Let $n$ be an odd integer. If $\sin \,\,n{\rm{\theta }} = {\rm{\;}}\mathop \sum \limits_{r = 0}^n \,\,{b_r}\,\,{\sin ^r}\,\,{\rm{\theta \;}}$ for all real ${\rm{\theta }}$, then

${b_0} = 0,\,{b_1} = {n^2} - 3n - 3$

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JEE • MATHEMATICS • MULTIPLE AND SUBMULTIPLE ANGLES

Multiple and submultiple angles practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 4

Analyzing +4 | -1

For

x \in R

\tan \,\,x + \frac{1}{2}\,\,\tan \,\,\frac{x}{2}\,\, + \,\,\frac{1}{{{2^2}}}\,\,\tan \,\,\frac{x}{{{2^2}}} + \ldots + \frac{1}{{{2^{n - 1}}}}\,\,\tan \,\,\left( {\frac{x}{{{2^{n - 1}}}}} \right)

is equal to

View answer and explanation

Correct option: B

\frac{1}{{{2^{n - 1}}}}\,\,\cot \,\,\left( {\frac{x}{{{2^{n - 1}}}}} \right) - 2\cot \,\,2x

Option B is correct because by using the identity

\tan x = \cot x - 2\cot 2x

repeatedly, we can express the given sum as a telescoping series, which simplifies to the given expression.

2 Hard

2 of 4

Understanding +4 | -1

Let

n

be an odd integer. If

\sin \,\,n\mathrm{\theta } = \mathrm{\;}\sum \limits_{r = 0}^n \,\,{b_r}\,\,{\sin ^r}\,\,\mathrm{\theta \;}

for all real

\mathrm{\theta }

, then

View answer and explanation

Correct option: B

{b_{0 = }}0,\,\;{b_1} = n

Option B (

{b_{0 = }}0,\;{b_1} = n

) is correct because when

\mathrm{\theta } = 0

\sin n\mathrm{\theta } = 0

and the derivative at

\mathrm{\theta } = 0

gives

{b_1} = n

3 Hard

3 of 4

Analyzing +4 | -1

\cos \,\,\frac{x}{2}\,\,.\,\,\cos \,\,\frac{x}{{{2^2}}} \ldots ..\cos \,\,\frac{x}{{{2^n}}} = \frac{{\sin \,\,x}}{{{2^n}\,\,\sin \;\,\,\frac{x}{{{2^n}}}}}

, then

\frac{1}{2}\,\,\tan \,\,\frac{x}{2} + \frac{1}{{{2^2}}}\,\,\tan \,\,\frac{x}{{{2^2}}} + ... + \frac{1}{{{2^n}}}\,\,\tan \,\,\frac{x}{{{2^n}}}

View answer and explanation

Correct option: B

\frac{1}{{{2^n}}}\,\,\cot \,\,\left( {\frac{x}{{{2^n}}}} \right) - \cot \,\,x

Option B (1/2^n cot(x/2^n) - cot x) is correct because by repeatedly using the identity tan x/2 = cot x/2 - 2cot x, the expression simplifies to the given form.

4 Hard

4 of 4

Analyzing +4 | -1

The maximum value of

(\cos \,\,{\alpha _1})\;(\cos \,\,{\alpha _2})\,\, \ldots \,\,(\cos \,\,{\alpha _n})\;\;

under the restriction

0\,\, \le \,\,{\alpha _{1,}}{\alpha _2}, \ldots ,{\alpha _n}\,\, \le \,\,\frac{\pi }{2}

and

(\cos \,\,{\alpha _1})\,(\cos \,\,{\alpha _2}) \ldots (\cos \,\,{\alpha _n}) = 1\;\;

, is

View answer and explanation

Correct option: A

\frac{1}{{{2^{n/2}}}}

Option A (

\frac{1}{{{2^{n/2}}}}

) is correct because the maximum value is achieved when all angles are equal, leading to the given result using the AM-GM inequality or similar arguments.

Revision notes and key takeaways

Review MULTIPLE AND SUBMULTIPLE ANGLES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 4 single-correct-answer MCQs on MULTIPLE AND SUBMULTIPLE ANGLES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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