Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.4 Questions and Answers.
Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.4
Question 1.
Express the following as a sum or difference of two trigonometric functions.
i. 2sin 4x cos 2x
ii. 2sin \(\frac{2 \pi}{3}\) cos \(\frac{\pi}{2}\)
iii. 2cos 4θ cos 2θ
iv. 2cos 35° cos 75°
Solution:
i. 2sin 4x cos 2x = sin(4x + 2x ) + sin (4x – 2x)
= sin 6x + sin 2x
ii.
[Note: Answer given in the textbook is sin \(\frac{7 \pi}{12}\) + sin \(\frac{\pi}{12}\) However, as per our calculation it is sin \(\frac{7 \pi}{6}\) + sin \(\frac{\pi}{6}\)
iii. 2cos 4θ cos 2θ = cos(4θ + 2θ)+cos (4θ – 2θ)
= cos 6θ + cos 2θ
iv. 2cos 35° cos75°
= cos(35° + 75°) + cos (35° – 75°)
= cos 110° + cos (-40)°
= cos 110° + cos 40° … [∵ cos(-θ) = cos θ]
Question 2.
Prove the following:
i. \(\frac{\sin 2 x+\sin 2 y}{\sin 2 x-\sin 2 y}=\frac{\tan (x+y)}{\tan (x-y)}\)
Solution:
ii. sin 6x + sin 4x – sin 2x = 4 cos x sin 2x cos 3x
Solution:
L.H.S. = sin 6x + sin 4x — sin 2x
= 2sin \(\left(\frac{6 x+4 x}{2}\right)\) cos \(\left(\frac{6 x-4 x}{2}\right)\) – 2 sin x cos x
= 2 sin 5x cos x — 2 sin x cos x
= 2 cos x (sin 5x — sin x)
= 2 cos \(\left[2 \cos \left(\frac{5 x+x}{2}\right) \sin \left(\frac{5 x-x}{2}\right)\right]\)
= 2 cos x (2 cos 3x sin 2x)
= 4 cos x sin 2x cos 3x
= R.H.S.
[Note: The question has been modified.]
iii. \(\frac{\sin x-\sin 3 x+\sin 5 x-\sin 7 x}{\cos x-\cos 3 x-\cos 5 x+\cos 7 x}\) = cot 2x
Solution:
iv. sin 18° cos 39° + sin 6° cos 15° = sin 24° cos 33°
Solution:
L.H.S. = sin 18°.cos 39° + sin 6°.cos 15°
= \(\frac{1}{2}\) (2 cos 39°sin 18° + 2.cos 15°.sin 6°)
= \(\frac{1}{2}\)[sin(39° + 18°) — sin(39° — 18°) + sin (15° + 6°) — sin (15° — 6°)]
= \(\frac{1}{2}\)(sin57° – sin21° + sin 21°- sin9°)
= \(\frac{1}{2}\)(sin57° – sin9°)
= \(\frac{1}{2}\) x 2. cos \(\left(\frac{57^{\circ}+9^{\circ}}{2}\right) \cdot \sin \left(\frac{57^{\circ}-9^{\circ}}{2}\right)\)
= cos 33° .sin 24°
= sin 24°. cos 33°
= R.H.S.
v. cos 20° cos 40° cos 60°cos 80° = 1/16
Solution:
L.H.S. = cos 20°.cos 40°.cos 60°.cos 80°
= cos 20°.cos 40°.\(\frac{1}{2}\) .cos 80°
= \(\frac{1}{2 \times 2}\)(2 cos 40°.cos 20°).cos 80°
= \(\frac{1}{4}\)[cos(40° + 20°) + cos(40°- 20°)].cos80°
= \(\frac{1}{4}\)(cos 60° + cos 20°) cos 80°
=\(\frac{1}{4}\)cos 60°. cos 80° + \(\frac{1}{4}\) cos 20°. cos 80°
= \(\frac{1}{4}\left(\frac{1}{2}\right) \cos 80^{\circ}+\frac{1}{2 \times 4}\left(2 \cos 80^{\circ} \cos 20^{\circ}\right)\)
= \(\frac{1}{8}\) cos 80° + \(\frac{1}{8}\)[cos (80° + 20°) + cos (80° — 20°)]
= \(\frac{1}{8}\)cos 80° + \(\frac{1}{8}\)(cos 100° + cos 60°)
= \(\frac{1}{8}\) cos 80° + \(\frac{1}{8}\)cos 100° + \(\frac{1}{8}\)cos 60°
= \(\frac{1}{8}\) cos 80° = \(\frac{1}{8}\) cos (180° – 80°) + \(\frac{1}{8} \times \frac{1}{2}\)
= \(\frac{1}{8}\) cos 80° – \(\frac{1}{8}\) cos 80° + \(\frac{1}{16}\) … [∵ cos (180 – θ) = – cos θ]
= \(\frac{1}{16}\) = R.H.S
vi. sin 20° sin 40° sin 60° sin 80° = 3/16
Solution:
