ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Very short answer/objective questions (1 to 3) :

Question 1.
If sin x = \(\frac{2}{3}\), find the value of cos 2x.
Solution:
Given sin x = \(\frac{2}{3}\)
∴ cos 2x = 1 – 2 sin² x
= 1 – 2 \(\left(\frac{2}{3}\right)^2\)
= \(1-2 \times \frac{4}{9}\)
= \(1-\frac{8}{9}=\frac{1}{9}\)

Question 2.
If cos x = – \(\frac{2}{5}\), find the value of cos 2x.
Solution:
Given cos x = – \(\frac{2}{5}\)
Now we know that,
cos 2x = 2 cos² x – 1
cos 2x = \(2\left(-\frac{2}{5}\right)^2-1\)
= \(\frac{8}{25}-1\)
= \(\frac{8-25}{25}\)
= \(-\frac{17}{25}\)

Question 3.
If tan x = \(\frac{1}{2}\) find the values of :
(i) tan 2x
(ii) sin 2x
(iii) cos 2x
Solution:
Given tan x = \(\frac{1}{2}\)

(i) tan 2x = \(\frac{2 \tan x}{1-\tan ^2 x}\)
= \(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\)
= \(\frac{1}{1-\frac{1}{4}}\)
= \(\frac{1}{\frac{3}{4}}=\frac{4}{3}\)

(ii) sin 2x = \(\frac{2 \tan x}{1+\tan ^2 x}\)
= \(\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}\)
= \(\frac{1}{1+\frac{1}{4}}\)
= \(\frac{1}{\frac{5}{4}}=\frac{4}{5}\)

(iii) cos 2x = \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\)
= \(\frac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2}\)
= \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\)
= \(\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{5}\)

Question 4.
If sin x = \(\frac{2}{3}\) find the value of sin 3x.
Solution:
Given sin x = \(\frac{2}{3}\)
Now, sin 3x = 3 sin x – 4 sin 3x
= \(3 \times \frac{2}{3}-4\left(\frac{2}{3}\right)^3\)
= \(2-4 \times \frac{8}{27}\)
= 2 – \(\frac{32}{27}\)
= \(\frac{54-32}{27}=\frac{22}{27}\)

Question 5.
If cos x = – \(\frac{2}{5}\), find the value of cos 3x.
Solution:
Given cos x = – \(\frac{2}{5}\)
We know that,
cos 3x = 4 cos³ x – 3 cos x
∴ cos 3x = \(4\left(-\frac{2}{5}\right)^3-3\left(-\frac{2}{5}\right)\)
= \(-\frac{32}{125}+\frac{6}{5}\)
= \(\frac{-32+150}{125}=\frac{118}{125}\)

Question 6.
Prove that :
(i) cos 2x + 2 sin² x = 1
(ii) (cos x- sin x)² = 1 – sin 2x
(iii) \(\frac{\cos 2 x}{\cos x-\sin x}\) = cos x + sin x
Solution:
(i) cos 2x = cos² x – sin² x
= 1 – sin² x – sin² x
= 1 – 2 sin2x
= cos² x + 2 sin² x = 1

(ii) LH.S. = (cos x – sin x)²
= cos² x + sin² x – 2 sin x cos x
= 1 – sin 2x
= R.H.S.

(iii) L.H.S. = \(\frac{\cos 2 x}{\cos x-\sin x}\)
= \(\frac{\cos ^2 x-\sin ^2 x}{\cos x-\sin x}\)
= \(\frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x}\)
= cos x + sin x
= R.H.S.

Question 7.
Prove that :
(i) \(\frac{\sin 2 x}{1+\cos 2 x}\) = tan x
(ii) \(\frac{\sin 2 x}{1-\cos 2 x}\) = cot x
(iii) \(\frac{1-\cos 2 x}{1+\cos 2 x}\) = tan² x
(iv) \(\frac{1+\sin 2 x-\cos 2 x}{1+\sin 2 x+\cos 2 x}\) = tan x
Solution:
(i) L.H.S. = \(\frac{\sin 2 x}{1+\cos 2 x}\)
= \(\frac{2 \sin x \cos x}{1+2 \cos ^2 x-1}\)
= \(\frac{2 \sin x \cos x}{2 \cos ^2 x}\)
= tan x
= R.H.S.

(ii) L.H.S. = \(\frac{\sin 2 x}{1-\cos 2 x}\)
= \(\frac{2 \sin x \cos x}{1-\left(1-2 \sin ^2 x\right)}\)
= \(\frac{2 \sin x \cos x}{2 \sin ^2 x}\)
= cot x
= R.H.S.

