Formulas

Formulas for $\sin 2A$, $\cos 2A$, $\tan 2A$, $\sin 3A$, $\cos 3A$, $\tan 3A$. For instance, $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$.

Sum and Product Formulas (Transformation Formulas)

These are used to convert sums/differences of sines/cosines into products, and vice versa. Examples: $\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$ and $2\sin A \cos B = \sin(A+B) + \sin(A-B)$.

Pythagorean Identities

Fundamental relations: $\sin^2 x + \cos^2 x = 1$, $1 + \tan^2 x = \sec^2 x$, $1 + \cot^2 x = \csc^2 x$. These are often used to simplify expressions.

Strategic Approach to Miscellaneous Problems

1. Understand the Goal — Before starting, clearly identify what you need to prove or solve. Is it an identity (LHS = RHS)? Or is it to find the value of an expression? This helps in choosing the right path.
2. Simplify to Basic Ratios (if complex) — If the expression involves $\tan, \cot, \sec, \csc$, often converting everything to $\sin$ and $\cos$ can reveal simplification opportunities. For example, replace $\tan x$ with $\frac{\sin x}{\cos x}$.
3. Look for Known Identity Patterns — Scan the expression for parts that resemble standard identities. Do you see $\sin C + \sin D$, $1 - \cos 2x$, $\sin 2x$, or compound angles? Immediately apply the relevant formula.
4. Work from Both Sides (for Identities) — If proving an identity, sometimes it's easier to simplify both the Left Hand Side (LHS) and Right Hand Side (RHS) independently until they become equal, rather than trying to transform one directly into the other.
5. Factorisation and Algebraic Manipulation — Don't forget basic algebra. Common factoring, taking LCM, or rationalising denominators can often lead to the desired form after applying trigonometric identities.
6. Check Your Steps — Trigonometry problems can be lengthy. Review each step to avoid calculation errors or incorrect application of identities. Ensure domain and range considerations (though less frequent in Class 11 miscellaneous). If stuck, try an alternative identity or strategy.

Fully Worked Examples

• Example 1: Prove that $\left(\cos x + \cos y\right)^2 + \left(\sin x - \sin y\right)^2 = 4\cos^2\left(\frac{x+y}{2}\right)$ Step 1: Expand the squares on the LHS. $\text{LHS} = (\cos^2 x + 2\cos x \cos y + \cos^2 y) + (\sin^2 x - 2\sin x \sin y + \sin^2 y)$ Step 2: Group terms using $\sin^2 \theta + \cos^2 \theta = 1$. $\text{LHS} = (\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\cos x \cos y - \sin x \sin y)$ $\text{LHS} = 1 + 1 + 2(\cos x \cos y - \sin x \sin y)$ Step 3: Apply the compound angle formula for $\cos(A+B)$. We know that $\cos(A+B) = \cos A \cos B - \sin A \sin B$. So, $\text{LHS} = 2 + 2\cos(x+y)$ Step 4: Use the double angle formula for $\cos 2A = 2\cos^2 A - 1$. Let $A = \frac{x+y}{2}$. Then $2A = x+y$. So, $\cos(x+y) = 2\cos^2\left(\frac{x+y}{2}\right) - 1$. Step 5: Substitute and simplify. $\text{LHS} = 2 + 2\left[2\cos^2\left(\frac{x+y}{2}\right) - 1\right]$ $\text{LHS} = 2 + 4\cos^2\left(\frac{x+y}{2}\right) - 2$ $\text{LHS} = 4\cos^2\left(\frac{x+y}{2}\right)$ Final Answer: Thus, $\text{LHS} = \text{RHS}$, and the identity is proven. Example 2: Prove that $\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x - \sin 3x)$ Step 1: Simplify LHS using sum-to-product formula. Recall $\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$. $\text{LHS} = \cot 4x (2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right))$ $\text{LHS} = \cot 4x (2\sin 4x \cos x)$ Step 2: Convert $\cot 4x$ to $\frac{\cos 4x}{\sin 4x}$. $\text{LHS} = \frac{\cos 4x}{\sin 4x} (2\sin 4x \cos x)$ $\text{LHS} = 2\cos 4x \cos x$ Step 3: Simplify RHS using sum-to-product formula. Recall $\sin C - \sin D = 2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$. $\text{RHS} = \cot x (2\cos\left(\frac{5x+3x}{2}\right)\sin\left(\frac{5x-3x}{2}\right))$ $\text{RHS} = \cot x (2\cos 4x \sin x)$ Step 4: Convert $\cot x$ to $\frac{\cos x}{\sin x}$. $\text{RHS} = \frac{\cos x}{\sin x} (2\cos 4x \sin x)$ $\text{RHS} = 2\cos x \cos 4x$ Final Answer: Since $\text{LHS} = 2\cos 4x \cos x$ and $\text{RHS} = 2\cos x \cos 4x$, we have $\text{LHS} = \text{RHS}$, and the identity is proven.

Exam Tips for Trigonometric Miscellaneous Problems

These problems are often high-scoring questions, so mastering them is crucial. Firstly, memorise all identities thoroughly; a slight error in a formula can derail the entire solution. Practice derivation of identities to understand their origin, which helps in recall. Secondly, always start by writing down the LHS and RHS separately if you are proving an identity. Don't try to manipulate the entire equation at once, which can lead to confusion. Thirdly, be patient with algebraic steps. Common mistakes include sign errors, incorrect factoring, or misapplying square root properties. Finally, if a problem seems too difficult, try working backward from the RHS or simplify both sides simultaneously to meet in the middle. Time management is also key; these problems can be lengthy, so allocate sufficient time and don't get stuck on one problem for too long.

Practice Questions with Solutions

• Q: Prove that $\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$. A: Step 1: Start with $\cos 6x = \cos(2 \cdot 3x)$ or $\cos(3 \cdot 2x)$. Let's use $\cos 6x = \cos(3 \cdot 2x)$. Step 2: Apply the triple angle formula $\cos 3A = 4\cos^3 A - 3\cos A$, where $A=2x$. $\cos 6x = 4\cos^3(2x) - 3\cos(2x)$ Step 3: Substitute $\cos 2x = 2\cos^2 x - 1$. $\cos 6x = 4(2\cos^2 x - 1)^3 - 3(2\cos^2 x - 1)$ Step 4: Expand $(2\cos^2 x - 1)^3$ using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. $(2\cos^2 x - 1)^3 = (2\cos^2 x)^3 - 3(2\cos^2 x)^2(1) + 3(2\cos^2 x)(1)^2 - 1^3$ $= 8\cos^6 x - 3(4\cos^4 x) + 6\cos^2 x - 1$ $= 8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1$ Step 5: Substitute this back into the expression for $\cos 6x$ and simplify. $\cos 6x = 4(8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1) - 3(2\cos^2 x - 1)$ $\cos 6x = 32\cos^6 x - 48\cos^4 x + 24\cos^2 x - 4 - 6\cos^2 x + 3$ $\cos 6x = 32\cos^6 x - 48\cos^4 x + (24-6)\cos^2 x + (-4+3)$ Final answer: $\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$. Hence proved.
• Q: Find the value of $\tan(\frac{\pi}{8})$. A: Step 1: Use the half-angle identity for $\tan$. We know $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$. Let $2\theta = \frac{\pi}{4}$, so $\theta = \frac{\pi}{8}$. Then $\tan(\frac{\pi}{4}) = \frac{2\tan(\frac{\pi}{8})}{1-\tan^2(\frac{\pi}{8})}$. Step 2: Let $y = \tan(\frac{\pi}{8})$. We know $\tan(\frac{\pi}{4}) = 1$. $1 = \frac{2y}{1-y^2}$ Step 3: Rearrange into a quadratic equation. $1-y^2 = 2y \implies y^2 + 2y - 1 = 0$ Step 4: Solve the quadratic equation for $y$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4+4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2}$ $y = -1 \pm \sqrt{2}$ Step 5: Determine the correct sign. Since $\frac{\pi}{8}$ is in the first quadrant ($0 < \frac{\pi}{8} < \frac{\pi}{2}$), $\tan(\frac{\pi}{8})$ must be positive. Final answer: $\tan(\frac{\pi}{8}) = \sqrt{2} - 1$.
• Q: Prove that $(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$. A: Step 1: Expand the terms on the LHS. $\text{LHS} = \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$ Step 2: Rearrange terms to group product-to-sum possibilities. $\text{LHS} = (\cos 3x \cos x + \sin 3x \sin x) + (\sin^2 x - \cos^2 x)$ Step 3: Apply compound angle identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$. $\cos 3x \cos x + \sin 3x \sin x = \cos(3x - x) = \cos 2x$ Step 4: Apply double angle identity $\cos 2x = \cos^2 x - \sin^2 x = -(\sin^2 x - \cos^2 x)$. So, $(\sin^2 x - \cos^2 x) = -\cos 2x$. Step 5: Substitute back into LHS. $\text{LHS} = \cos 2x + (-\cos 2x)$ Final answer: $\text{LHS} = 0$. Hence proved.
• Q: Prove that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$. A: Step 1: Group terms on the LHS to apply sum-to-product formulas. Group $\cos 4x + \cos 2x$ and $\sin 4x + \sin 2x$. $\text{LHS} = \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}$ Step 2: Apply $\cos C + \cos D = 2\cos(\frac{C+D}{2})\cos(\frac{C-D}{2})$ and $\sin C + \sin D = 2\sin(\frac{C+D}{2})\cos(\frac{C-D}{2})$. For numerator: $\cos 4x + \cos 2x = 2\cos(\frac{4x+2x}{2})\cos(\frac{4x-2x}{2}) = 2\cos 3x \cos x$ For denominator: $\sin 4x + \sin 2x = 2\sin(\frac{4x+2x}{2})\cos(\frac{4x-2x}{2}) = 2\sin 3x \cos x$ Step 3: Substitute these back into the LHS. $\text{LHS} = \frac{2\cos 3x \cos x + \cos 3x}{2\sin 3x \cos x + \sin 3x}$ Step 4: Factor out common terms $\cos 3x$ from the numerator and $\sin 3x$ from the denominator. $\text{LHS} = \frac{\cos 3x (2\cos x + 1)}{\sin 3x (2\cos x + 1)}$ Step 5: Cancel out the common factor $(2\cos x + 1)$. Final answer: $\text{LHS} = \frac{\cos 3x}{\sin 3x} = \cot 3x$. Hence proved.

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