Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(d)

I.

Question 1.
Simplify
(i) \(\frac{\sin 2 \theta}{1+\cos 2 \theta}\)
Solution:
\(\frac{\sin 2 \theta}{1+\cos 2 \theta}\)
= \(\frac{2 \sin \theta \cos \theta}{2 \cos ^{2} \theta}\)
= tan θ

(ii) \(\frac{3 \cos \theta+\cos 3 \theta}{3 \sin \theta-\sin 3 \theta}\)
Solution:
\(\frac{3 \cos \theta+\cos 3 \theta}{3 \sin \theta-\sin 3 \theta}\)
= \(\frac{3 \cos \theta+4 \cos ^{3} \theta-3 \cos \theta}{3 \sin \theta-\left(3 \sin \theta-4 \sin ^{3} \theta\right)}\)
= \(\frac{4 \cos ^{3} \theta}{4 \sin ^{3} \theta}\)
= cot3θ

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d)

Question 2.
Evaluate the following.
(i) 6 sin 20° – 8 sin3 20°
Solution:
6 sin 20° – 8 sin320°
= 2(3 sin 20° – 4 sin320°)
= 2 sin (3 × 20)
= 2 sin 60°
= 2 \(\left[\frac{\sqrt{3}}{2}\right]\)
= √3

(ii) cos272° – sin254°
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q2(ii)

(iii) sin242° – sin212°
Solution:
sin (42 + 12) sin (42 – 12)
= sin 54° . sin 30°
= \(\left[\frac{\sqrt{5}+1}{4}\right] \frac{1}{2}\)
= \(\frac{\sqrt{5}+1}{8}\)

Question 3.
(i) Express \(\frac{\sin 4 \theta}{\sin \theta}\) interms of cos3θ and cos θ.
Solution:
consider sin 4θ = sin(3θ + θ)
= sin 3θ cos θ + cos 3θ sin θ
= (3 sin θ – 4 sin3θ) cos θ + (4 cos3θ – 3 cos θ) sin θ
= 3 sin θ cos θ – 4 sin3θ cos θ + 4 cos3θ sin θ – 3 cos θ sin θ
= 4 cos3θ sin θ – 4 sin3θ cos θ
= sin θ (4 cos3θ – 4 sin2θ cos θ)
\(\frac{\sin 4 \theta}{\sin \theta}=\frac{\sin \theta\left(4 \cos ^{3} \theta-4 \sin ^{2} \theta \cos \theta\right)}{\sin \theta}\)
= 4 cos3θ – 4(1 – cos2θ) cos θ
= 4 cos3θ – 4 cos θ + 4 cos3θ
= 8 cos3θ – 4 cos θ

(ii) Express cos6A + sin6A in terms of sin 2A.
Solution:
cos6A + sin6A
= (cos2A)3 + (sin2A)3
= (cos2A + sin2A)3 – 3 cos2A sin2A (cos2A + sin2A)
= 1 – 3 cos2A sin2A
= 1 – \(\frac{3}{4}\) (4 cos2A sin2A)
= 1 – \(\frac{3}{4}\) (sin22A)

(iii) Express \(\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}\) in terms of tan θ/2
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q3(iii)

Question 4.
(i) If sin α = \(\frac{3}{5}\), where \(\frac{\pi}{2}\) < α < π, evaluate cos 3α and tan 2α.
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q4(i)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q4(i).1

(ii) If cos A = \(\frac{7}{25}\) and \(\frac{3 \pi}{2}\) < A < 2π, then find the value of cot A/2.
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q4(ii)

(iii) If 0 < θ < \(\frac{\pi}{8}\), show that \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 \theta)}}}=2 \cos (\theta / 2)\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q4(iii)

Question 5.
Find the extreme values of
(i) cos 2x + cos2x
Solution:
cos 2x + cos2x
= 2 cos2x – 1 + cos2x
= 3 cos2x – 1
-1 ≤ cos x ≤ 1
0 ≤ cos2x ≤ 1
0 ≤ 3 cos2x ≤ 3
-1 ≤ 3 cos2x – 1 ≤ 2
maximum value = 2
minimum value = -1

(ii) 3 sin2x + 5 cos2x
Solution:
3 sin2x + 5 cos2x
= 3(1 – cos2x) + 5 cos2x
= 3 – 3 cos2x + 5 cos2x
= 3 + 2 cos2x
-1 ≤ cos x ≤ 1
0 ≤ cos2x ≤ 1
0 ≤ 2 cos2x ≤ 2
3 ≤ 3 + 2 cos2x ≤ 5
maximum value = 5
minimum value = 3

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d)

