RD Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Ex 9.1 Solutions
Question - 11 : - (cos α + cos β) 2 +(sin α + sin β) 2 = 4 cos2 (α – β)/2
Answer - 11 : -
Let us consider LHS:
(cos α + cos β)2 + (sin α + sin β)2
Upon expansion, we get,
(cos α + cos β)2 + (sin α + sin β)2 =
= cos2 α + cos2 β + 2cos α cos β + sin2 α + sin2 β + 2 sin α sin β
= 2 + 2 cos α cos β + 2 sin α sin β
= 2 (1 + cos α cos β + sin α sin β)
= 2 (1 + cos (α – β)) [since, cos (A – B) = cos A cosB + sin A sin B]
= 2 (1 + 2 cos2 (α – β)/2 – 1) [since,cos2x = 2cos2 x – 1]
= 2 (2 cos2 (α – β)/2)
= 4 cos2 (α – β)/2
= RHS
Hence Proved.
Question - 12 : - sin2 (π/8+ x/2) – sin2 (π/8 – x/2) = 1/√2 sin x
Answer - 12 : -
Let us consider LHS:
sin2 (π/8 + x/2) – sin2 (π/8– x/2)
we know, sin2 A – sin2 B= sin (A+B) sin (A-B)
so,
sin2 (π/8 + x/2) – sin2 (π/8– x/2) = sin (π/8 + x/2 + π/8 – x/2) sin (π/8 + x/2 – (π/8 – x/2))
= sin (π/8 + π/8) sin (π/8 + x/2 – π/8 + x/2)
= sin π/4 sin x
= 1/√2 sin x [since, since π/4 = 1/√2]
= RHS
Hence proved.
Question - 13 : - 1 + cos2 2x= 2 (cos4 x + sin4 x)
Answer - 13 : -
Let us consider LHS:
1 + cos2 2x
We know, cos2x = cos2 x – sin2 x
cos2 x + sin2 x = 1
so,
1 + cos2 2x = (cos2 x+ sin2 x) 2 + (cos2 x –sin2 x) 2
= (cos4 x + sin4 x + 2cos2 x sin2 x) + (cos4 x + sin4 x– 2 cos2 x sin2 x)
= cos4 x + sin4 x +cos4 x + sin4 x
= 2 cos4 x + 2 sin4 x
= 2 (cos4 x + sin4 x)
= RHS
Hence proved.
Question - 14 : - cos3 2x+ 3 cos 2x = 4 (cos6 x – sin6 x)
Answer - 14 : -
Let us consider RHS:
4 (cos6 x – sin6 x)
Upon expansion we get,
4 (cos6 x – sin6 x) =4 [(cos2 x)3 – (sin2 x)3]
= 4 (cos2 x – sin2 x)(cos4 x + sin4 x + cos2 x sin2 x)
By using the formula,
a3 – b3 = (a-b) (a2 +b2 + ab)
= 4 cos 2x (cos4 x + sin4 x+ cos2 x sin2 x + cos2 x sin2 x– cos2 x sin
We know, cos 2x = cos2 x – sin2 x
So,
= 4 cos 2x (cos4 x + sin4 x+ 2 cos2 x sin2 x – cos2 x sin2 x)
= 4 cos 2x [(cos2 x)2 +(sin2 x)2 + 2 cos2 x sin2 x– cos2 x sin2 x]
We know, a2 + b2 + 2ab= (a + b)2
= 4 cos 2x [(1)2 – 1/4 (4 cos2 xsin2 x)]
= 4 cos 2x [(1)2 – 1/4 (2 cos x sin x)2]
We know, sin 2x = 2sin x cos x
= 4 cos 2x [(12) – 1/4 (sin 2x)2]
= 4 cos 2x (1 – 1/4 sin2 2x)
We know, sin2 x = 1 – cos2 x
= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]
= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]
= 4 cos 2x [3/4 + 1/4 cos2 2x]
= 4 (3/4 cos 2x + 1/4 cos3 2x)
= 3 cos 2x + cos3 2x
= cos3 2x + 3 cos 2x
= LHS
Hence proved.
Question - 15 : - (sin 3x +sin x) sin x + (cos 3x – cos x) cos x = 0
Answer - 15 : -
Let us consider LHS:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= (sin 3x) (sin x) + sin2 x + (cos 3x)(cos x) – cos2 x
= [(sin 3x) (sin x) + (cos 3x) (cos x)] + (sin2 x– cos2 x)
= [(sin 3x) (sin x) + (cos 3x) (cos x)] – (cos2 x– sin2 x)
= cos (3x – x) – cos 2x
We know, cos 2x = cos2 x –sin2 x
cos A cos B + sin A sin B = cos(A – B)
So,
= cos 2x – cos 2x
= 0
= RHS
Hence Proved.
Question - 16 : - cos2 (π/4– x) – sin2 (π/4 – x) = sin 2x
Answer - 16 : -
Let us consider LHS:
cos2 (π/4 – x) – sin2 (π/4– x)
We know, cos2 A – sin2 A= cos 2A
So,
cos2 (π/4 – x) – sin2 (π/4– x) = cos 2 (π/4 – x)
= cos (π/2 – 2x)
= sin 2x [since, cos (π/2 – A) = sin A]
= RHS
Hence proved.
