Class 11-NCERT Solutions-Chapter-03-Trigonometric Functions

Answer:

We know that \( 180^{\circ} = \pi \text{ radian} \). So, \( 1^{\circ} = \frac{\pi}{180} \text{ radian} \).

1. \( 25^{\circ} = 25 \times \frac{\pi}{180} = \frac{5\pi}{36} \) radian.
2. \( -47^{\circ}30' = -\left(47 + \frac{30}{60}\right)^{\circ} = -47.5^{\circ} = -\frac{95}{2}^{\circ} \).
   In radians: \( -\frac{95}{2} \times \frac{\pi}{180} = -\frac{19\pi}{72} \) radian.
3. \( 240^{\circ} = 240 \times \frac{\pi}{180} = \frac{4\pi}{3} \) radian.
4. \( 520^{\circ} = 520 \times \frac{\pi}{180} = \frac{26\pi}{9} \) radian.

Answer:

We know that \( \pi \text{ radian} = 180^{\circ} \). So, \( 1 \text{ radian} = \frac{180}{\pi}^{\circ} \).

1. \( \frac{11}{16} \text{ rad} = \frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180}{22/7} = \frac{11 \times 180 \times 7}{16 \times 22} = \frac{315}{8}^{\circ} \)
   \( = 39^{\circ} + \left(\frac{3}{8} \times 60\right)' = 39^{\circ} 22' + \left(\frac{1}{2} \times 60\right)'' = 39^{\circ} 22' 30'' \).
2. \( -4 \text{ rad} = -4 \times \frac{180}{\pi} = -4 \times \frac{180 \times 7}{22} = -\frac{2520}{11}^{\circ} \)
   \( = -229^{\circ} + \left(\frac{1}{11} \times 60\right)' = -229^{\circ} 5' + \left(\frac{5}{11} \times 60\right)'' \approx -229^{\circ} 5' 27'' \).
3. \( \frac{5\pi}{3} \text{ rad} = \frac{5\pi}{3} \times \frac{180}{\pi} = 5 \times 60 = 300^{\circ} \).
4. \( \frac{7\pi}{6} \text{ rad} = \frac{7\pi}{6} \times \frac{180}{\pi} = 7 \times 30 = 210^{\circ} \).

Answer:

Revolutions in 1 minute (60 seconds) = 360.

Revolutions in 1 second = \( \frac{360}{60} = 6 \).

Angle turned in 1 revolution = \( 2\pi \) radians.

Angle turned in 6 revolutions = \( 6 \times 2\pi = 12\pi \) radians.

Answer:

Formula: \( \theta = \frac{l}{r} \), where \( \theta \) is in radians.

\( l = 22 \) cm, \( r = 100 \) cm.

\( \theta = \frac{22}{100} \) radian.

In degrees: \( \frac{22}{100} \times \frac{180}{\pi} = \frac{22}{100} \times \frac{180 \times 7}{22} = \frac{180 \times 7}{100} = \frac{126}{10} = 12.6^{\circ} \).

\( 12.6^{\circ} = 12^{\circ} + (0.6 \times 60)' = 12^{\circ} 36' \).

Answer:

Diameter = 40 cm \( \Rightarrow \) Radius \( r = 20 \) cm.

Chord length = 20 cm.

Since radius = chord = 20 cm, the triangle formed by two radii and the chord is an equilateral triangle.

Angle at center \( \theta = 60^{\circ} = \frac{\pi}{3} \) radian.

Arc length \( l = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \) cm.

Answer:

Let radii be \( r_1, r_2 \) and arc length be \( l \).

\( \theta_1 = 60^{\circ} = \frac{\pi}{3} \). \( \theta_2 = 75^{\circ} = 75 \times \frac{\pi}{180} = \frac{5\pi}{12} \).

Since \( l = r_1 \theta_1 = r_2 \theta_2 \):

\( r_1 \left(\frac{\pi}{3}\right) = r_2 \left(\frac{5\pi}{12}\right) \)

\( \frac{r_1}{r_2} = \frac{5\pi/12}{\pi/3} = \frac{5}{12} \times \frac{3}{1} = \frac{5}{4} \).

