Utilizing S Chand Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(c) as a study aid can enhance exam preparation.
S Chand Class 11 ICSE Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(c)
Evaluate:
(i) 2 sin 15° cos 15°
(ii) 1 – 2 sin² 22.5°
(iii) 2 cos² 157.5° – 1
(iv) cos² \(\frac { π }{ 2 }\) – sin² \(\frac { π }{ 2 }\)
(v) \(\frac { 1 }{ 2 }\) – sin² \(\frac { 7π }{ 12 }\)
(vi) cos \(\frac { π }{ 8 }\) sin \(\frac { π }{ 8 }\)
(vii) \(\frac { 1 }{ 2 }\)
(viii) 8 cos³ \(\frac { π }{ 9 }\) – 6 cos \(\frac { π }{ 9 }\)
Solution:
(i) 2 sin 15° cos 15° = sin (2 x 15°)
= sin 30° = \(\frac { 1 }{ 2 }\) [∵ 2 sin θ cos θ = sin 2θ]
(ii) 1 – 2 sin² 22.5° = cos (2 x 22.5°)
= cos 45° = \(\frac{1}{\sqrt{2}}\) [∵ cos 2θ =1 – 2 sin² θ]
(iii) 2 cos² 157.5° – 1 = cos (2 x 157.5°)
= cos (315°)
= cos (360° – 45°) [∵ cos 2θ = 2 cos² θ – 1]
= cos 45° = \(\frac{1}{\sqrt{2}}\)
(iv) cos² \(\frac { π }{ 12 }\) – sin² \(\frac { π }{ 12 }\) = cos (2 x \(\frac { π }{ 12 }\))
= cos \(\frac { π }{ 6 }\) = \(\frac{\sqrt{3}}{2}\) [∵ cos² θ – sin² θ = cos 2θ]
Question 2.
Find the values of sin 2θ, cos 2θ, and tan 2θ, given :
(i) sin θ = \(\frac { 3 }{ 5 }\), θ in Quadrant I.
(ii) sin θ = \(\frac { 3 }{ 5 }\), θ in Quadrant II.
(iii) sin θ = – \(\frac { 1 }{ 2 }\), θ in Quadrant IV.
(iv) tan θ = – \(\frac { 1 }{ 5 }\), θ in Quadrant II.
Solution:
(i) Given sin θ = \(\frac { 3 }{ 5 }\)
since θ lies in first quadrant ∴ cos θ > 0
∴ θ = \(\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
∴ sin 2θ = 2 sin θ cos θ = 2 x \(\frac { 3 }{ 5 }\) x \(\frac { 4 }{ 5 }\) = \(\frac { 24 }{ 25 }\)
Thus, cos 2θ = 1 – 2 sin² θ
= 1 – 2(\(\frac { 3 }{ 5 }\))² = 1 – 2 x \(\frac { 9 }{ 25 }\) = \(\frac { 7 }{ 25 }\)
and tan 2θ = \(\frac{\sin 2 \theta}{\cos 2 \theta}=\frac{\frac{24}{25}}{\frac{7}{25}}=\frac{24}{7}\)
(ii) Given sin θ = \(\frac { 3 }{ 5 }\)
since θ lies in 2nd quadrant ∴ cos θ < 0
(iii) given sin θ = – \(\frac { 1 }{ 2 }\) since θ lies in IVth quadrant ∴ cos θ > 0
(iv) Given tan θ = – \(\frac { 1 }{ 5 }\)
since θ lies in 2nd quadrant,
sin θ > 0 and cos θ < 0
Question 3.
ABC is an acute-angled triangle inscribed in a circle of radius 5 cm and centre O. The sine of angle A is equal to \(\frac { 3 }{ 5 }\).
Calculate without using tables :
(i) the length of BC
(ii) sin OBC
(iii) sin BOC
(iv) cos BOC.
Solution:
(i) Since ∠BOC = 2 x ∠BAC = 2A
[We know that angle subtended by an arc at the centre is twice the angle subtended by it on circumference of a circle]
Since AD ⊥ BC ∴ OD ⊥ BC
Thus, ∠BOD = ∠COD = A
In ∆BOD, \(\frac { BD }{ OB }\) = sin A
⇒ BD = 5 sin A
We know that ⊥ drawn from centre of circle bisects the chord.
∴ BC = 2BD = 2 x 5 sin A
⇒ BC = 10 sin A = 10 x \(\frac { 3 }{ 5 }\) = 6
(ii) From figure ; B = 90° – A
∴ sin ∠OBC = sin (90° – A) = cos A
= + \(\sqrt{1-\sin ^2 A}\)
Thus sin ∠BOC = \(\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
(iii) sin ∠BOC = sin 2A = 2 sin A cos A
= 2 x \(\frac { 3 }{ 5 }\) x \(\frac { 4 }{ 5 }\) = \(\frac { 24 }{ 25 }\)
(iv) cos BOC = cos 2A = 1 – 2 sin² A
= 1 – 2 x (\(\frac { 3 }{ 5 }\))²
= 1 – 2 x \(\frac { 9 }{ 25 }\) = \(\frac { 7 }{ 25 }\)
Question 4.
Derive functions of 120° from functions of 60° and check by using relations between functions of supplementary angles.
Solution:
sin 120° = sin (2 x 60°)
= 2 sin 60° cos 60°
= 2 x \(\frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\sqrt{3}}{2}\) [∵ sin 2θ = 2 sin θ cos θ]
Also sin (120°) = sin (180° – 60°) = sin 60° = \(\frac{\sqrt{3}}{2}\)
cos 120° = cos(2 x 60°) = 1 – 2 sin² 60°
= 1 – 2(\(\frac{\sqrt{3}}{2}\))²
= 1 – 2 x \(\frac { 3 }{ 4 }\) = 1 – \(\frac { 3 }{ 2 }\) = – \(\frac { 1 }{ 2 }\)
Also cos (120°) = cos (180° – 60°)
= – cos 60° = – \(\frac { 1 }{ 2 }\)
This verifies the result.
tan 120° = tan(2 x 60°) = \(\frac{2 \tan 60^{\circ}}{1-\tan ^2 60^{\circ}}\)
= \(\frac{2 \times \sqrt{3}}{1-(\sqrt{3})^2}=\frac{2 \sqrt{3}}{1-3}=+\frac{2 \sqrt{3}}{-2}=-\sqrt{3}\)
Also tan 120°= tan (180° – 60°)
= – tan 60° = – 6\(\sqrt{3}\)
This verifies the result.
This verifies the result.
Question 5.
If sin θ = a and sin 2θ = b, find an expression for cos 6 in terms of a and b. hence find a relation between a and b not involving θ.
Solution:
Given sin θ = a and sin 2θ = b …(1)
⇒ b = 2 sin θ cos θ = 2 x a cos θ [using (1)]
⇒ cos θ = \(\frac { b }{ 2a }\) … (2)
squaring and adding eqn. (1) and (2); we have
sin² θ + cos² θ = a² + (\(\frac { b }{ 2a }\))² = a² + \(\frac { b² }{ 4a² }\)
⇒ 1 = a² + \(\frac { b² }{ 4a² }\) ⇒ 4a4 + b² = 4a²
which is the required relation between a and b.
Question 6.
(i) Given that tan A = \(\frac { 1 }{ 5 }\), find the values of tan 2A, tan 4A and tan (45° – 4A).
(ii) If A is an obtuse angle whose sine is \(\frac { 5 }{ 13 }\) and B is an acute angle whose tangent is \(\frac { 3 }{ 4 }\), without using tables find the values of
(a) sin 2B, (b) tan (A – B).
Solution:
(i) Given tan A = \(\frac { 1 }{ 5 }\)
(ii) Given sin A = \(\frac { 5 }{ 13 }\) and tan B = \(\frac { 3 }{ 4 }\)
since B be an acute angle
∴ sin B, cos B > 0
(a) sin 2B = 2 sin B cos B = 2 x \(\frac { 4 }{ 5 }\) x \(\frac { 3 }{ 5 }\) = \(\frac { 24 }{ 25 }\)
(b) since A be an obtuse angle ∴ tan A < 0 and cos A < 0
Question 7.
Express
(i) cos 6α in terms of functions of 5α ;
(ii) sin 10 in terms of functions of 5θ ;
(iii) tan 8α in terms of tan 4α;
(iv) cos 2θ in terms of cos 4θ ;
(v) tan 4Φ in terms of cos 8Φ;
(vi) sin\(\frac { 5θ }{ 2 }\) in terms of cos 5θ ;
(vii) cos 20θ in terms of sin 5θ.
Solution:
Question 8.
Using the half angle formulas, find the exact value of (i) sin 15° (ii) sin 292 \(\frac { 1° }{ 2 }\).
Solution:
Question 9.
In the triangle ABC, in which C is the right angle, prove that:
sin 2A = \(\frac { 2ab }{ c² }\), cos 2A = \(\frac { b² – a² }{ c² }\),
sin \(\frac { A }{ 2 }\) = \(\sqrt{\frac{c-b}{2 c}}\), cos \(\frac { A }{ 2 }\) = \(\sqrt{\frac{c+b}{2 c}}\).