(iii) L.H.S. = \(\frac{1-\cos 2 x}{1+\cos 2 x}\)
= \(\frac{1-\left(1-2 \sin ^2 x\right)}{1+2 \cos ^2 x-1}\)
= \(\frac{2 \sin ^2 x}{2 \cos ^2 x}\)
= tan^{2</sup x
= R.H.S.}

(iv) L.H.S. = \(\frac{1+\sin 2 x-\cos 2 x}{1+\sin 2 x+\cos 2 x}\)

Question 8.
Prove that :
(i) \(\frac{\cos ^3 x-\sin ^3 x}{\cos x-\sin x}=\frac{1}{2}\) (2 + sin 2x)
(ii) \(\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}\) = tan x
Solution:
(i) L.H.S. = \(\frac{\cos ^3 x-\sin ^3 x}{\cos x-\sin x}\)
= \(\frac{(\cos x-\sin x)\left(\cos ^2 x+\cos x \sin x+\sin ^2 x\right)}{\cos x-\sin x}\)
= cos² x + sin² x + sin x cos x
= 1 + \(\frac{1}{2}\) (2 sin x cos x)
= 1 + \(\frac{1}{2}\) sin 2x
= \(\frac{1}{2}\) (2 + sin 2x)
= R.H.S.

(ii) L.H.S. = \(\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}\)
= \(\frac{2 \sin ^2 x+\sin x}{2 \sin x \cos x+\cos x}\)
= \(\frac{\sin x(2 \sin x+1)}{\cos x(2 \sin x+1)}\)
= \(\frac{\sin x}{\cos x}\)
= tan x
= R.H.S.

Question 9.
(i) \(2 \cos 22 \frac{1}{2}^{\circ} \sin 22 \frac{1}{2}^{\circ}\)
(ii) 2 cos² 15° – 1
(iii) 8 cos³ 20° – 6 cos 20°
(iv) 3 sin 40° – 4 sin³ 40°
Solution:
(i) \(2 \cos 22 \frac{1}{2}^{\circ} \sin 22 \frac{1}{2}^{\circ}\)
= sin (2 × 22 \(\frac{1}{2}^{\circ}\))
= sin 45°
= \(\frac{1}{\sqrt{2}}\)

(ii) 2 cos² 15° – 1
= cos (2 × 15°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)

(iii) 8 cos3 20° – 6 cos 20° 2 [4 cos3 20° – 3 cos 20°]
= 2 cos (3 × 20°)
= 3 cos 60°
= \(\frac{3}{2}\)

(iv) 3 sin 40° – 4 sin³ 40°
= sin (3 × 40°)
= sin 120°
= sin (180° – 60°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)

Question 10.
Prove that :
(i) cos \(\frac{\pi}{5}\) + cos \(\frac{3 \pi}{5}\) = \(\frac{1}{2}\)
(ii) sin² 24° – sin² 6° = \(\frac{\sqrt{5}-1}{8}\)
(iii) sin² 72° – sin² 60° = \(\frac{\sqrt{5}-1}{8}\)
(iv) sin² 72° – cos² 30° = \(\frac{\sqrt{5}-1}{8}\)
Solution:
(i) cos \(\frac{\pi}{5}\) + cos \(\frac{3 \pi}{5}\)
= cos 36° + cos 108°
= cos 36° + sin (90° + 18°)
= cos 36° – sin 18°
[∵ cos (90° + θ) = – sin θ]
= \(\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-1}{4}\)
= \(\frac{\sqrt{5}+1-\sqrt{5}+1}{4}\)
= \(\frac{2}{4}=\frac{1}{2}\)

(ii) sin² 24° – sin² 6° = sin (24° + 6°) sin (24° – 6°)
[∵ sin (A + B) sin (A – B) = sin² A – sin² B]
= sin 30° sin 18°
= \(\frac{1}{2} \times \frac{\sqrt{5}-1}{4}\)
= \(\frac{\sqrt{5}-1}{8}\)

(iii) L.H.S. = sin² 72° – sin² 60°

(iv) L.H.S. = sin² 72° – cos² 30°

Long answer questions (11 to 26) :

Question 11.
Prove that cos x cos 2x cos 4x cos 8x = \(\frac{\sin 16 x}{16 \sin x}\)
Solution:
L.HS.= cos x cos 2x cos 4x cos 8x
= \(\frac{1}{2 \sin x}\) [2sin x cos x] cos 2x cos 4x cos 8x
= \(\frac{1}{2 \sin x}\) (sin 2x cos 2x) cos 4x cos 8x
= \(\frac{1}{2^2 \sin x}\) (2 sin 2x cos 2x) cos 4x cos 8x
= \(\frac{1}{4 \sin x}\) (sin 4x cos 4x) cos Sx
= \(\frac{1}{8 \sin x}\) (2 sin 4x cos 4x) cos 8x
= \(\frac{1}{8 \sin x}\) (sin 8x cos 8x)
[∵ sin 2θ = 2 sin θ cos θ]
= \(\frac{1}{16 \sin 2 x}\) (2 sin 8x cos 8x)
= \(\frac{\sin 16 x}{16 \sin x}\)
= R.H.S.

