Practicing S Chand Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(d) is the ultimate need for students who intend to score good marks in examinations.
S Chand Class 11 ICSE Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(d)
Question 1.
Express the following as functions of angles less than 45° :
(i) sin 194°
(ii) sin 348°
(iii) cos 189°
(iv) sin (- 1785°)
(v) tan (3598°)
(vi) cot (- 1952°)
(vii) cosec (- 7498°).
Solution:
(i) sin 194° = sin (180° + 14°) = – sin 14° [∵ sin (180° + θ ) = – sin θ]
(ii) Sin 348° = sin (360° – 12°) = – sin 12° [∵ sin (360° – θ ) = – sin θ]
(iii) cos 189° = cos (180° + 9°) = – cos 9° [∵ cos (180° + θ ) = – cos θ]
(iv) sin(- 1785°) = – sin (1785°) [∵ sin (- θ ) = – sin θ ]
= – sin (360° x 4 + 345°) = – sin (345°) [∵ sin (360° + θ) = sin θ]
= – sin (360°- 15°) = sin 15° [∵ sin (360° – θ ) = – sin θ ]
(v) tan (3598°) = tan (3600 – 2°) = tan (10 x 360° – 2°)
= tan (- 2°) [∵ tan (10 x 360° – θ ) = tan (- θ )]
= – tan 2°
(vi) cot (- 1952°) = – cot 1952° = – cot (1800° + 152°) = – cot (5 x 360° + 152°)
= – cot (152°) [∵ cot (2nπ + θ ) = cot θ ∀ n ∈ N]
= – cot (180°- 28°)
= cot 28° [∵ cot (180° – θ ) = – cot θ ]
(vii) cosec (- 7498°) = – cosec (7498°) [∵ cosec (- θ ) = – cosec θ ]
= – cosec [7200° + 298°] = – cosec [20 x 360° + 298°]
= – cosec (298°) = – cosec (360° – 62°)
= cosec 62° [∵ cosec (360° – θ ) = – cosec θ ]
= cosec (90° – 28°)
= sec 28°
Question 2.
Without using tables, give the value of each of the following :
(i) sin 120°
(ii) cot 330°
(iii) sec 210°
(iv) cos 315°,
(v) cosec 675°
(vi) cos 855°
(vii) sin 4530°
(viii) sec \(\frac { 15π }{ 4 }\)
Solution:
(i) sin 120° = sin (180° – 60°) = sin 60° = \(\frac{\sqrt{3}}{2}\) [∵ sin (180°- θ ) = sin θ ]
(ii) cos 330 = cot (360° – 30°) = – cot 30° = \(\sqrt{3}\) [∵ cot (360° – θ ) = – cot θ ]
(iii) sec 210° = sec (180° + 30°) = – sec 30° = \(\frac{-2}{\sqrt{3}}\) [∵ sec (180° + θ ) = – sec θ ]
(iv) cos 315° = cos (360° – 45°) = cos 45° = \(\frac{1}{\sqrt{2}}\) [∵ cos (360° – θ ) = cos θ ]
(v) cosec 675° = cosec (360° + 315°) = cosec (315°) = cosec (360° – 45°)
= – cosec 45° = – \(\sqrt{2}\) [∵ cosec (360° – θ ) = – cosec θ]
(vi) cos 855° = cos (720° + 135°) = cos 135° = cos (180° – 45°) = – cos 45°
= – \(\frac{1}{\sqrt{2}}\) [∵ cos (180° – θ ) = – cos θ]
(vii) sin 4530° = sin (4320° + 210°) = sin (12 x 360° + 210°)
= sin 210° [∵ sin (2nπ + θ) = sin θ]
= sin (180°+ 30°) = – sin 30° = – \(\frac { 1 }{ 2 }\) [∵ sin (180° + θ ) = – sin θ]
(viii) \(\sec \left(\frac{15 \pi}{4}\right)=\sec \left(4 \pi-\frac{\pi}{4}\right)=\sec \left(-\frac{\pi}{4}\right)=\sec \frac{\pi}{4}=\sqrt{2}\)
Question 3.
