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# RD Chapter 5- Trigonometric Functions Ex-5.3 Interview Questions Answers

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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

(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

(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)

(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 + cotx + 2 cot x – cosecx

We know that 1 + cotx = cosecx

cosecx + 2 cot x – cosecx

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

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.

krishan

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