Question 12.
Prove that \(\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}\) = 2 tan 2x.
Solution:

Question 13.
Prove that : 1 + cos² 2x = 2 (cos⁴ x + sin⁴ x).
Solution:
L.H.S. = 1 + cos² 2x
= 1 + (cos 2x)²
= 1 + (cos² x – sin² x)²
= (cos² x + sin² x)² + (cos² x – sin² x)²
= cos⁴ x + sin⁴ x + 2 sin² x cos² x + cos⁴ x + sin⁴ x – 2 sin² x cos² x
= 2 (cos⁴ x + sin⁴ x)
= R.H.S.

Question 14.
Prove that :
(i) cot \(\frac{x}{2}\) – tan \(\frac{x}{2}\) = 2 cot x
(ii) \(\sqrt{\frac{1+\sin x}{1-\sin x}}\) = tan (\(\left(\frac{\pi}{4}+\frac{x}{2}\right)\))
(iii) tan (\(\left(\frac{\pi}{4}+\frac{x}{2}\right)\)) = tan x + sec x
(iv) \(\frac{\cos 2 x}{1+\sin 2 x}\)
Solution:
(i) L.H.S. = cot \(\frac{x}{2}\) – tan \(\frac{x}{2}\)

(ii)

(iii) By using (ii),

(iv) L.H.S. = \(\frac{\cos 2 x}{1+\sin 2 x}\)

Question 15.
Prove that tan x + cot x = 2 cosec 2x and deduce that tan 75° + cot 75° = 4.
Solution:
L.H.S. = tan x + cot x
= \(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\)
= \(\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\)
= \(\frac{1}{\sin x \cos x}\)
= \(\frac{2}{2 \sin x \cos x}\)
= \(\frac{2}{\sin 2 x}\)
= 2 cosec 2x
= R.H.S.
Thus, tan x + cot x = 2 cosec 2x …………………(1)
putting x = 75° in eqn. (1) ; we have
tan 75° + cot 75° = 2 cosec (2 × 750)
= 2 cosec (150°)
= 2 cosec (180° – 30°)
= 2 cosec 30°
= 2 × 2 = 4

Question 16.
Prove that :
(i) tan x + \(\tan \left(\frac{\pi}{3}+x\right)+\tan \left(\frac{2 \pi}{3}+x\right)\) = 3 tan 3x
(ii) cos x \(\cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)\) = \(\frac{1}{4}\) cos 3x.
Solution:
(i) L.H.S. = tan x + \(\tan \left(\frac{\pi}{3}+x\right)+\tan \left(\frac{2 \pi}{3}+x\right)\)

(ii) L.H.S. = cos x \(\cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)\)

Question 17.
(i) If \(\frac{\sin x}{a}=\frac{\cos x}{b}\), prove that a sin 2x + b cos 2x = b.
(ii) If tan² x = 2 tan² y + 1, show that cos 2x + sin² y = 0.
Solution:
Given \(\frac{\sin x}{a}=\frac{\cos x}{b}\)
⇒ tan x = \(\frac{a}{b}\) …………………(1)
L.H.S. = a sin 2x + b cos 2x

(ii) Given tan² x = 2 tan² y + 1 ………………(1)
L.H.S. = cos 2x + sin² y

Question 18.
If 2 cos y = x + \(\frac{1}{x}\), prove that cos 2y = \(\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)\).
Solution:
Given 2 cos y = x + \(\frac{1}{x}\) ………………….(1)
L.H.S. = cos 2y
= 2 cos² y – 1

Question 19.
If tan y = 3 tan x, prove that tan (x + y) = \(\frac{2 \sin 2 y}{1+2 \cos 2 y}\).
Solution:
Given tan y = 3 tan x …………………….(1)
L.H.S. = tan (x + y)

Question 20.
Prove that :
(i) cos 4x = 1 – 8 cos 2x + 8 cos 4x
(ii) sin 5x = 5 sin x – 20 sin³ x + 16 sin⁵ x.
Solution:
(i) L.H.S. = cos 4x
= cos (2 × 2x)
= 2 cos² 2x – 1
[∵ cos 2θ = 2 cos² θ – 1]
= 2 [(2 cos² x – 1)²]
= 2 [4 cos⁴ x ± 1 – 4 cos² x] – 1
= 8 cos⁴ x + 2 – 8 cos² x – 1
= 8 cos⁴ x – 8 cos² x + 1
= R.H.S.