Ratio is 5:4.

Answer:

Length of pendulum is the radius \( r = 75 \) cm.

1. \( l = 10 \) cm. \( \theta = \frac{l}{r} = \frac{10}{75} = \frac{2}{15} \) radian.
2. \( l = 15 \) cm. \( \theta = \frac{15}{75} = \frac{1}{5} \) radian.
3. \( l = 21 \) cm. \( \theta = \frac{21}{75} = \frac{7}{25} \) radian.

Answer:

In 3rd quadrant, only tan and cot are positive; others are negative.

\( \sin^2 x + \cos^2 x = 1 \Rightarrow \sin^2 x = 1 - \left(-\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \).

\( \sin x = -\frac{\sqrt{3}}{2} \) (negative in 3rd).

\( \sec x = \frac{1}{\cos x} = -2 \).

\( \csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}} \).

\( \tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3} \).

\( \cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}} \).

Answer:

In 2nd quadrant, sin and csc are positive; others are negative.

\( \cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25} \).

\( \cos x = -\frac{4}{5} \) (negative in 2nd).

\( \csc x = \frac{5}{3} \).

\( \sec x = -\frac{5}{4} \).

\( \tan x = \frac{3/5}{-4/5} = -\frac{3}{4} \).

\( \cot x = -\frac{4}{3} \).

Answer:

In 3rd quadrant, tan and cot are positive.

\( \tan x = \frac{4}{3} \).

\( \sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9} \).

\( \sec x = -\frac{5}{3} \) (negative in 3rd).

\( \cos x = -\frac{3}{5} \).

\( \sin x = \tan x \cdot \cos x = \frac{4}{3} \times \left(-\frac{3}{5}\right) = -\frac{4}{5} \).

\( \csc x = -\frac{5}{4} \).

Answer:

In 4th quadrant, cos and sec are positive.

\( \cos x = \frac{5}{13} \).

\( \sin^2 x = 1 - \frac{25}{169} = \frac{144}{169} \).

\( \sin x = -\frac{12}{13} \) (negative in 4th).

\( \csc x = -\frac{13}{12} \).

\( \tan x = \frac{-12/13}{5/13} = -\frac{12}{5} \).

\( \cot x = -\frac{5}{12} \).

Answer:

\( \cot x = -\frac{12}{5} \).

\( \sec^2 x = 1 + \frac{25}{144} = \frac{169}{144} \).

\( \sec x = -\frac{13}{12} \) (negative in 2nd).

\( \cos x = -\frac{12}{13} \).

\( \sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right)\left(-\frac{12}{13}\right) = \frac{5}{13} \).

\( \csc x = \frac{13}{5} \).

Answer:

\( 765^{\circ} = 2 \times 360^{\circ} + 45^{\circ} \).

\( \sin(2 \times 360^{\circ} + 45^{\circ}) = \sin 45^{\circ} = \frac{1}{\sqrt{2}} \).

Answer:

\( \csc(-1410^{\circ}) = -\csc(1410^{\circ}) \).

\( 1410^{\circ} = 4 \times 360^{\circ} - 30^{\circ} \) or \( 3 \times 360^{\circ} + 330^{\circ} \).

Let's use \( 4 \times 360^{\circ} - 30^{\circ} \).
\( \csc(1410^{\circ}) = \csc(4 \times 360^{\circ} - 30^{\circ}) = \csc(-30^{\circ}) = -\csc 30^{\circ} = -2 \).

So, \( -\csc(1410^{\circ}) = -(-2) = 2 \).

Answer:

\( \frac{19\pi}{3} = 6\pi + \frac{\pi}{3} \).

\( \tan\left(6\pi + \frac{\pi}{3}\right) = \tan \frac{\pi}{3} = \sqrt{3} \).

Answer:

\( \sin(-\theta) = -\sin \theta \).

\( -\sin \frac{11\pi}{3} = -\sin\left(4\pi - \frac{\pi}{3}\right) \).

