RD Chapter 5 Trigonometric Functions Ex 5.3 Solutions
Question - 1 : - Find the values of the following trigonometric ratios:
(i) sin 5π/3
(ii) sin 17π
(iii) tan 11π/6
(iv) cos (-25π/4)
(v) tan 7π/4
(vi) sin 17π/6
(vii) cos 19π/6
(viii) sin (-11π/6)
(ix) cosec (-20π/3)
(x) tan (-13π/4)
(xi) cos 19π/4
(xii) sin 41π/4
(xiii) cos 39π/4
(xiv) sin 151π/6
Answer - 1 : -
(i) sin 5π/3
5π/3 = (5π/3 × 180)o
= 300o
= (90×3 + 30)o
Since, 300o lies in IV quadrant inwhich sine function is negative.
sin 5π/3 = sin (300)o
= sin (90×3 + 30)o
= – cos 30o
= – √3/2
(ii) sin 17π
Sin 17π = sin 3060o
= sin (90×34 + 0)o
Since, 3060o lies in the negativedirection of x-axis i.e., on boundary line of II and III quadrants.
Sin 17π = sin (90×34 + 0)o
= – sin 0o
= 0
(iii) tan 11π/6
tan 11π/6 = (11/6 × 180)o
= 330o
Since, 330o lies in the IV quadrant inwhich tangent function is negative.
tan 11π/6 = tan (300)o
= tan (90×3 + 60)o
= – cot 60o
= – 1/√3
(iv) cos (-25π/4)
cos (-25π/4) = cos (-1125)o
= cos (1125)o
Since, 1125o lies in the I quadrant inwhich cosine function is positive.
cos (1125)o = cos (90×12 + 45)o
= cos 45o
= 1/√2
(v) tan 7π/4
tan 7π/4 = tan 315o
= tan (90×3 + 45)o
Since, 315o lies in the IV quadrant inwhich tangent function is negative.
tan 315o = tan (90×3 + 45)o
= – cot 45o
= -1
(vi) sin 17π/6
sin 17π/6 = sin 510o
= sin (90×5 + 60)o
Since, 510o lies in the II quadrant inwhich sine function is positive.
sin 510o = sin (90×5 + 60)o
= cos 60o
= 1/2
(vii) cos 19π/6
cos 19π/6 = cos 570o
= cos (90×6 + 30)o
Since, 570o lies in III quadrant inwhich cosine function is negative.
cos 570o = cos (90×6 + 30)o
= – cos 30o
= – √3/2
(viii) sin (-11π/6)
sin (-11π/6) = sin (-330o)
= – sin (90×3 + 60)o
Since, 330o lies in the IV quadrant inwhich the sine function is negative.
sin (-330o) = – sin (90×3 + 60)o
= – (-cos 60o)
= – (-1/2)
= 1/2
(ix) cosec (-20π/3)
cosec (-20π/3) = cosec (-1200)o
= – cosec (1200)o
= – cosec (90×13 + 30)o
Since, 1200o lies in the II quadrantin which cosec function is positive.
cosec (-1200)o = – cosec (90×13 + 30)o
= – sec 30o
= -2/√3
(x) tan (-13π/4)
tan (-13π/4) = tan (-585)o
= – tan (90×6 + 45)o
Since, 585o lies in the III quadrantin which the tangent function is positive.
tan (-585)o = – tan (90×6 + 45)o
= – tan 45o
= -1
(xi) cos 19π/4
cos 19π/4 = cos 855o
= cos (90×9 + 45)o
Since, 855o lies in the II quadrant inwhich the cosine function is negative.
cos 855o = cos (90×9 + 45)o
= – sin 45o
= – 1/√2
(xii) sin 41π/4
sin 41π/4 = sin 1845o
= sin (90×20 + 45)o
Since, 1845o lies in the I quadrant inwhich the sine function is positive.
