Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise Textbook Exercise Questions and Answers.
Rajasthan Board RBSE Solutions for Class 11 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 11. Students can also read RBSE Class 11 Maths Important Questions for exam preparation. Students can also go through RBSE Class 11 Maths Notes to understand and remember the concepts easily.
Question 1.
Prove that 2cosπ13cos9π13 + cos3π13 + cos5π13 = 0.
Answer:
L.H.S = 2cosπ13cos9π13 + cos3π13 + cos5π13
Question 2.
Prove that:
(sin 3x + sin x)sin x + (cos 3x - cos x) cos x = 0
Answer:
L.H.S. = (sin 3x + sin x) sin x +(cos 3x - cos x) cos x
= sin 3x sin x + sin2 x + cos3x cosx - cos2 x
= cos 3x cos x + sin 3x sin x + sin2x - cos2 x
= cos(3x - x) - (cos2 x - sin2 x)
=cos2x - cos2x
(Using formula cos 2A = cos2A - sin2A)
= O = R.H.S. Hence Proved
Question 3.
Prove that:
(cos x + cos y)2 + (sin x - sin y)2 = 4cos2x+y2
L,H.S. = (cos x + cos y)2 +(sin x - sin y)2
= cos2 x + cos2 y + 2cosxcosy + sin2x + sin2y - 2sinxsiny
= (cos2x + sin2x) + (cos2y + sin2y) + 2cos x cos y - 2sinx siny
= 1 + 1 + 2(cos x cosy - sin x sin y) {∵ cos2θ + sin2θ = 1)
= 2 + 2cos (x + y)
= 2[1 + cos(x + y)]
= 2[1 + 2cos2(x+y2) - 1]
[Using formula cos x = 2 cos2x2 - 1]
= 2 × 2 cos2x+y2
= 4 cos2x+y2
= R.H.S
Hence Proved
Question 4.
Prove that
(cos x - cos y)2 + (sin x - sin y)2 = 4sin2
Answer:
L.H.S. = (cos x - cos y)2 ± (sin x - sin y)2
= cos x + cos y - 2cosx cos y + sin2x + sin2y - 2sinx siny
=(cos2x + sin2x) + (cos2y + sin2y) - 2cosx cosy - 2sinx sin y
= 1 + 1 - 2[cosx cosv+ sin x sin y] {From cos2θ + sin2θ = 1)
= 2 - 2 [cos (x - y)]
= 2[1 - cos(x - y)]
Question 5.
Prove that
sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
Answer:
L.H.S = sin x + sin 3x + sin 5x + sin 7x
= (sin 7x + sin x) + (sin 5x + sin 3x)
= 2sin7x+x2cos7x−x2 + 2sin5x+3x2cos5x−3x2
[Using formula sin A + sin B = 2 sinA+B2cosA−B2]
= 2sin8x2cos6x2 + 2sin8x2cos2x2
= 2sin 4x.cos 3x + 2sin 4x cos x
= 2sin 4x[cos 3x + cos x]
= 2sin 4x[2 cos3x+x2cos3x+x2]
[Using formula cosx + cosy = 2cosx+y2cosx−y2]
= 2sin 4x[2 cos4x2cos2x2]
= 4sin 4x cos 2x cos x
= 4 cos x cos 2x sin 4x
= R.H.S
Hence Proved.
Question 6.
Prove that:
(sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x) = tan 6x
Answer:
Question 7.
Prove that:
sin 3x + sin 2x - sin x = 4sin x cos x2 cos 3x2
Answer:
L.H.S = sin 3x + sin 2x - sin x
Find sinx2, cosx2 and tanx2 in each of the following question.
Question 8.
tan x = -43, x is second quadrant.
Answer:
tan x = -43, x lies second quadrant.
90° < x < 180° ⇒ 45° < x2 < 90°
Thus, x2 will lies in first quadrant so, all trigonometric ratios of x2 will be positive.
Thus required value of tanx2 will be 2, since x2 will lie in first quadrant in which all trigonometric ratios are positive so tan x2=−12 is negligible.
Required value of = 2
Thus, sin x = 42 which is positive since. It is given that x lies in second quadrant in which sin x is positive.
(Since, x lies in second quadrant in which cos x is negative)
Only positive value will be considered since x2 lies in first quadrant.
Thus, sin x2=2√5, cos x2=1√5 and tan x2 = 2 are required values.
Question 9.
cos x = -13, x in third quadrant.
Answer:
cos x = -13
180° < x < 270°
[∵ given that x lies in third quadrant. In this will be positive.]
Thus, x2 will be in second quadrant is negligible since x2 lies in second quadrant so cos x2 will be negative.
[Since x2 lies in second quadrant so sin x2 will be positive]
Since, x lies in third quadrant so value of x lies between π and 3π2.
i.e., π < x < 3π2
Then, value of x2 will be between π2 and 3π2.
i.e π2< x2 <3π4
Thus, will lies in second quadrant so that sin will be +ve and cosx2. and tan x2 will be negative.
then sin x2=√63, and cos x2=−√33
and tanx2 = -√2
Thus,sin x2=√63, cos x2=−√33 and tan x2 = -√2
Question 10.
sin x = 14, x in second quadrant.
Answer:
sin x = 14, [Using formula sin2x + cos2x = 1]
Since x2 will lie in first quadrant.
Since, x lie is second quadrant.
Thus π2 < x < π
π4<x2<π2
x2 will lie in first quadrant then all trigonometric ratio will be always positive.
Thus, sinx2=√8+2√154, cosx2=√8−2√154 and tanx2=4+√15 are required values.