With the help of tables, find the values, correct to four places of decimals, of each of the following :
(i) cos 116°
(ii) sin 267°
(iii) sin (- 263°)
(iv) cos 280° 10′
(v) tan (2015° 24′).
Solution:
(i) cos 116° = cos (180° – 64°) = – cos 64° = – 0.4384
(ii) sin 267° = sin (270° – 3°) = – cos 3° = – 0.9986
(iii) sin (- 263°) = – sin (263°) = – sin (270° – 7°) = cos 7° = 0.9925 [∵ sin (270° – θ ) = – cos θ ]
(iv) cos (280° 10′) = cos (270° + 10° 10′) = sin (10° 10′) = 0.1766
(v) tan (2015° 24′) = tan (1800° + 215° 24′) = tan (215° 24′) = tan (180° + 35° 24′)
= tan 35° 24′ [∵ tan (180° + θ ) – tan θ ]
= 0.7107
Question 4.
Find the value of sin 750° cos 300° + cos 1470° sin (- 1020°).
Solution:
sin 750° = sin (720° + 30°) = sin 30° = \(\frac { 1 }{ 2 }\)
cos 300° = cos (360° – 60°) = cos 60° = \(\frac { 1 }{ 2 }\)
cos 1470° = cos (1440° + 30°) = cos (360° x 4 + 30°) = cos 30° = \(\frac{\sqrt{3}}{2}\)
sin (- 1020°) = – sin 1020° = – sin (720° + 300°) = – sin 300°
= – sin (360° – 60°) = sin 60° = \(\frac{\sqrt{3}}{2}\)
∴ sin 750° cos 300° + cos 1470° sin (- 1020°) = \(\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}\) = 1 = R.H.S
Question 5.
Evaluate \(\frac{\cos 3 \theta-2 \cos 4 \theta}{\sin 3 \theta+2 \sin 4 \theta}\), when θ = 150 .
Solution:
Given θ =150°
∴ cos 3θ = cos 450° = cos (360° + 90°) = cos 90° = θ
cos 4θ = cos 600° = cos (360° + 240°) = cos 240° = cos (180° + 60°) = – cos 60° = – \(\frac { 1 }{ 2 }\)
sin 3θ = sin 450° = sin (360° + 90°) = sin 90° = 1
sin 4θ = sin 600° = sin (360° + 240°) = sin (180° + 60°) = – sin 60° = – \(\frac{\sqrt{3}}{2}\)
∴\(\frac{\cos 3 \theta-2 \cos 4 \theta}{\sin 3 \theta+2 \sin 4 \theta}=\frac{0-2\left(-\frac{1}{2}\right)}{1+2\left(-\frac{\sqrt{3}}{2}\right)}=\frac{-1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=-\frac{1}{2}(\sqrt{3}+1)\)
Question 6.
Simplify : \(\frac{\cos (-\theta)}{\sin \left(90^{\circ}+\theta\right)}\).
Solution:
Now, \(\frac{\cos (-\theta)}{\sin \left(90^{\circ}+\theta\right)}=\frac{\cos \theta}{\cos \theta}\) = 1 = R.H.S. [∵ sin (90° + 0) = – cos θ]
Question 7.
\(\frac{\tan (-\theta)}{\sin \left(540^{\circ}+\theta\right)}\).
Solution:
\(\frac{\tan (-\theta)}{\sin \left(540^{\circ}+\theta\right)}=\frac{-\tan \theta}{\sin \left(360^{\circ}+180^{\circ}+\theta\right)}=\frac{-\tan \theta}{\sin \left(180^{\circ}+\theta\right)}=\frac{-\tan \theta}{-\sin \theta}=\frac{1}{\cos \theta}=\sec \theta\)
Question 8.
