NCERT Solutions Class 11 Maths Chapter 3 Miscellaneous Exercise

🚀 Level Up: The Final Challenge

The Miscellaneous Exercise is where the real game begins! These questions test your ability to mix and match formulas creatively.

🧠 Arsenal of Formulas (Cheat Sheet):

• The "CD" Formulas (Your Best Friends):

- \(\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\)

- \(\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}\)

- \(\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}\) (⚠️ Watch the minus!)

• Product to Sum:

- \(2 \cos A \cos B = \cos(A+B) + \cos(A-B)\)

- \(2 \sin A \sin B = \cos(A-B) - \cos(A+B)\)

• Half-Angle Formulas (Crucial for Q8-Q10):

- \(\sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}}\)

- \(\cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}}\)

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🔥 Detailed Solutions

🧩 Question 1: Prove that \(2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0\).

💡 Solution:

We start with the first term. Use the product-to-sum formula: \(2 \cos A \cos B = \cos(A+B) + \cos(A-B)\).

Let \(A = \frac{9\pi}{13}\) and \(B = \frac{\pi}{13}\).

LHS:

\[ \left[ \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} \right] + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} \]

Now, observe the angles. Note that \(\frac{10\pi}{13} = \pi - \frac{3\pi}{13}\) and \(\frac{8\pi}{13} = \pi - \frac{5\pi}{13}\).

We know \(\cos(\pi - \theta) = -\cos \theta\).

\[ = \left( -\cos \frac{3\pi}{13} \right) + \left( -\cos \frac{5\pi}{13} \right) + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} \]

\[ = 0 \]

LHS = RHS. Proved. ✅

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🧩 Question 3: Prove that \((\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2 \frac{x+y}{2}\).

💡 Solution:

Expand the squares using \((a+b)^2\) and \((a-b)^2\).

LHS:

\[ (\cos^2 x + \cos^2 y + 2\cos x \cos y) + (\sin^2 x + \sin^2 y - 2\sin x \sin y) \]

Group the terms: \((\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)\)

\[ = 1 + 1 + 2 \cos(x+y) \]

\[ = 2 + 2 \cos(x+y) \]

\[ = 2 [ 1 + \cos(x+y) ] \]

Use Half-Angle Formula: \(1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}\).

\[ = 2 \left[ 2 \cos^2 \frac{x+y}{2} \right] \]

\[ = 4 \cos^2 \frac{x+y}{2} \]

LHS = RHS. Proved. ✅

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🧩 Question 6: Prove \(\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x\).

💡 Solution:

Apply the CD formulas to the pairs in brackets.

Numerator (Sin C + Sin D):

\[ 2 \sin 6x \cos x + 2 \sin 6x \cos 3x \]

Factor out \(2 \sin 6x\):

\[ = 2 \sin 6x (\cos x + \cos 3x) \]

Denominator (Cos C + Cos D):

\[ 2 \cos 6x \cos x + 2 \cos 6x \cos 3x \]

Factor out \(2 \cos 6x\):

\[ = 2 \cos 6x (\cos x + \cos 3x) \]

Final Fraction:

\[ \frac{2 \sin 6x (\cos x + \cos 3x)}{2 \cos 6x (\cos x + \cos 3x)} \]

Cancel common terms:

\[ = \frac{\sin 6x}{\cos 6x} = \tan 6x \]

LHS = RHS. Proved. ✅

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🧩 Question 8: Find \(\sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}\) if \(\tan x = -\frac{4}{3}\), \(x\) in quadrant II.

💡 Solution:

Step 1: Determine the Quadrant of \(x/2\).

Since \(x\) is in Quadrant II:

\[ \frac{\pi}{2} < x < \pi \]

Divide by 2:

\[ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \]

So, \(\frac{x}{2}\) lies in Quadrant I. (All values will be positive! 🌟)

Step 2: Find \(\cos x\).

Given \(\tan x = -\frac{4}{3}\).

Using \(\sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9}\).

So, \(\sec x = -\frac{5}{3}\) (Negative in QII).

\(\implies \cos x = -\frac{3}{5}\).

Step 3: Apply Half-Angle Formulas.

\(\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}\)

\[ = \sqrt{\frac{1 - (-3/5)}{2}} = \sqrt{\frac{1 + 3/5}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]

\(\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}}\)

\[ = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

\(\tan \frac{x}{2} = \frac{\sin x/2}{\cos x/2}\)

\[ = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2 \]

Answer:

\[ \sin \frac{x}{2} = \frac{2\sqrt{5}}{5}, \cos \frac{x}{2} = \frac{\sqrt{5}}{5}, \tan \frac{x}{2} = 2 \]

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🧩 Question 9: Find \(\sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}\) if \(\cos x = -\frac{1}{3}\), \(x\) in quadrant III.

💡 Solution:

Step 1: Determine Quadrant of \(x/2\).

Since \(x\) is in Quadrant III:

\[ \pi < x < \frac{3\pi}{2} \]

Divide by 2:

\[ \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4} \]

So, \(\frac{x}{2}\) lies in Quadrant II.

(Signs: Sin is (+), Cos is (-), Tan is (-))

Step 2: Apply Half-Angle Formulas.

Given \(\cos x = -\frac{1}{3}\).

\(\sin \frac{x}{2} = \sqrt{\frac{1 - (-1/3)}{2}} = \sqrt{\frac{4/3}{2}} = \sqrt{\frac{2}{3}}\)

(Positive in QII)

\(\cos \frac{x}{2} = -\sqrt{\frac{1 + (-1/3)}{2}} = -\sqrt{\frac{2/3}{2}} = -\sqrt{\frac{1}{3}}\)

(Negative in QII)

\(\tan \frac{x}{2} = \frac{\sqrt{2}/\sqrt{3}}{-1/\sqrt{3}} = -\sqrt{2}\)

Answer:

\[ \sin \frac{x}{2} = \frac{\sqrt{6}}{3}, \cos \frac{x}{2} = -\frac{\sqrt{3}}{3}, \tan \frac{x}{2} = -\sqrt{2} \]

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🎓 Expert Tips for the Exam:

🛑 The "Half-Angle" Trap:

In Q8, Q9, and Q10, the most common mistake is getting the signs wrong. Always divide the inequality range of \(x\) by 2 to find exactly where \(\frac{x}{2}\) lands.

- If \(x\) is QII, \(x/2\) is QI.

- If \(x\) is QIII, \(x/2\) is QII.

🚀 Grouping Strategy:

In questions like Q6 (Sin 7x + Sin 5x + ...), always pair the angles such that their average or difference creates a common factor.

Example: Pair \(7x\) with \(5x\) (average \(6x\)) and \(9x\) with \(3x\) (average \(6x\)).