NCERT Solutions Class 11 Mathematics Trigonometric-Functions Exercise 3-Misc

Question 5:
Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 * cos x * cos 2x * sin 4x

Answer:
LHS:
     sin x + sin 3x + sin 5x + sin 7x
 = 2 * sin(x + 3x)/2 * cos(x – 3x)/2 + 2 * sin(5x + 7x)/2 * cos(5x – 7x)/2
 = 2 * sin(4x/2) * cos(-2x/2) + 2 * sin(12x/2) * cos(-2x/2)
 = 2 * sin 2x * cos(-x) + 2 * sin 6x * cos(-x)
 = 2 * sin 2x * cos x + 2 * sin 6x * cos x
 = 2 * cos x * (sin 2x + sin 6x)
 = 2 * cos x * {sin(2x + 6x)/2 * cos(2x – 6x)/2}
 = 2 * cos x * {2 * sin(8x/2) * cos(-4x/2)}
 = 2 * cos x * {2 * sin 4x * cos(-2x)}
 = 4 * cos x * sin 4x * cos 2x
 = 4 * cos x * cos 2x * sin 4x
= RHS
 
 
 

Question 6:
Prove that: {(sin 7x + sin 5x) + (sin 9x + sin 3x)}/{(cos 7x + cos 5x) + (cos 9x + cos 3x)} = tan 6x

Answer:
We know that
sin A + sin B = 2sin (A + B)/2  * cos (A – B)/2
and cos A + cos B = 2cos (A + B)/2  * cos (A – B)/2
LHS:
    {(sin 7x + sin 5x) + (sin 9x + sin 3x)}/{(cos 7x + cos 5x) + (cos 9x + cos 3x)}
= [2sin (7x + 5x)/2 * cos (7x – 5x)/2 + 2sin (9x + 3x)/2 * cos (9x – 3x)/2]/[ 2cos (7x + 5x)/2 * cos
(7x – 5x)/2 + 2cos (9x + 3x)/2 * cos (9x – 3x)/2]
= [2 * sin 6x * cos x + 2 * sin 6x * cos 3x]/ [2 * cos 6x * cos x + 2 * cos 6x * cos 3x]
= [2 * sin 6x(cos x + cos 3x)]/ [2 * sin 6x(cos x + cos 3x)]
= tan 6x
= RHS

Question 7:
Prove that: sin 3x + sin 2x – sin x = 4 * sin x * cos(x/2) * cos(3x/2)

Answer:
LHS:
   sin 3x + sin 2x – sin x
= sin 3x – sin x + sin 2x
= 2 * cos(3x + x)/2 * sin(3x – x)/2 + sin 2x
= 2 * cos 2x * sin x + 2 * sin x * cos x                      (sin 2x = 2 * sin x * cos x)
= 2 *sin x * (cos 2x + cos x)
= 2 * sin x * 2 * cos(2x + x)/2 *cos(2x – x)/2
= 4 * sin x * cos(3x/2) * cos(x/2)
= RHS

Question 8:
Find sin x/2, cos x/2 and tan x/2, if tan x = −4/3, x in quadrant II

Answer:
Here, x is in quadrant II.
i.e π/2 < x < π
⇒ π/4 < x/2 < π/2
Therefore, sin x/2, cos x/2 and tan x/2 are lies in first quadrant.
It is given that
tan x = -4/3
Now, sec² x = 1 + tan² x = 1 + (-4/3)² = 1 + 16/9 = 25/9
So, cos² x = 9/25
⇒ cos x = ± 3/5
Since x is in quadrant II, cos x is negative.
So, cos x = -3/5
Now, cos x = 2cos² x/2 – 1
⇒ -3/5 = 2cos² x/2 – 1
⇒ 2cos² x/2 = 1 – 3/5
⇒ 2cos² x/2 = 2/5
⇒ cos² x/2 = 1/5
⇒ cos x = 1/√5             [Since cos x is positive]
Again sin² x/2 + cos² x/2 = 1
⇒ sin² x/2 + (1/√5)² = 1
⇒ sin² x/2 + 1/5 = 1
⇒ sin² x/2 = 1 – 1/5
⇒ sin² x/2 = 4/5
⇒ sin x/2 = 2/√5              [Since sin x/2 is positive]
tan x/2 = (sin x/2)/(cos x/2) = (2/√5)/(1/√5) = 2
Thus, the respective values of sin x/2, cos x/2 and tan x/2 are 2√5/5, √5/5 and 2.

Question 9:
Find sin x/2, cos x/2 and tan x/2 for cos x = −1/3, x in quadrant II

Answer:
Here, x is in quadrant II.
i.e π < x < 3π/2
⇒ π/2 < x/2 < 3π/4
Therefore, cos x/2 and tan x/2 are negative whereas sin x/2 is positive.
Given, cos x = -1/3
Now, cos x = 1 – 2 sin² x/2
⇒ sin² x/2 = (1 – cos x)/2
                    = {1 – (-1/3)}/2
                    = (1 + 1/3)/2
                    = (4/3)/2
                    = 2/3
⇒ sin x/2 = √(2/3)
⇒ sin x/2 = (√2/√3) * (√3/√3)
⇒ sin x/2 = √6/3
       cos x = 2 cos² x/2 – 1
⇒ cos² x/2 = (1 + cos x)/2                  
                    = {1 + (-1/3)}/2
                    = (1 – 1/3)/2
                    = (2/3)/2
                    = 1/3
⇒ cos x/2 = -1/√3                         [Since cos x/2 is negative]
⇒ cos x/2 = (-1/√3) * (√3/√3)
⇒ cos x/2 = -√3/3
tan x/2 = (sin x/2)/(cos x/2) = (√2/√3)/( -1/√3) = -√2
Thus, the values of sin x/2, cos x/2 and tan x/2 are: √6/3, -√3/3, -√2

Question 10:
Find sin x/2, cos x/2 and tan x/2 for sin x = 1/4, x in quadrant II

Answer:
Here, x is in quadrant II.
i.e π/2 < x < π
⇒ π/4 < x/2 < π/2
Therefore, sin x/2, cos x/2 and tan x/2 are all positive.
Given, sin x = 1/4
cos² x = 1 – sin² x
            = 1 – (1/4)²
            = 1 – 1/16
            = 15/16
cos x = -√15/4                                          [since cos x is negative in quadrant II]
Now, sin² x/2 = (1 – cos x)/2
                         = {1 – (-√15/4)}/2
                         = (1 + √15/4)/2
                         = (4 + √15)/8
⇒ sin x/2 = √[(4 + √15)/8]                     [Since sin x/2 is positive]
⇒ sin x/2 = √[{(4 + √15)/8} * (2/2)]
⇒ sin x/2 = √[(8 + 2√15)/16]
⇒ sin x/2 = √{(8 + 2√15)}/4
Again, cos² x/2 = (1 + cos x)/2
                         = {1 + (-√15/4)}/2
                         = (1 – √15/4)/2
                         = (4 – √15)/8
⇒ cos x/2 = √[(4 – √15)/8]                          [Since cos x/2 is positive]
⇒ cos x/2 = √[{(4 – √15)/8} * (2/2)]
⇒ cos x/2 = √[(8 – 2√15)/16]
⇒ cos x/2 = √{(8 – 2√15)}/4
Now, tan x/2 = (sin x/2)/(cos x/2)
                        = [√{(8 + 2√15)}/4]/[ √{(8 – 2√15)}/4]
                        = [√(8 + 2√15)]/[ √(8 – 2√15)]
                        = √[{(8 + 2√15)]/(8 – 2√15)} * {(8 + 2√15)]/(8 + 2√15)}]
                        = √[(8 + 2√15)²]/(64 – 60)]  
                        = (8 + 2√15)/2
                        = 4 + √15      
Therefore, value of sin x/2, cos x/2 and tan x/2 are: √{(8 + 2√15)}/4, √{(8 – 2√15)}/4 and 4 + √15