Solution:
From right angled ∆ACB, we have
Question 10.
If cos α = \(\frac { 3 }{ 5 }\), cos ß = \(\frac { 4 }{ 5 }\), find the value of cos \(\frac { α-ß }{ 2 }\), assuming α and ß to be acute angles.
Solution:
Given cos α = \(\frac { 3 }{ 5 }\), cos ß = \(\frac { 4 }{ 5 }\)
since α, ß are acute angles
∴ sin α, sin ß > 0
Question 11.
Given that cos\(\frac { A }{ 2 }\) = \(\frac { 12 }{ 13 }\), calculate without the use of tables, the values of sin A, cos A and tan A.
Solution:
Question 12.
Given that tan x = \(\frac { 12 }{ 5 }\), cos y = \(\frac { -3 }{ 5 }\), and the angles x and y are in the same quadrants, calculate without the use of tables the value of
(i) sin(x + y).
(ii) cos \(\frac { y }{ 2 }\).
Solution:
Given tan x = \(\frac { 12 }{ 5 }\) ; cosy = – \(\frac { 3 }{ 5 }\)
⇒ \(\sin y=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}\)
since x and y lies in same quadrant
∴ x and y lies in Illrd quadrant. [ ∵ tan x > 0]
∴ sin x < 0 and cos x < 0
Question 13.
Given tan sin² = sin α cos α, show that cos 2ß = 2 cos² [\(\frac { π }{ 4 }\) + α].
Solution:
Given sin² ß = sin α cos α
⇒ \(\frac{1-\cos 2 \beta}{2}\) = sin α cos α
⇒ 1 – cos 2ß = 2 sin α cos α
⇒ 1 – cos 2ß = sin 2α
⇒ cos 2ß = 1 – sin 2α …(1)
Also 2 cos²\(\left(\frac{\pi}{4}+\alpha\right)=2\left[\frac{1+\cos 2\left(\frac{\pi}{4}+\alpha\right)}{2}\right]\) [∵ cos² α = \(\frac{1+\cos 2 \theta}{2}\) ]
= 1 + cos [\(\frac { π }{ 2 }\) + 2α] = 1 – sin 2α … (2)
∴ from eqn. (1) and eqn. (2); we have
cos 2ß = 2 cos² (\(\frac { π }{ 4 }\) + α)
Question 14.
Derive formulas for the following in terms of functions of 2θ and then of θ.
(i) sin 4θ, (ii) cos 4θ, (iii) tan 4θ.
Solution:
(i) sin 4θ = sin (2 x 2θ) = 2 sin 2θ cos 2θ
= 2 (2 sin θ cos θ) (cos² θ – sin² θ)
= 4 sin θ cos θ (cos² θ – sin² θ)
(ii) cos 4θ = cos (2 x 2θ) = 2 cos² 2θ – 1
= [2 (2 cos² θ – 1)² – 1]
= 2 (4 cos4 θ – 4 cos² θ + 1) – 1
= 8 cos4 θ – 8 cos² θ + 1
(iii) tan 4θ = tan (2 x 2θ) = \(\frac{2 \tan 2 \theta}{1-\tan ^2 2 \theta}\)
= \(\frac{2\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)}{1-\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)^2}\)
= \(\frac{4 \tan \theta\left(1-\tan ^2 \theta\right)}{\left(1-\tan ^2 \theta\right)^2-4 \tan ^2 \theta}\)
= \(\frac{4 \tan \theta\left(1-\tan ^2 \theta\right)}{\tan ^4 \theta-2 \tan ^2 \theta+1-4 \tan ^2 \theta}\)
= \(\frac{4 \tan \theta\left(1-\tan ^2 \theta\right)}{\tan ^4 \theta-6 \tan ^2 \theta+1}\)
Question 15.
If sin α = \(\frac { 3 }{ 5 }\), find the value of
(i) sin 3α, (ii) cos 3α, (iii) tan 3α.
Solution:
Given α = \(\frac { 3 }{ 5 }\)
(ii) When α lies in first quadrant
(iii) When a lies in first quadrant ∴ tan α > 0
Question 16.
If 2 cos θ = x + \(\frac { 1 }{ x }\), prove that 2 cos 3θ = x³ + \(\frac { 1 }{ x³ }\).
Solution:
Question 17.
Calculate without using tables tan 20° tan 40° tan 80°.
Solution:
