NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

Get Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise PDF in Hindi and English Medium. Trigonometric Functions Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Trigonometric Functions Miscellaneous Exercise Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 3 Class 11 Trigonometric Functions Miscellaneous Exercise provided in NCERT Textbook.

• Trigonometric Functions Class 11 Ex 3.1
• Trigonometric Functions Class 11 Ex 3.2
• Trigonometric Functions Class 11 Ex 3.3
• Trigonometric Functions Class 11 Ex 3.4
• Trigonometric Functions Class 11 Miscellaneous Exercise
• त्रिकोणमितीय फलन प्रश्नावली 3.1 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.2 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.3 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.4 का हल हिंदी में
• त्रिकोणमितीय फलन विविध प्रश्नावली का हल हिंदी में
• Trigonometry Formulas
• Trigonometry Functions Class 11 Notes
• NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions
• Trig Cheat Sheet
• JEE Main Trigonometry Previous Year Questions

Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

Miscellaneous Exercise Class 11 Maths 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
Ans:

Hence proved.
Miscellaneous Exercise Class 11 Maths Question 2:
Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0.
Ans:

L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= sin 3x sin x + sin ² x + cos 3x cos x – cos ² x
= cos 3x cos x + sin 3x sin x – (cos ² x – sin ² x)
= cos (3x – x) – cos 2x
[∵ cos(A – B) = cos A cos B + sin A sin B]
= cos 2x – cos 2x = 0
=R.H.S.
Hence proved.

Miscellaneous Exercise Class 11 Maths Question 3:

Prove that:
(cos x + cos y) ² + (sin x – sin y) ² = 4 cos ² \(\left(\frac{x+y}{2}\right)\)

Ans:

L.H.S.= (cos x + cos y) ² + (sin x – sin y) ²
= cos ² x + cos ² y + 2 cos x cos y + sin ² x + sin ² y – 2 sin x sin y
= (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)
[∵ cos (A + B) = (cos A cos B – sin A sin B)]
= 2 + 2 cos (x + y)
= 2 [1 + cos (x + y)]
= 2[1 + \(2 \cos ^{2}\left(\frac{x+y}{2}\right)\) – 1]
[∵ cos 2A = 2 cos ² A – 1]
= 4 c0s ² \(\left(\frac{x+y}{2}\right)\)
= R.H.S.
Hence proved.

Miscellaneous Exercise Class 11 Maths Question 4:
Prove that:
(cos x – cos y) ² + (sin x – sin y) ² = 4 sin ² \(\frac{x-y}{2}\)
Ans:

L.H.S.= (cos x – cos y) ² + (sin x – sin y) ²
= cos ² x + cos ² y – 2 cos x cos y + sin ² x + sin ² y – 2 sin x sin y
= (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 [1 – {1 – 2 sin ² \(\left(\frac{x-y}{2}\right)\)}]
[∵ cos 2A = 1 – 2 sin ² A]
= 4 sin ² \(\left(\frac{x-y}{2}\right)\)
= R.H.S.
Hence proved.

Miscellaneous Exercise Class 11 Maths Question 5:
Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)\)
∴ L.H.S. = (sin x + sin 3x) + (sin 5x + sin 7x)
= (sin x + sin 5x) + (sin 3x + sin 7x)
= \(2 \sin \left(\frac{x+5 x}{2}\right)\) . \(\cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)\)
= 2 sin 3x cos (- 2x) + 2 sin 5x cos (- 2x)
= 2 sin 3x cos 2x + 2 sin 5x cos 2x
= 2 cos 2x [sin 3x + sin 5x]
= 2 cos 2x [latex]2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)[/latex]
= 2 cos 2x [2 sin 4x . cos (- x)]
= 4 cos 2x sin 4x cos x
= R.H.S.
Hence proved.

Miscellaneous Exercise Class 11 Maths Question 6:
Prove that: \(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\) = tan 6x
Ans:
It is known that

Hence proved.

Miscellaneous Exercise Class 11 Maths Question 7:
Prove that: sin 3x + sin 2x – sin x = 4 sin x cos \(\frac{x}{2}\) cos \(\frac{3 x}{2}\).
Ans:
L.H.S. = sin 3x + sin 2x – sin x
= sin 3x + (sin 2x – sin x)

Miscellaneous Exercise Class 11 Maths Question 8:
Find sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) for tan x = – \(\frac{4}{3}\), x in quadrant II.
Ans:

Miscellaneous Exercise Class 11 Maths Question 9:
Find sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) for cos x = – \(\frac{1}{3}\), x in quadrant III.
Ans:


Thus, the respective values of sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) are \(\frac{\sqrt{6}}{5}\), \(\frac{\sqrt{3}}{3}\) and – √2.

Miscellaneous Exercise Class 11 Maths Question 10:
Find sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) for sin x = \(\frac{1}{4}\), x in quadrant II.
Ans:

Thus, the respective values of sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) are \(\sqrt{\frac{8+2 \sqrt{15}}{4}}\), \(\sqrt{\frac{8-2 \sqrt{15}}{4}}\) and 4 + √15.

त्रिकोणमितीय फलन विविध प्रश्नावली का हल हिंदी मे

प्रश्न 2.
सिद्ध कीजिए : (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0.
हल :
बायाँ पक्ष = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= sin 3x sin x + sin2x + cos 3x cos x – cos²x
= (cos 3x cos x + sin 3x sin x) – (cos² x – sin² x)
= cos 2x – cos 2x [cos (A – B) = cos A cos B + sin A sin B]
= 0
= दायाँ पक्ष।

निम्नलिखित प्रत्येक प्रश्न में sin \(\frac { x }{ 2 }\) , cos \(\frac { x }{ 2 }\) और tan \(\frac { x }{ 2 }\) ज्ञात कीजिए।

प्रश्न 8.
tan x = \(\frac { -4 }{ 3 }\) , x द्वितीय चतुर्थाश में हैं।

प्रश्न 9.
cosx = \(\frac { -1 }{ 2 }\) , x तृतीय चतुर्थाश में है।

प्रश्न 10.
sin x = \(\frac { 1 }{ 4 }\) , x द्वितीय चतुर्थाश में है।