\(\frac{\sin \left(90^{\circ}-\theta\right) \sec \left(180^{\circ}-\theta\right) \sin (-\theta)}{\sin \left(180^{\circ}+\theta\right) \cot \left(360^{\circ}-\theta\right) {cosec}\left(90^{\circ}+\theta\right)}\)
Solution:
\(\frac{\sin \left(90^{\circ}-\theta\right) \sec \left(180^{\circ}-\theta\right) \sin (-\theta)}{\sin \left(180^{\circ}+\theta\right) \cot \left(360^{\circ}-\theta\right) {cosec}\left(90^{\circ}+\theta\right)}=\frac{\cos \theta(-\sec \theta)(-\sin \theta)}{(-\sin \theta)(-\cot \theta)(\sec \theta)}=\frac{\cos \theta}{\cot \theta}=\sin \theta\)
Question 9.
\(\frac{\sin 150^{\circ}-5 \cos 300^{\circ}+7 \tan 225^{\circ}}{\tan 135^{\circ}+3 \sin 210^{\circ}}\)
Solution:
Question 10.
If sin (7Φ + 9°) = cos 2Φ, find a value of Φ.
Solution:
Given sin (7Φ + 9°) = cos 2Φ = sin (\(\frac { π }{ 2 }\) – 2Φ) ⇒ 7Φ + 9° = \(\frac { π }{ 2 }\) – 2Φ ⇒ 9Φ = 81° ⇒ Φ = 9°
Question 11.
Find the values of lying between 0° and 360° when
(i) sin θ = \(\frac { 1 }{ 2 }\),
(ii) tan θ = – 1
(iii) sec θ = – 2
(iv) sin θ = sin 21°
(v) tan θ = – 2.0145
(vi) sin θ = cos 317°
(vii) cos θ = sin 285°.
Solution:
(i) sin θ = \(\frac { 1 }{ 2 }\) = sin 30° or sin (180°- 30°); θ ≤ θ ≤ 360° ⇒ θ = 30° or 150°
(iv) Given sin θ = sin 21° or sin (180° – 21°) ⇒ θ = 21°, 180° – 21° ⇒ θ = 21°, 159°
(v) Given tan θ = – 2.0145 = tan (π – 63° 36′) or tan (2π – 63° 36′)
⇒ θ = 180°- 63° 36′ or 360° – 63° 36′
∴ θ = 116° 24′, 296° 24′
(vi) Given sin θ = cos 317° = cos (270° + 47°)
⇒ sin θ = sin 47° or sin (180° – 47°) i.e. θ = 47° or 180° – 47°
⇒ θ = 47° or 133°
(vii) Given cos θ = sin 285° = sin (270° + 15°)
⇒ cos θ = – cos 15° = cos (180° – 15°) or cos (180° + 15°)
⇒ cos θ = cos 165° or cos 195°
⇒ θ = 165°, 195°
Question 12.
If θ ° < θ < 90° and cos θ = \(\frac { 4 }{ 5 }\), find the values of
(i) cos (90° + θ)
(ii) cosec (180° + θ)
(iii) tan (360° – θ)
(iv) sin (270° – θ).
Solution:
Question 13.
Find six angles for which sin θ = – \(\frac{\sqrt{3}}{2}\).
Solution:
Given sin θ = – \(\frac{\sqrt{3}}{2}\) = – sin 60° = sin (180° + 60°), sin (360° – 60°)
∴ sin θ = sin (180° + 60°), sin (360° – 60°) ⇒ θ = 240°, 300°
Also, sin (600°) = sin (360° + 240°) = sin 240° = sin (180° + 60°) = – sin 60° = – \(\frac{\sqrt{3}}{2}\)
sin (660°) = sin (720° – 60°) = sin (- 60°) = – sin 60° = – \(\frac{\sqrt{3}}{2}\)
sin (960°) = sin (720° + 240°) = sin 240° = – \(\frac{\sqrt{3}}{2}\)
sin (1020°) = sin (720° + 300°) = sin 300° = sin (360° – 60°) = – sin 60° = – \(\frac{\sqrt{3}}{2}\)
Hence the required six angles for which sin θ = – \(\frac{\sqrt{3}}{2}\) are :
240°, 300°, 600°, 660°, 960°, 1020°.
Question 14.
Find all the angles between θ° and 720° whose tangent is – \(\frac{1}{\sqrt{3}}\).
Solution:
Question 15.
Find the values of θ between 0° and 360° which satisfy the equations
(i) sin² θ = \(\frac { 3 }{ 4 }\)
(ii) cos 3θ = \(\frac { 1 }{ 2 }\).
Solution:
(i) sin² θ = \(\frac { 3 }{ 4 }\) ⇒ sin θ = ± \(\frac{\sqrt{3}}{2}\)
Case-I.
When sin θ = \(\frac{\sqrt{3}}{2}\) = sin \(\frac { π }{ 3 }\) or sin (π – \(\frac { π }{ 3 }\))
⇒ sin θ = sin \(\frac { π }{ 3 }\), sin(\(\frac { 2π }{ 3 }\)) ⇒ θ = \(\frac { π }{ 3 }\), \(\frac { 2π }{ 3 }\)
Case-II.
When sin θ = – \(\frac{\sqrt{3}}{2}\) = – sin \(\frac { π }{ 3 }\) = – sin \(\frac { π }{ 3 }\)
= sin(π + \(\frac { π }{ 3 }\)) or sin (2π – \(\frac { π }{ 3 }\))
⇒ θ = π + \(\frac { π }{ 3 }\), 2π – \(\frac { π }{ 3 }\) ⇒ θ = \(\frac { 4π }{ 3 }\), \(\frac { 5π }{ 3 }\)
Thus, the required values of θ are \(\frac { π }{ 3 }\), \(\frac { 2π }{ 3 }\), \(\frac { 4π }{ 3 }\), \(\frac { 5π }{ 3 }\)
(ii) Given cos 3θ = \(\frac { 1 }{ 2 }\) = cos \(\frac { π }{ 3 }\) = cos(2nπ ± \(\frac { π }{ 3 }\))
Hence the required values of θ are \(\frac{\pi}{9}, \frac{7 \pi}{9}, \frac{5 \pi}{9}, \frac{13 \pi}{9}, \frac{11 \pi}{9}, \frac{17 \pi}{9}\)
i.e. 20°, 140°, 100°, 260°, 220°, 340°.
Question 16.
If tan θ = 0.4, when θ lies between 0° and 360°, write down the possible values of θ and sin θ.
Solution:
Given tan θ = 0.4 = tan (21° 48′) or tan (180° + 21° 48′)
⇒ θ = 21° 48′ or 180° + 21° 48′ ⇒ θ = 21° 48′, 201° 48′ where θ ∈ (0, 360°)
∴ sin θ = sin (21° 48′) = 0.3714
and sin (201° 48′) = sin (180° + 21° 48′) = – sin 21° 48′ = – 0.3714
Question 17.
If cos x° = sin 200°, find the possible values of x between – 180° and 360°.
Solution:
Given cos x° = sin 200 = sin (270° – 70°)
⇒ cos x° = – cos 70° = cos (180° – 70°) or cos (180° + 70°) or cos (- 180° + 70°)
⇒ cos x° = cos 110°, cos 250°, cos (- 110°) [∵ – 180° < x < 360°]
⇒ x° = ± 110°, 250°
Question 18.
If A, B, C are the angles of a triangle, prove that cos C = – cos (A + B).
Solution:
Since A, B and C are angles of triangle
∴ A + B + C = 180° … (1)
∴ cos C = cos (180° – \(\overline{A+B}\)) = – cos (A + B) [using eqn. (1)]
Question 19.
tan \(\frac{B+C}{2}\) = cot \(\frac { A }{ 2 }\).
Solution:
\(\tan \frac{\mathrm{B}+\mathrm{C}}{2}=\tan \left(\frac{180^{\circ}-\mathrm{A}}{2}\right)=\tan \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)=\cot \frac{\mathrm{A}}{2}\)
Question 20.
\(\frac{\tan (B+C)+\tan (C+A)+\tan (A+B)}{\tan (\pi-A)+\tan (2 \pi-B)+\tan (3 \pi-C)}\) = 1.
Solution:
Question 21.
If A, B, C, D are the angles of a quadrilateral, prove that
cos\(\frac { 1 }{ 2 }\) (A + B) + cos\(\frac { 1 }{ 2 }\) (C + D) = 0.
Solution:
since A, B, C and D are the angles of quadrilateral