\( \sin(4\pi - \frac{\pi}{3}) = \sin(-\frac{\pi}{3}) = -\sin \frac{\pi}{3} \).

So, \( -(-\sin \frac{\pi}{3}) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).

Answer:

\( \cot(-\theta) = -\cot \theta \).

\( -\cot \frac{15\pi}{4} = -\cot\left(4\pi - \frac{\pi}{4}\right) \).

\( \cot(4\pi - \frac{\pi}{4}) = \cot(-\frac{\pi}{4}) = -\cot \frac{\pi}{4} = -1 \).

So, \( -(-1) = 1 \).

Answer:

L.H.S. \( = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2 \)

\( = \frac{1}{4} + \frac{1}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2} = \) R.H.S.

Answer:

\( \csc \frac{7\pi}{6} = \csc (\pi + \frac{\pi}{6}) = -\csc \frac{\pi}{6} = -2 \).

L.H.S. \( = 2\left(\frac{1}{2}\right)^2 + (-2)^2 \left(\frac{1}{2}\right)^2 \)

\( = 2\left(\frac{1}{4}\right) + 4\left(\frac{1}{4}\right) = \frac{1}{2} + 1 = \frac{3}{2} = \) R.H.S.

Answer:

\( \csc \frac{5\pi}{6} = \csc (\pi - \frac{\pi}{6}) = \csc \frac{\pi}{6} = 2 \).

L.H.S. \( = (\sqrt{3})^2 + 2 + 3\left(\frac{1}{\sqrt{3}}\right)^2 \)

\( = 3 + 2 + 3\left(\frac{1}{3}\right) = 3 + 2 + 1 = 6 = \) R.H.S.

Answer:

\( \sin \frac{3\pi}{4} = \sin(\pi - \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).

L.H.S. \( = 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2(2)^2 \)

\( = 2(\frac{1}{2}) + 2(\frac{1}{2}) + 2(4) = 1 + 1 + 8 = 10 = \) R.H.S.

Answer:

(i) \( \sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) \)

\( = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} \)

\( = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}} \).

(ii) \( \tan 15^{\circ} = \tan(45^{\circ} - 30^{\circ}) \)

\( = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} \)

\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2 - \sqrt{3} \).

Answer:

Using \( \cos A \cos B - \sin A \sin B = \cos(A+B) \):

L.H.S. \( = \cos\left[ \left(\frac{\pi}{4}-x\right) + \left(\frac{\pi}{4}-y\right) \right] \)

\( = \cos\left[ \frac{\pi}{2} - (x+y) \right] \)

\( = \sin(x+y) = \) R.H.S.

Answer:

\( \tan(\frac{\pi}{4}+x) = \frac{1+\tan x}{1-\tan x} \)

\( \tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x} \)

L.H.S. \( = \frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}} = \left( \frac{1+\tan x}{1-\tan x} \right)^2 = \) R.H.S.

Answer:

L.H.S. \( = \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \)

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

Answer:

L.H.S. \( = (\sin x)(\cos x) [ \tan x + \cot x ] \)

\( = \sin x \cos x \left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right) \)

\( = \sin x \cos x \left( \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \right) \)

\( = 1 = \) R.H.S.

Answer:

Use \( \cos(A-B) = \cos A \cos B + \sin A \sin B \).

Here \( A = (n+1)x \) and \( B = (n+2)x \).

L.H.S. \( = \cos[(n+1)x - (n+2)x] \)

\( = \cos[nx + x - nx - 2x] = \cos(-x) = \cos x = \) R.H.S.

Answer:

Use \( \cos A - \cos B = -2\sin\frac{A+B}{2} \sin\frac{A-B}{2} \).

L.H.S. \( = -2 \sin\left(\frac{3\pi}{4}\right) \sin(x) \)

\( = -2 \left(\frac{1}{\sqrt{2}}\right) \sin x = -\sqrt{2} \sin x = \) R.H.S.

Answer:

L.H.S. \( = (\sin 6x - \sin 4x)(\sin 6x + \sin 4x) \)

\( = (2\cos 5x \sin x)(2\sin 5x \cos x) \)

\( = (2\sin x \cos x)(2\sin 5x \cos 5x) = \sin 2x \sin 10x = \) R.H.S.