(ii) L.H.S. = sin 5x
= sin (2x + 3x)
= sin 2x cos 3x + cos 2x sin 3x
= 2 sin x cos x (4 cos³ x – 3 cos x) + (1 – 2 sin² x) (3 sin x – 4 sin³ x)
= 2 sin x cos² x (4 cos² x – 3) + (1 – 2 sin² x) (3 sin x – 4 sin³ x)
= 2 sin x (1 – sin² x) [4 (1 – sin² x) – 3] + (1 – 2 sin² x) (3 sin x-4 sin³ x)
= 2 sin x (1 – sin² x) (1 – 4 sin² x) + (1 – 2 sin² x) (3 sin x – 4 sin³ x)
= 2 sin x (1 – 5 sin² x + 4sin⁴ x) + (3 sin x – 10 sin³ x + 8 sin⁵ x)
= 5 sin x – 20 sin³ x + 16 sin⁵ x
= R.H.S.

Question 21.
Prove that cos³ x + cos³ (\(\frac{2 \pi}{3}\) + x) + cos³ (\(\frac{2 \pi}{3}\) – x) = \(\frac{3}{4}\) cos 3x.
Solution:
L.H.S. = cos³ x + cos³ (\(\frac{2 \pi}{3}\) + x) + cos³ (\(\frac{2 \pi}{3}\) – x)
[since cos 3x = 4 cos³ x – 3 cos x
⇒ cos³ x = \(\frac{1}{4}\) (cos 3x + 3 cos x)]

Question 22.
Prove that :
(i) sin 6° sin 42° sin 66° sin 78° = \(\frac{1}{16}\)
(ii) \(\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}=-\frac{1}{16}\)
Solution:
(i) L.H.S. = sin 6° sin 42° sin 66° sin 78°
= \(\frac{1}{4}\) (2 sin 66° sin 6°) (2 sin 78° sin 42°)
= \(\frac{1}{4}\) [cas (66° – 6°) – cos (66° + 6°)] [cos (78° – 42°) – cos (78° + 42°)]
= \(\frac{1}{4}\) [cos 60° – cos 72°] [cos 36° – cos 120°]
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – cos (90° – 18°)] [cos 36° – cos (180° – 60°)]
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – sin 18°] [cos 36° + cos 60°]
= \(\frac{1}{4}\left[\frac{1}{2}-\frac{\sqrt{5}-1}{4}\right]\left[\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right]\)
= \(\frac{1}{4}\left[\frac{2-\sqrt{5}+1}{4}\right]\left[\frac{\sqrt{5}+1+2}{4}\right]\)
= \(\frac{1}{4}\left[\frac{3-\sqrt{5}}{4}\right]\left[\frac{3+\sqrt{5}}{4}\right]\)
= \(\frac{1}{64}\) (9 – 5)
= \(\frac{4}{64}=\frac{1}{16}\)
= R.H.S.

(ii) L.H.S. = \(\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\)
= = cos 36° cos 72° cos 144° cos 288°
[∵ \(\frac{\pi}{5}\) = 36°]
= cos 36° cos (90° – 18°) cos (180° – 36°) cos (270° + 1 8°)
= cos 36° sin 18° (- cos 36°) (sin 18°)
= – cos² 36° sin² 18°
= \(-\left(\frac{\sqrt{5}+1}{4}\right)^2\left(\frac{\sqrt{5}-1}{4}\right)^2\)
= – \(\frac{[(\sqrt{5}+1)(\sqrt{5}-1)]^2}{16 \times 16}\)
= – \(\frac{(5-1)^2}{16 \times 16}\)
= – \(\frac{1}{16}\)
= R.H.S.

Aliter:

Question 23.
Given that cos \(\frac{x}{2}=\frac{12}{13}\) and x lies in first quadrant, calculate without the use of tables, the values of sin x, cos x and tan x.
Solution:
Given cos \(\frac{x}{2}=\frac{12}{13}\)
We know that, sin² \(\frac{x}{2}\) + cos² \(\frac{x}{2}\) = 1
⇒ sin \(\frac{x}{2}\) = ± \(\sqrt{1-\cos ^2 \frac{x}{2}}\)
⇒ sin \(\frac{x}{2}\) = ± \(\sqrt{1-\frac{144}{169}}= \pm \frac{5}{13}\)
Since x lies in first quadrant.
∴ \(\frac{x}{2}\) also lies in first quadrant
∴ sin \(\frac{x}{2}\) > 0
Thus, sin \(\frac{x}{2}=\frac{5}{13}\)
Now sin x = 2 sin \(\frac{x}{2}\) cos \(\frac{x}{2}\)
= 2 × \(\frac{5}{13} \times \frac{12}{13}=\frac{120}{169}\)
and cos x = 2 cos² \(\frac{x}{2}\) – 1
= 2 \(\left(\frac{12}{13}\right)^2\) – 1
= \(\frac{288}{169}\) – 1
= \(\frac{119}{169}\)
Thus, tan x = \(\frac{\sin x}{\cos x}\)
= \(\frac{\frac{120}{169}}{\frac{119}{169}}=\frac{120}{119}\)

Question 24.
Given that tan x = \(\frac{12}{5}\), cos y = \(-\frac{3}{5}\) and x, y are in the same quadrant, calculate without the use of tables the values of
(i) sin (x + y)
(ii) cos \(\frac{y}{2}\)
Solution:
(i) Given tan x= \(\frac{12}{5}\), which is positive.
and cos y = \(-\frac{3}{5}\), which is negative
Thus x and y lies in third quadrant [since x andy lies in same quadrant]
We know that,
sec² x = 1 + tan² x
∴ sec² x = 1 + \(\frac{144}{25}=\frac{169}{25}\)
⇒ sec x = ± \(\frac{13}{5}\)
since x lies in 3rd quadrant
∴ sec x < 0
Thus, sec x = – \(\frac{13}{5}\)
⇒ cos x = – \(\frac{5}{13}\)
∴ sin x = tan x . cos x
= \(\frac{12}{5} \times-\frac{5}{13}=-\frac{12}{13}\)
Thus sin y = – \(\sqrt{1-\cos ^2 y}\)
= – \(\sqrt{1-\frac{9}{25}}=-\frac{4}{5}\)
[∵ y lies in 3rd quad.
∴ sin y < 0]

(ii) sin (x + y) = sin x cos y + cos x sin y
= \(\left(-\frac{12}{13}\right)\left(-\frac{3}{5}\right)+\left(-\frac{5}{13}\right)\left(-\frac{4}{5}\right)\)
= \(\frac{36}{65}+\frac{20}{65}=\frac{56}{65}\)

(iii) since y lies in third quadrant thus \(\frac{y}{2}\) lies in 2nd quadrant
∴ cos \(\frac{y}{2}\) < 0
Now cos \(\frac{y}{2}\) = – \(\sqrt{\frac{1+\cos y}{2}}\)
= – \(\sqrt{\frac{1+\left(-\frac{3}{5}\right)}{2}}\)
= \(-\sqrt{\frac{1}{5}}=-\frac{1}{\sqrt{5}}\)

Question 25.
Find sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) if
(i) tan x = – \(\frac{4}{3}\), x lies in second quadrant
(ii) cos x = – \(\frac{1}{3}\) x lies in third quadrant
Solution:
(i) Since 90° ≤ x ≤ 180° 45° ≤ \(\frac{x}{2}\) ≤ 90°
Thus \(\frac{x}{2}\) lies in first quadrant.
∴ sin \(\frac{x}{2}\), cos \(\frac{x}{2}\), tan \(\frac{x}{2}\) > 0
Given tan x = – \(\frac{4}{3}\)
since sec² x = 1 + tan² x
sec² x = 1 + \(\frac{16}{25}\)
= \(\frac{25}{9}\)
⇒ sec x = ± \(\frac{5}{3}\)
since x lies in 2nd quadrant
∴ sec x < 0
Thus, sec x = – \(\frac{5}{3}\)
⇒ cos x = – = ± \(\frac{3}{5}\)
∴ sin x = tan x cos x
= \(-\frac{4}{3} \times\left(-\frac{3}{5}\right)=\frac{4}{5}\)

(ii) since x lies in 3rd quadrant
∴ sin x < 0

Question 26.
If tan \(\frac{x}{2}=\sqrt{\frac{a-b}{a+b}} \tan \frac{\alpha}{2}\), prove that cos x = \(\frac{a \cos \alpha+b}{a+b \cos \alpha}\).
Solution:
Given tan \(\frac{x}{2}=\sqrt{\frac{a-b}{a+b}} \tan \frac{\alpha}{2}\) …………………….(1)