--Advertisement--

Class 11 Maths NCERT Solutions for Chapter 3 Trigonometric Functions Miscellaneous Exercise

Class 11 Maths NCERT Solutions for Chapter 3 Trigonometric Functions Miscellaneous Exercise

Trigonometric Functions Miscellaneous Exercise Solutions

1. Prove that : 2 cos(π/3) cos(9π/13) + cos(3π/13) + cos (5π/13) = 0 

Solution

L.H.S 

= 0 = R.H.S 

2. Prove that: (sin 3+ sin x) sin + (cos 3– cos x) cos = 0
Solution
L.H.S 
= (sin 3x + sin x)sin x + (cos 3x - cos x) cos x 
= sin 3x sin x + sin2x + cos3x cosx – cos2
= cos 3x cos x + sin 3x sin x - (cos2x - sin2x) 
= cos(3x - x) - cos2x  [cos (A - B) = cosA cosB + sin A sinB] 
= cos 2x - cos 2x 
= 0
R.H.S

3. Prove that (cos x + cos y)2 + (sin x - sin y)2 = 4 cos2[(x+y)/2] 
Solution
L.H.S = (cos x + cos y)2 + (sin x - sin y)2  
= cos2x + cos2y + 2 cosx cosy + sin2x + sin2y - 2sin x siny 
= (cos2x + sin2x) + (cos2y + sin2y) + 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 + 2cos (x + y)
= 2[1 + cos (x + y)] 
= 2[1 + 2 cos2{(x+y)/2}- 1]  [cos2A = 2 cos2A - 1]
= 4 cos2{(x + y)/2} = R.H.S 

4. Prove that : (cos x - cos y)2 + (sin x - sin y)2  = 4 sin2{(x - y)/2}
Solution
L.H.S = (cos x - cos y)2  + (sin x - sin y)2 
= cos2x + cos2y - 2 cos x cos y + sin2x + sin2y - 2sinx sin y
= (cos2x + sin2x) + (cos2y + sin2y)  - 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[1 - cos (x - y)] 
= 2[1 - {1 - 2sin2{(x -y)/2}]  [cos2A = 1 - 2 sin2 A]
= 4sin2{(x - y)/2} = R.H.S. 

5.  Prove that sin x + sin 3x + sin 5x + sin 7x = 4 cosx cos2x sin4x
Solution
It is known that sin A + sin B = 2sin[(A + B)/2].cos [(A-B)/2]
∴ L.H.S = sin x + sin 3x + sin 5x + sin 7x 
= (sin x + sin 5x) + (sin 3x + sin 7x) 
= 2sin [(x + 5x)/2] .cos [(x - 5x)/2] + 2 sin [(3x + 7x)/2] cos [(3x - 7x)/2] 
= 2 sin 3x cos(-2x) + 2 sin 5x cos (-2x) 
= 2 sin 3x cos 2x + 2sin 5x cos 2x 
= 2 cos 2x[sin 3x + sin 5x] 
= 2 cos 2x[2sin{(3x + 5x)/2}. cos{(3x - 5x)/2}
= 2cos 2x[2sin4x. cos(-x)] 
= 4 cos 2x sin 4x cos x = R.H.S.

6. Prove that : [(sin 7x + sin 5x) + (sin 9x + sin 3x)]/[(cos 7x + cos 5x)+(cos 9x + cos 3x)] = tan 6x 
Solution
It is known that 

= tan 6x 
R.H.S 

7. Prove that : sin 3x + sin 2x - sin x = 4sin x cos (x/2) cos (3x/2) 
Solution
L.H.S. = sin 3x + sin 2x - sin x 
= sin 3x + (sin 2x - sin x) 

 8. tan x = -4/3, x in quadrant II
Solution
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. 
It is given that tan x = -4/3 
sec2 x = 1 + tan2 x = 1 + (-4/3)2 = 1 + 16/9 = 25/9 
∴ cos2 x = 9/25 
⇒ cos x = ± (3/5) 
As x is in quadrant II, cos x is negative. 
∴ cos x = -3/5 
Now, cos x = 2 cos2(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/2) = 1/√5 [∵ cos(x/2) is positive] 
∴ cos(x/2) = √5/5 
sin2(x/2) + cos2(x/2) = 1 
⇒ sin2 (x/2) + (1/√5)2  = 1 
⇒ sin2 (x/2) = 1 - 1/5 = 4/5 
⇒ sin (x/2) = 2/√5  [∵sin (x/2) is positive]
∴ sin (x/2) = 2√5/5 

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are 2√5/5, √5/5, and 2. 

9. Find sin(x/2), cos (x/2) and tan(x/2) for cos x = -1/3, x in quadrant III 
Solution
Here, x is in quadrant III. 
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. 
It is given that cos x = -1/3 . 
cos x = 1 - 2sin2(x/2) 
⇒ sin2(x/2) = (1 - cos x)/2 

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are √6/3, -√3/2 and -√2. 

10. Find, sin(x/2) cos(x/2) and tan(x/2) for sin x = 1/4 , x in quadrant II 
Solution 
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. 
It is given that sin x = 1/4. 
cos2x = 1 - sin2x  = 1 - (1/4)2 = 1 - 1/16 = 15/16 
⇒ cos x = -√15/4  [cos x is negative in quadrant II] 

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are and  4+√15.
Previous Post Next Post

--Advertisement--

--Advertisement--