Answer:

L.H.S. \( = (\cos 2x - \cos 6x)(\cos 2x + \cos 6x) \)

\( = [2\sin 4x \sin 2x] [2\cos 4x \cos 2x] \)

\( = (2\sin 4x \cos 4x)(2\sin 2x \cos 2x) = \sin 8x \sin 4x = \) R.H.S.

Answer:

L.H.S. \( = (\sin 6x + \sin 2x) + 2\sin 4x \)

\( = 2\sin 4x \cos 2x + 2\sin 4x \)

\( = 2\sin 4x (\cos 2x + 1) \)

\( = 2\sin 4x (2\cos^2 x) = 4\cos^2 x \sin 4x = \) R.H.S.

Answer:

L.H.S. \( = \frac{\cos 4x}{\sin 4x} (2\sin 4x \cos x) = 2\cos 4x \cos x \).

R.H.S. \( = \frac{\cos x}{\sin x} (2\cos 4x \sin x) = 2\cos 4x \cos x \).

L.H.S. = R.H.S.

Answer:

L.H.S. \( = \frac{-2\sin 7x \sin 2x}{2\cos 10x \sin 7x} = -\frac{\sin 2x}{\cos 10x} = \) R.H.S.

Answer:

L.H.S. \( = \frac{2\sin 4x \cos x}{2\cos 4x \cos x} = \frac{\sin 4x}{\cos 4x} = \tan 4x = \) R.H.S.

Answer:

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

Answer:

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

Answer:

L.H.S. \( = \frac{-(\sin 3x - \sin x)}{-(\cos^2 x - \sin^2 x)} = \frac{2\cos 2x \sin x}{\cos 2x} = 2\sin x = \) R.H.S.

Answer:

Group first and last terms:

L.H.S. \( = \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x} \)

\( = \frac{2\cos 3x \cos x + \cos 3x}{2\sin 3x \cos x + \sin 3x} \)

\( = \frac{\cos 3x(2\cos x + 1)}{\sin 3x(2\cos x + 1)} = \cot 3x = \) R.H.S.

Answer:

We know \( 3x = 2x + x \). So \( \cot 3x = \cot(2x + x) = \frac{\cot 2x \cot x - 1}{\cot 2x + \cot x} \).

\( \cot 3x (\cot 2x + \cot x) = \cot 2x \cot x - 1 \)

\( \cot 3x \cot 2x + \cot 3x \cot x = \cot 2x \cot x - 1 \)

Rearranging gives: \( \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1 \).

Answer:

\( \tan 4x = \tan 2(2x) = \frac{2\tan 2x}{1-\tan^2 2x} \).

Substitute \( \tan 2x = \frac{2\tan x}{1-\tan^2 x} \) and simplify.

Result matches R.H.S.

Answer:

\( \cos 4x = \cos 2(2x) = 1 - 2\sin^2 2x \).

\( = 1 - 2(2\sin x \cos x)^2 \)

\( = 1 - 2(4\sin^2 x \cos^2 x) = 1 - 8\sin^2 x \cos^2 x = \) R.H.S.

Answer:

\( \cos 6x = \cos 3(2x) = 4\cos^3 2x - 3\cos 2x \).

Substitute \( \cos 2x = 2\cos^2 x - 1 \):

\( = 4(2\cos^2 x - 1)^3 - 3(2\cos^2 x - 1) \).

Expand using \( (a-b)^3 \) and simplify to get R.H.S.

Answer:

L.H.S. \( = \cos(\frac{10\pi}{13}) + \cos(\frac{8\pi}{13}) + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} \)

Note \( \frac{10\pi}{13} = \pi - \frac{3\pi}{13} \) and \( \frac{8\pi}{13} = \pi - \frac{5\pi}{13} \).

\( = -\cos \frac{3\pi}{13} - \cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0 = \) R.H.S.

Answer:

L.H.S. \( = \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x \)

\( = (\cos 3x \cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x) \)

\( = \cos(3x-x) - \cos 2x = \cos 2x - \cos 2x = 0 = \) R.H.S.