Question 6.
If a ≤ cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6 ≤ b, then find the largest value of a and smallest value of b.
Solution:
a ≤ cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6 ≤ b
consider cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6
= cos θ + 3√2[sin θ cos \(\frac{\pi}{4}\) + cos θ sin \(\frac{\pi}{4}\)] + 6
= cos θ + 3√2 sin θ \(\frac{1}{\sqrt{2}}\) + 3√2 cos θ \(\frac{1}{\sqrt{2}}\) + 6
= cos θ + 3 sin θ + 3 cos θ + 6
= 4 cos θ + 3 sin θ + 6
∴ a = 4, b = 3, c = 6
maximum value = c + \(\sqrt{a^{2}+b^{2}}\)
= 6 + \(\sqrt{16+9}\)
= 6 + 5
= 11
minimum value = c – \(\sqrt{a^{2}+b^{2}}\)
= 6 – 5
= 1

Question 7.
Find the periods for the following functions.
(i) cos4x
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q7(i)

(ii) \(2 \sin \left[\frac{\pi x}{4}\right]+3 \cos \left[\frac{\pi x}{3}\right]\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q7(ii)

(iii) sin2x + 2 cos2x
Solution:
Let f(x) = sin2x + 2 cos2x
= 1 – cos2x + 2 cos2x
= 1 + cos2x
= 1 + \(\frac{1+\cos 2 x}{2}\)
∴ period of cos 2x = \(\frac{2 \pi}{2}\) = π
∴ period of f(x) = π

(iv) \(2 \sin \left[\frac{\pi}{4}+x\right] \cos x\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) I Q7(iv)

(v) \(\frac{5 \sin x+3 \cos x}{4 \sin 2 x+5 \cos x}\)
Solution:
Let f(x) = \(\frac{5 \sin x+3 \cos x}{4 \sin 2 x+5 \cos x}\)
period of sin x = 2π
period of cos x = 2π
period of sin 2x = \(\frac{2 \pi}{2}\) = π
period of cos x = 2π
L.C.M. of (2π, 2π, π, 2π) = 2π
∴ period of f(x) = 2π

II.

Question 1.
(i) If 0 < A < \(\frac{\pi}{4}\) and cos A = \(\frac{4}{5}\), find the values of sin 2A and cos 2A.
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q1(i)

(ii) For what values of A in the first quadrant, the expression \(\frac{\cot ^{3} A-3 \cot A}{3 \cot ^{2} A-1}\) is positive?
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q1(ii)

(iii) Prove that \(\frac{\cos 3 A+\sin 3 A}{\cos A-\sin A}=1+2 \sin A\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q1(iii)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q1(iii).1

Question 2.
(i) Prove that \(\cot \left[\frac{\pi}{4}-\theta\right]=\frac{\cos 2 \theta}{1-\sin 2 \theta}\) and hence find the value of cot 15°.
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q2(i)

(ii) If θ lies in third quadrant and sin θ = \(\frac{-4}{5}\), find the values of cosec (θ/2) and tan (θ/2).
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q2(ii)

(iii) If 450° < θ < 540° and sin θ = \(\frac{12}{13}\), then calculate sin (θ/2) and cos (θ/2)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q2(iii)

(iv) Prove that \(\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}}=\frac{4}{\sqrt{3}}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q2(iv)

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d)

Question 3.
Prove that
(i) \(\frac{\sin 2 A}{(1-\cos 2 A)} \cdot \frac{(1-\cos A)}{\cos A}=\tan \frac{A}{2}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q3(i)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q3(i).1

(ii) \(\frac{\sin 2 x}{(\sec x+1)} \cdot \frac{\sec 2 x}{(\sec 2 x+1)}\) = \(\tan \frac{x}{2}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q3(ii)

(iii) \(\frac{\left(\cos ^{3} \theta-\cos 3 \theta\right)}{\cos \theta}+\frac{\left(\sin ^{3} \theta+\sin 3 \theta\right)}{\sin \theta}=\mathbf{3}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q3(iii)

Question 4.
(i) Show that cos A = \(\frac{\cos 3 A}{(2 \cos 2 A-1)}\). Hence find the value of cos 15°.
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q4(i)

(ii) Show that sin A = \(\frac{\sin 3 A}{1+2 \cos 2 A}\). Hence find the value of sin 15°.
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q4(ii)

(iii) Prove that tan α = \(\frac{\sin 2 \alpha}{1+\cos 2 \alpha}\) and hence deduce the values tan 15° and tan 22\(\frac{1^{\circ}}{2}\).
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q4(iii)

Question 5.
Prove that
(i) \(\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=4\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q5(i)

(ii) √3 cosec 20° – sec 20° = 4
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q5(ii)

(iii) tan 9° – tan 27° – cot 27° + cot 9° = 4
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q5(iii)

(iv) If \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\), then prove that a sin 2α + b cos 2α = b
Solution:
Given \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\)
b sin α = a cos α
L.H.S. = a sin 2α + b cos 2α
= a 2 sin α cos α + b (1 – 2 sin2α)
= 2 sin α (a cos α) + b – 2b sin2α
= 2 sin α (b sin α) + b – 2b sin2α
= 2b sin2α + b – 2b sin2α
= b

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d)

Question 6.
(i) In a ∆ABC; if \(\tan \frac{A}{2}=\frac{5}{6}\) and tan\(\tan \frac{B}{2}=\frac{20}{37}\), then show that \(\tan \frac{C}{2}=\frac{2}{5}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q6(i)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q6(i).1

(ii) If cos θ = \(\frac{5}{13}\) and 270° < θ < 360°, evaluate sin (θ/2) and cos (θ/2).
Solution:
cos θ = \(\frac{5}{13}\)
given 270 < θ < 360°
⇒ 135 < \(\frac{\theta}{2}\) < 180°
∴ θ lies in the fourth quadrant
θ/2 lies is second quadrant
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q6(ii)

(iii) If 180° < θ < 270° and sin θ = \(\frac{-4}{5}\) calculate sin[θ/2] cos[θ/2]
Solution:
Given sin θ = \(\frac{-4}{5}\)
⇒ cos θ = \(\frac{-3}{5}\)
given 180 < θ < 270
∴ θ in the III quadrant
Now 90 < \(\frac{\theta}{2}\) < 135
∴ \(\frac{\theta}{2}\) is in II quadrant
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q6(iii)

Question 7.
(i) Prove that \(\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}\) = 2
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q7(i)

(ii) Show that \(\cos ^{4}\left(\frac{\pi}{8}\right)+\cos ^{4}\left(\frac{3 \pi}{8}\right)+\cos ^{4}\left(\frac{5 \pi}{8}\right)+\cos ^{4}\left(\frac{7 \pi}{8}\right)\) = \(\frac{3}{2}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q7(ii)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) II Q7(ii).1

III.

Question 1.
(i) If tan x + tan(x + \(\frac{\pi}{3}\)) + tan(x + \(\frac{2 \pi}{3}\)) = 3, show that tan 3x = 1
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q1(i)

(ii) Prove that \(\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \cdot \sin \frac{3 \pi}{5} \cdot \sin \frac{4 \pi}{5}\) = \(\frac{5}{16}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q1(ii)

(iii) Show that \(\cos ^{2}\left(\frac{\pi}{10}\right)+\cos ^{2}\left(\frac{2 \pi}{5}\right)+\cos ^{2}\left(\frac{3 \pi}{5}\right)\) + \(\cos ^{2}\left(\frac{9 \pi}{10}\right)\) = 2
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q1(iii)

Question 2.
(i) \(\frac{1-\sec 8 \alpha}{1-\sec 4 \alpha}=\frac{\tan 8 \alpha}{\tan 2 \alpha}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q2(i)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q2(i).1

(ii) \(\left[1+\cos \frac{\pi}{10}\right]\left[1+\cos \frac{3 \pi}{10}\right]\left[1+\cos \frac{7 \pi}{10}\right]\) \(\left[1+\cos \frac{9 \pi}{10}\right]=\frac{1}{16}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q2(ii)

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d)

Question 3.
(i) Prove that \(\cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7} \cdot \cos \frac{8 \pi}{7}=\frac{1}{8}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q3(i)

(ii) \(\cos \frac{\pi}{11} \cdot \cos \frac{2 \pi}{11} \cdot \cos \frac{3 \pi}{11} \cdot \cos \frac{4 \pi}{11}\) \(\cos \frac{5 \pi}{11}=\frac{1}{32}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q3(ii)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q3(ii).1

Question 4.
(i) If cos α = \(\frac{3}{5}\) and cos β = \(\frac{5}{13}\) and α, β are acute angles, then prove that
(a) \(\sin ^{2}\left[\frac{\alpha-\beta}{2}\right]=\frac{1}{65}\)
(b) \(\cos ^{2}\left[\frac{\alpha+\beta}{2}\right]=\frac{16}{65}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q4(i)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q4(i).1

(ii) If A is not an integral multiple of π, prove that cos A . cos 2A . cos 4A . cos 8A = \(\frac{\sin 16 A}{16 \sin A}\) and hence deduce that \(\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}=\frac{1}{16}\)
Solution:
Take 16 sin A {cos A . cos 2A . cos 4A . cos 8A}
= 8(2 sin A . cos A) cos 2A . cos 4A . cos 8A
= 8 sin 2A . cos 2A . cos 4A . cos 8A
= 4(2 sin 2A . cos 2A) . cos 4A . cos 8A
= 4 sin 4A . cos 4A . cos 8A
= 2 (2 sin 4A . cos 4A) . cos 8A
= 2 sin 8A . cos 8A
= sin (16A)
∴ 16 sin A {cos A . cos 2A . cos 4A . cos 8A} = sin (16A)
∴ cos A . cos 2A . cos 4A . cos 8A = \(\frac{\sin (16 A)}{16 \sin A}\)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(d) III Q4(ii)