sin 1845o = sin (90×20 + 45)o
= sin 45o
= 1/√2
(xiii) cos 39π/4
cos 39π/4 = cos 1755o
= cos (90×19 + 45)o
Since, 1755o lies in the IV quadrantin which the cosine function is positive.
cos 1755o = cos (90×19 + 45)o
= sin 45o
= 1/√2
(xiv) sin 151π/6
sin 151π/6 = sin 4530o
= sin (90×50 + 30)o
Since, 4530o lies in the III quadrantin which the sine function is negative.
sin 4530o = sin (90×50 + 30)o
= – sin 30o
= -1/2
Question - 2 : - provethat:
(i) tan 225o cot405o + tan 765o cot 675o = 0
(ii) sin8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2
(iii) cos24o + cos 55o + cos 125o + cos204o + cos 300o = 1/2
(iv) tan(-125o) cot (-405o) – tan (-765o) cot (675o)= 0
(v) cos570o sin 510o + sin (-330o) cos(-390o) = 0
(vi) tan11π/3 – 2 sin 4π/6 – 3/4 cosec2 π/4 + 4 cos2 17π/6= (3 – 4√3)/2
(vii) 3sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1
Answer - 2 : -
(i) tan 225o cot 405o +tan 765o cot 675o = 0
Let us consider LHS:
tan 225° cot 405° + tan 765° cot 675°
tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8+ 45°) cot (90° × 7 + 45°)
We know that when n is odd, cot → tan.
tan 45° cot 45° + tan 45° [-tan 45°]
tan 45° cot 45° – tan 45° tan 45°
1 × 1 – 1 × 1
1 – 1
0 = RHS
∴ LHS = RHS
Hence proved.
(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2
Let us consider LHS:
sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6
sin 480° cos 690° + cos 780° sin 1050°
sin (90° × 5 + 30°) cos (90° × 7 + 60°) + cos (90° × 8+ 60°) sin (90° × 11 + 60°)
We know that when n is odd, sin → cos andcos → sin.
cos 30° sin 60° + cos 60° [-cos 60°]
√3/2 × √3/2 – 1/2 × 1/2
3/4 – 1/4
2/4
1/2
= RHS
∴ LHS = RHS
Hence proved.
(iii) cos 24o + cos 55o +cos 125o + cos 204o + cos 300o =1/2
Let us consider LHS:
cos 24o + cos 55o +cos 125o + cos 204o + cos 300o
cos 24° + cos (90° × 1 – 35°) + cos (90° × 1 + 35°) +cos (90° × 2 + 24°) + cos (90° × 3 + 30°)
We know that when n is odd, cos → sin.
cos 24° + sin 35° – sin 35° – cos 24° + sin 30°
0 + 0 + 1/2
1/2
= RHS
∴ LHS = RHS
Hence proved.
(iv) tan (-125o) cot (-405o) – tan(-765o) cot (675o) = 0
Let us consider LHS:
tan (-125o) cot (-405o) – tan(-765o) cot (675o)
We know that tan (-x) = -tan (x) and cot (-x) = -cot(x).
[-tan(225°)] [-cot (405°)] – [-tan (765°)] cot (675°)
tan (225°) cot (405°) + tan (765°) cot (675°)
tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8+ 45°) cot (90° × 7 + 45°)
tan 45° cot 45° + tan 45° [-tan 45°]
1 × 1 + 1 × (-1)
1 – 1
0
= RHS
∴ LHS = RHS
Hence proved.
(v) cos 570o sin 510o +sin (-330o) cos (-390o) = 0
Let us consider LHS:
cos 570o sin 510o +sin (-330o) cos (-390o)
We know that sin (-x) = -sin (x) and cos (-x) = +cos(x).
cos 570o sin 510o +[-sin (330o)] cos (390o)
cos 570o sin 510o –sin (330o) cos (390o)
cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3+ 60°) cos (90° × 4 + 30°)
We know that cos is negative at 90° + θ i.e. in Q2 andwhen n is odd, sin → cos and cos → sin.
-cos 30° cos 60° – [-cos 60°] cos 30°
-cos 30° cos 60° + cos 60° cos 30°
0
= RHS
∴ LHS = RHS
Hence proved.
(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec2 π/4+ 4 cos2 17π/6 = (3 – 4√3)/2
Let us consider LHS:
tan 11π/3 – 2 sin 4π/6 – 3/4 cosec2 π/4+ 4 cos2 17π/6
tan (11 × 180o)/3 – 2 sin (4 × 180o)/6– 3/4 cosec2 180o/4 + 4 cos2 (17 ×180o)/6
tan 660o – 2 sin 120o –3/4 (cosec 45o)2 + 4 (cos 510o)2
tan (90° × 7 + 30°) – 2 sin (90° × 1 + 30°) – 3/4[cosec 45°]2 + 4 [cos (90° × 5 + 60°)]2
We know that tan and cos is negative at 90° + θ i.e.in Q2 and when n is odd, tan → cot,sin → cos and cos → sin.
[-cot 30°] –2 cos 30° – 3/4 [cosec 45°]2 + [-sin 60°]2
– cot 30° – 2 cos 30° – 3/4 [cosec 45°]2 +[sin 60°]2
-√3 – 2√3/2 – 3/4 (√2)2 + 4 (√3/2)2
-√3 – √3 – 6/4 + 12/4
(3 – 4√3)/2
= RHS
∴ LHS = RHS
Hence proved.
(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1
Let us consider LHS:
3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4
3 sin 180o/6 sec 180o/3 – 4 sin5(180o)/6 cot 180o/4
3 sin 30° sec 60° – 4 sin 150° cot 45°
3 sin 30° sec 60° – 4 sin (90° × 1 + 60°) cot 45°
We know that when n is odd, sin → cos.
3 sin 30° sec 60° – 4 cos 60° cot 45°
3 (1/2) (2) – 4 (1/2) (1)
3 – 2
1
= RHS
∴ LHS = RHS
Hence proved.
Question - 3 : - Provethat:
(i) 
(ii) 
(iii)
(iv)
(v)
Answer - 3 : -
(i)
1 = RHS
∴ LHS = RHS
Hence proved.
(ii)
1 + 1
2 = RHS
∴ LHS = RHS
Hence proved.
(iii)
1 = RHS
∴ LHS = RHS
Hence proved.
(iv)
{1 + cot x – (-cosec x)} {1 + cot x + (-cosec x)}
{1 + cot x + cosec x} {1 + cot x – cosec x}
{(1 + cot x) + (cosec x)} {(1 + cot x) – (cosec x)}
By using the formula, (a + b) (a – b) = a2 –b2
(1 + cot x)2 – (cosec x)2
1 + cot2 x + 2 cot x – cosec2 x
We know that 1 + cot2 x = cosec2 x
cosec2 x + 2 cot x – cosec2 x
2 cot x = RHS
∴ LHS = RHS
Hence proved.
(v)
1 = RHS
∴ LHS = RHS
Hence proved.
Question - 4 : - Prove that: sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 = 2
Answer - 4 : -
Let us consider LHS:
sin2 π/18 + sin2 π/9 +sin2 7π/18 + sin2 4π/9
sin2 π/18 + sin2 2π/18+ sin2 7π/18 + sin2 8π/18
sin2 π/18 + sin2 2π/18+ sin2 (π/2 – 2π/18) + sin2 (π/2 – π/18)
We know that when n is odd, sin → cos.
sin2 π/18 + sin2 2π/18+ cos2 2π/18 + cos2 2π/18
when rearranged,
sin2 π/18 + cos2 2π/18+ sin2 π/18 + cos2 2π/18
We know that sin2 + cos2x =1.
So,
1 + 1
2 = RHS
∴ LHS = RHS
Hence proved.