Question - 17 : - cos 4x =1 – 8 cos2 x + 8 cos4 x
Answer - 17 : -
Let us consider LHS:
cos 4x
We know, cos 2x = 2 cos2 x – 1
So,
cos 4x = 2 cos2 2x – 1
= 2(2 cos2 2x – 1)2 –1
= 2[(2 cos2 2x) 2 +12 – 2×2 cos2 x] – 1
= 2(4 cos4 2x + 1 – 4 cos2 x)– 1
= 8 cos4 2x + 2 – 8 cos2 x– 1
= 8 cos4 2x + 1 – 8 cos2 x
= RHS
Hence Proved.
Question - 18 : - sin 4x =4 sin x cos3 x – 4 cos x sin3 x
Answer - 18 : -
Let us consider LHS:
sin 4x
We know, sin 2x = 2 sin x cos x
cos 2x = cos2 x – sin2 x
So,
sin 4x = 2 sin 2x cos 2x
= 2 (2 sin x cos x) (cos2 x – sin2 x)
= 4 sin x cos x (cos2 x – sin2 x)
= 4 sin x cos3 x – 4 sin3 xcos x
= RHS
Hence proved.
Question - 19 : - 3(sin x –cos x) 4 + 6 (sin x + cos x) 2 + 4(sin6 x + cos6 x) = 13
Answer - 19 : -
Let us consider LHS:
3(sin x – cos x) 4 + 6 (sin x +cos x) 2 + 4 (sin6 x + cos6 x)
We know, (a + b)2 = a2 +b2 + 2ab
(a – b)2 = a2 + b2 –2ab
a3 + b3 = (a + b) (a2 +b2 – ab)
So,
3(sin x – cos x) 4 + 6 (sin x +cos x) 2 + 4 (sin6 x + cos6 x)= 3{(sin x – cos x) 2}2 + 6 {(sin x)2 +(cos x)2 + 2 sin x cos x)} + 4 {(sin2 x)3 +(cos2 x)3}
= 3{(sin x) 2 + (cos x)2 –2 sin x cos x)}2 + 6 (sin2 x + cos2 x+ 2 sin x cos x) + 4{(sin2 x + cos2 x) (sin4 x+ cos4 x – sin2 x cos2 x)}
= 3(1 – 2 sin x cos x) 2 + 6 (1 +2 sin x cos x) + 4{(1) (sin4 x + cos4 x – sin2 xcos2 x)}
We know, sin2 x + cos2 x= 1
So,
= 3{12 + (2 sin x cos x) 2 –4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2 x)2 +(cos2 x)2 + 2 sin2 x cos2 x– 3 sin2 x cos2 x)}
= 3{1 + 4 sin2 x cos2 x– 4 sin x cos x} + 6 (1 + 2 sin x cos x) + 4{(sin2 x + cos2 x) 2 –3 sin2 x cos2 x)}
= 3 + 12 sin2 x cos2 x– 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)2 – 3 sin2 xcos2 x)}
= 9 + 12 sin2 x cos2 x+ 4(1 – 3 sin2 x cos2 x)
= 9 + 12 sin2 x cos2 x+ 4 – 12 sin2 x cos2 x
= 13
= RHS
Hence proved.
Question - 20 : - 2(sin6 x+ cos6 x) – 3(sin4 x + cos4 x)+ 1 = 0
Answer - 20 : -
Let us consider LHS:
2(sin6 x + cos6 x) –3(sin4 x + cos4 x) + 1
We know, (a + b)2 = a2 +b2 + 2ab
a3 + b3 = (a + b) (a2 +b2 – ab)
So,
2(sin6 x + cos6 x) –3(sin4 x + cos4 x) + 1 = 2{(sin2 x) 3 +(cos2 x) 3} – 3{(sin2 x) 2 +(cos2 x) 2} + 1
= 2{(sin2 x + cos2 x)(sin4 x + cos4 x – sin2 x cos2 x}– 3{(sin2 x) 2 + (cos2 x) 2 +2sin2 x cos2 x – 2sin2 x cos2 x}+ 1
= 2{(1) (sin4 x + cos4 x+ 2 sin2 x cos2 x – 3 sin2 xcos2 x} – 3{(sin2 x + cos2 x) 2 –2sin2 x cos2 x} + 1
We know, sin2 x + cos2 x= 1
= 2{(sin2 x + cos2 x) 2 –3 sin2 x cos2 x} – 3{(1)2 –2sin2 x cos2 x} + 1
= 2{(1)2 – 3 sin2 xcos2 x} – 3(1 – 2sin2 x cos2 x)+ 1
= 2(1 – 3 sin2 x cos2 x)– 3 + 6 sin2 x cos2 x + 1
= 2 – 6 sin2 x cos2 x– 2 + 6 sin2 x cos2 x
= 0
= RHS
Hence proved.