Answer:

\( (\cos x + \cos y)^2 = \left(2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\right)^2 \)

\( (\sin x - \sin y)^2 = \left(2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\right)^2 \)

Add: \( 4\cos^2\frac{x+y}{2} (\cos^2\frac{x-y}{2} + \sin^2\frac{x-y}{2}) = 4\cos^2\frac{x+y}{2} \).

Answer:

Expand terms: \( (\cos^2 x + \cos^2 y - 2\cos x \cos y) + (\sin^2 x + \sin^2 y - 2\sin x \sin y) \)

\( = 2 - 2(\cos x \cos y + \sin x \sin y) = 2 - 2\cos(x-y) \)

\( = 2(1 - \cos(x-y)) = 2(2\sin^2 \frac{x-y}{2}) = 4\sin^2 \frac{x-y}{2} \).

Answer:

Group \( (\sin 7x + \sin x) + (\sin 5x + \sin 3x) \).

\( = 2\sin 4x \cos 3x + 2\sin 4x \cos x \)

\( = 2\sin 4x (\cos 3x + \cos x) \)

\( = 2\sin 4x (2\cos 2x \cos x) = 4\cos x \cos 2x \sin 4x \).

Answer:

Numerator: \( 2\sin 6x \cos x + 2\sin 6x \cos 3x = 2\sin 6x(\cos x + \cos 3x) \).

Denominator: \( 2\cos 6x \cos x + 2\cos 6x \cos 3x = 2\cos 6x(\cos x + \cos 3x) \).

Ratio: \( \frac{2\sin 6x}{2\cos 6x} = \tan 6x \).

Answer:

L.H.S. \( = (\sin 3x - \sin x) + \sin 2x \)

\( = 2\cos 2x \sin x + 2\sin x \cos x \)

\( = 2\sin x (\cos 2x + \cos x) \)

\( = 2\sin x (2\cos \frac{3x}{2} \cos \frac{x}{2}) = 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2} \).

Answer:

\( x \) in II implies \( \frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \) (Quadrant I, so all positive).

\( \sec^2 x = 1 + \frac{16}{9} = \frac{25}{9} \Rightarrow \sec x = -\frac{5}{3} \) (II quad) \( \Rightarrow \cos x = -\frac{3}{5} \).

\( \sin \frac{x}{2} = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1+3/5}{2}} = \frac{2}{\sqrt{5}} \).

\( \cos \frac{x}{2} = \sqrt{\frac{1+\cos x}{2}} = \sqrt{\frac{1-3/5}{2}} = \frac{1}{\sqrt{5}} \).

\( \tan \frac{x}{2} = 2 \).

Answer:

\( x \) in III implies \( \pi < x < \frac{3\pi}{2} \Rightarrow \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4} \) (Quadrant II: sin+, cos-).

\( \sin \frac{x}{2} = \sqrt{\frac{1 - (-1/3)}{2}} = \sqrt{\frac{2}{3}} \).

\( \cos \frac{x}{2} = -\sqrt{\frac{1 + (-1/3)}{2}} = -\sqrt{\frac{1}{3}} \).

\( \tan \frac{x}{2} = -\sqrt{2} \).

Answer:

\( x \) in II \( \Rightarrow \frac{x}{2} \) in I.

\( \cos x = -\frac{\sqrt{15}}{4} \).

\( \sin \frac{x}{2} = \sqrt{\frac{1 + \sqrt{15}/4}{2}} = \sqrt{\frac{4+\sqrt{15}}{8}} \).

\( \cos \frac{x}{2} = \sqrt{\frac{1 - \sqrt{15}/4}{2}} \).

Values simplify to \( \sin \frac{x}{2} = \frac{\sqrt{8+2\sqrt{15}}}{4} = \frac{\sqrt{5}+\sqrt{3}}{4} \), \( \cos \frac{x}{2} = \frac{\sqrt{5}-\sqrt{3}}{4} \), \( \tan \frac{x}{2} = 4+\sqrt{15} \).