11th NCERT/CBSE Trigonometric Functions Exercise 3.3 Questions 25
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Find the value of other five trignometric functions in Exercises 1 to 5

Question (1)

Prove that ${\sin ^2}\frac{\pi }{6} + {\cos ^2}\frac{\pi }{3} - {\tan ^2}\frac{\pi }{4} = - \frac{1}{2}$

Solution

\[LHS = {\sin ^2}\frac{\pi }{6} + {\cos ^2}\frac{\pi }{3} - {\tan ^2}\frac{\pi }{4}\] \[LHS = {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{{\rm{1}}}{2}} \right)^2} - {\left( 1 \right)^2}\] \[ = \frac{1}{4} + \frac{1}{4} - 1 = - \frac{1}{2} = RHS\]

Question (2)

Prove that $2{\sin ^2}\frac{\pi }{6} + \cos e{c^2}\frac{{7\pi }}{6}{\cos ^2}\frac{\pi }{3} = \frac{3}{2}$

Solution

\[LHS = 2{\sin ^2}\frac{\pi }{6} + \cos e{c^2}\frac{{7\pi }}{6}{\cos ^2}\frac{\pi }{3}\] \[ = 2{\left( {\frac{1}{2}} \right)^2} + \cos e{c^2}\left( {\pi + \frac{\pi }{6}} \right){\left( {\frac{1}{2}} \right)^2}\] \[ = 2{\left( {\frac{1}{2}} \right)^2} + {\left( { - \cos ec\frac{\pi }{6}} \right)^2}\left( {\frac{1}{4}} \right)\] \[ = \frac{1}{2} + {\left( { - 2} \right)^2}\left( {\frac{1}{4}} \right)\] \[ = \frac{1}{2} + \frac{4}{4} = \frac{3}{2}\] \[ = RHS\]

Question (3)

Prove that ${\cot ^2}\frac{\pi }{6} + \cos ec\frac{{5\pi }}{6} + 3{\tan ^2}\frac{\pi }{6} = 6$

Solution

\[LHS = {\cot ^2}\frac{\pi }{6} + \cos ec\frac{{5\pi }}{6} + 3{\tan ^2}\frac{\pi }{6}\] \[ = {\left( {\sqrt 3 } \right)^2} + \cos ec\left( {\pi - \frac{\pi }{6}} \right) + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2}\] \[ = 3 + \cos ec\frac{\pi }{6} + 3 \times \frac{1}{3}\] \[ = 3 + 2 + 1 = RHS\]

Question (4)

Prove that $2{\sin ^2}\frac{{3\pi }}{4} + 2{\cos ^2}\frac{\pi }{4} + 2{\sec ^2}\frac{\pi }{3} = 10$

Solution

\[LHS = 2{\sin ^2}\frac{{3\pi }}{4} + 2{\cos ^2}\frac{\pi }{4} + 2{\sec ^2}\frac{\pi }{3}\] \[LHS = 2{\left\{ {\sin \left( {\pi - \frac{\pi }{4}} \right)} \right\}^2} + 2{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2{\left( 2 \right)^2}\] \[ = 2{\left\{ {\sin \frac{\pi }{4}} \right\}^2} + 2 \times \frac{1}{2} + 8\] \[ = 2{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 1 + 8\] \[ = 1 + 1 + 8 = 10 = RHS\]

Question (5)

Find the value of :
(i) sin 75°     (ii) tan 15°

Solution

(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x + y) = sin x cos y + cos x sin y]
\[ = \left( {\frac{1}{{\sqrt 2 }}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) + \left( {\frac{1}{{\sqrt 2 }}} \right)\left( {\frac{1}{2}} \right)\] \[ = \frac{{\sqrt 3 }}{{2\sqrt 2 }} + \frac{1}{{2\sqrt 2 }} = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\]
\[ = \frac{{\sqrt 3 }}{{2\sqrt 2 }} + \frac{1}{{2\sqrt 2 }} = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\]
Use formula
\[\tan \left( {x - y} \right) = \frac{{\tan x - \tan y}}{{1 + \tan x\tan y}}\]
\[ = \frac{{\tan {{45}^o} - \tan {{30}^o}}}{{1 + \tan {{45}^o}\tan {{30}^o}}}\] \[ = \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + 1\left( {\frac{1}{{\sqrt 3 }}} \right)}} = \frac{{\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}{{\frac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}\] \[ = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\] \[ = \frac{{3 + 1 - 2\sqrt 3 }}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}}\] \[ = \frac{{4 - 2\sqrt 3 }}{{3 - 1}} = 2 - \sqrt 3 \]

Prove the following

Question (6)

\[\cos \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - y} \right) - \sin \left( {\frac{\pi }{4} - x} \right)\sin \left( {\frac{\pi }{4} - y} \right) = \sin \left( {x + y} \right)\]

Solution

\[\cos \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - y} \right) - \sin \left( {\frac{\pi }{4} - x} \right)\sin \left( {\frac{\pi }{4} - y} \right) \] Let $\frac{\pi }{4} - x = \alpha $ ; $\frac{\pi }{4} - y = \beta $
\[LHS = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] \[ = \cos \left( {\alpha + \beta } \right)\] \[ = \cos \left( {\frac{\pi }{4} - x + \frac{\pi }{4} - y} \right)\] \[ = \cos \left[ {\frac{\pi }{2} - \left( {x + y} \right)} \right]\] \[ = \sin \left( {x + y} \right) = RHS\]

Question (7)

\[\frac{{\tan \left( {\frac{\pi }{4} + x} \right)}}{{\tan \left( {\frac{\pi }{4} - x} \right)}} = {\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}\]

Solution

\[\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and
\[\tan \left( {A - B} \right) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]
\[LHS = \frac{{\tan \left( {\frac{\pi }{4} + x} \right)}}{{\tan \left( {\frac{\pi }{4} - x} \right)}}\] \[ = \frac{{\left( {\frac{{\tan \frac{\pi }{4} + \tan x}}{{1 - \tan \frac{\pi }{4}\tan x}}} \right)}}{{\left( {\frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}}} \right)}} = \frac{{\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)}}{{\left( {\frac{{1 - \tan x}}{{1 + \tan x}}} \right)}}\] \[ = {\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)^2} = RHS\]

Question (8)

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

Solution

\[LHS = \frac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\frac{\pi }{2} + x} \right)}}\] \[ = \frac{{\left[ { - \cos x} \right]\left[ {\cos x} \right]}}{{\left( {\sin x} \right)\left( { - \sin x} \right)}}\] \[ = \frac{{ - {{\cos }^2}x}}{{ - {{\sin }^2}x}} = {\cot ^2}x\] \[ = RHS\]

Question (9)

\[\cos \left( {\frac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\frac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right] = 1\]

Solution

\[\cos \left( {\frac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\frac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right] \] \[ = \sin x\cos x\left[ {\tan x + \cot x} \right]\] \[ = \sin x\cos x\left[ {\frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}}} \right]\] \[ = \sin x\cos x\left[ {\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x\cos x}}} \right]\] \[ = 1 = RHS\]

Question (10)

\[\sin \left( {n + 1} \right)x\sin \left( {n + 2} \right)x + \cos \left( {n + 1} \right)x\cos \left( {n + 2} \right)x = \cos x\]

Solution

\[\sin \left( {n + 1} \right)x\sin \left( {n + 2} \right)x + \cos \left( {n + 1} \right)x\cos \left( {n + 2} \right)x\] \[Let \; \left( {n + 1} \right)x = \alpha ;\left( {n + 2} \right)x = \beta \] \[LHS = \sin \alpha \sin \beta + \cos \alpha \cos \beta \] \[ = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] \[ = \cos \left( {\alpha - \beta } \right)\] \[ = \cos \left[ {\left( {n + 1} \right)x - \left( {n + 2} \right)x} \right]\] \[ = \cos \left( {\require{cancel}\cancel{nx} + x - \cancel{nx} - 2x} \right)\] \[ = \cos \left( { - x} \right)\] \[ = \cos x = RHS\]

Question (11)

\[\cos \left( {\frac{{3\pi }}{4} + x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right) = - \sqrt 2 \sin x\]

Solution

Use following formula
\[\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right) \cdot \sin \left( {\frac{{A - B}}{2}} \right)\]
\[LHS = \cos \left( {\frac{{3\pi }}{4} + x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)\] \[ = - 2\sin \left\{ {\frac{{\left( {\frac{{3\pi }}{4} + x} \right) + \left( {\frac{{3\pi }}{4} - x} \right)}}{2}} \right\} \cdot \sin \left\{ {\frac{{\left( {\frac{{3\pi }}{4} + x} \right) - \left( {\frac{{3\pi }}{4} - x} \right)}}{2}} \right\}\] \[ = - 2\sin \left( {\frac{{3\pi }}{4}} \right)\sin x\] \[ = - 2\sin \left( {\pi - \frac{\pi }{4}} \right)\sin x\] \[ = - 2\sin \frac{\pi }{4}\sin x\] \[ = - 2 \times \frac{1}{{\sqrt 2 }} \times \sin x\] \[ = - \sqrt 2 \sin x = RHS\]

Question (12)

\[{\sin ^2}6x - {\sin ^2}4x = \sin 2x\sin 10x\]

Solution

Use following formula
\[\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\]
\[LHS = {\sin ^2}2x - {\sin ^2}6x\] \[ = \left( {\sin 6x + \sin 4x} \right)\left( {\sin 6x - \sin 4x} \right)\] \[ = \left[ {2\sin \left( {\frac{{6x + 4x}}{2}} \right)\cos \left( {\frac{{6x - 4x}}{2}} \right)} \right]\left[ {2\cos \left( {\frac{{6x + 4x}}{2}} \right) \cdot \sin \left( {\frac{{6x - 4x}}{2}} \right)} \right]\] \[ = \left( {2\sin 5x\cos x} \right)\left( {2\cos 5x\sin x} \right)\] \[ = \left( {2\sin 5x\cos 5x} \right)\left( {2\sin x\cos x} \right)\] \[ = \sin 10x\sin 2x = RHS\]

Question (13)

\[{\cos ^2}2x - {\cos ^2}6x = \sin 4x\sin 8x\]

Solution

Use following formula
\[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\]
\[LHS = {\cos ^2}2x - {\cos ^2}6x\] \[ = \left( {\cos 2x + \cos 6x} \right)\left( {\cos 2x - \cos 6x} \right)\] \[ = \left[ {2\cos \left( {\frac{{2x + 6x}}{2}} \right)\cos \left( {\frac{{2x - 6x}}{2}} \right)} \right]\left[ { - 2\sin \left( {\frac{{2x + 6x}}{2}} \right)\sin \left( {\frac{{2x - 6x}}{2}} \right)} \right]\] \[ = \left[ {2\cos 4x\cos \left( { - 2x} \right)} \right]\left[ { - 2\sin 4x\sin \left( { - 2x} \right)} \right]\] \[ = \left[ {2\cos 4x\cos 2x} \right]\left[ { - 2\sin 4x\left( { - \sin 2x} \right)} \right]\] \[ = \left[ {2\sin 4x\cos 4x} \right]\left[ {2\sin 2x\cos 2x} \right]\] \[ = \sin 8x\sin 4x = RHS\]

Question (14)

\[\sin 2x + 2\sin 4x + \sin 6x = 4{\cos ^2}x\sin 4x\]

Solution

\[LHS = \sin 2x + 2\sin 4x + \sin 6x\] \[ = \left( {\sin 2x + \sin 6x} \right) + 2\sin 4x\]
Use formula
\[\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[ = \left[ {2\sin \left( {\frac{{2x + 6x}}{2}} \right)\cos \left( {\frac{{2x - 6x}}{2}} \right)} \right] + 2\sin 4x\] \[ = 2\sin 4x\cos \left( { - 2x} \right) + 2\sin 4x\] \[ = 2\sin 4x\cos 2x + 2\sin 4x\] \[ = 2\sin 4x\left( {\cos 2x + 1} \right)\]
Use following formula
\[1 + \cos A = 2{\cos ^2}\frac{A}{2}\]
\[ =2 \sin 4x\left( {2{{\cos }^2}\frac{{2x}}{2}} \right)\] \[ =2 \sin 4x\left( {2{{\cos }^2}x} \right)\] \[ = 4{\cos ^2}x\sin 4x = RHS\]

Question (15)

\[\cot 4x\left( {\sin 5x + \sin 3x} \right) = \cot x\left( {\sin 5x - \sin 3x} \right)\]

Solution

\[LHS = \cot 4x\left( {\sin 5x + \sin 3x} \right)\]
\[\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[ = \frac{{\cos 4x}}{{\sin 4x}}\left[ {2\sin \left( {\frac{{5x + 3x}}{2}} \right)\cos \left( {\frac{{5x - 3x}}{2}} \right)} \right]\] \[ = \left( {\frac{{\cos 4x}}{{\sin 4x}}} \right)\left( {2\sin 4x\cos x} \right)\] \[ = 2\cos 4x\cos x\]
\[RHS = \cot x\left( {\sin 5x - \sin 3x} \right)\]
\[\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\]
\[ = \frac{{\cos x}}{{\sin x}}\left[ {2\cos \left( {\frac{{5x + 3x}}{2}} \right)\sin \left( {\frac{{5x - 3x}}{2}} \right)} \right]\] \[ = \frac{{\cos x}}{{\sin x}}\left[ {2\cos 4x\sin x} \right]\] \[ = 2\cos 4x\cos x\] \[LHS = RHS\]

Question (16)

\[\frac{{\cos 9x - \cos 5x}}{{\sin 17x - \sin 3x}} = - \frac{{\sin 2x}}{{\cos 10x}}\]

Solution

\[LHS = \frac{{\cos 9x - \cos 5x}}{{\sin 17x - \sin 3x}}\]
\[\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\] and
\[\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\]
\[ = \frac{{ - 2\sin \left( {\frac{{9x + 5x}}{2}} \right) \cdot \sin \left( {\frac{{9x - 5x}}{2}} \right)}}{{2\cos \left( {\frac{{17x + 3x}}{2}} \right) \cdot \sin \left( {\frac{{17x - 3x}}{2}} \right)}}\] \[ = \frac{{ - 2\sin 7x \cdot \sin 2x}}{{2\cos 10x \cdot \sin 7x}}\] \[ = \frac{{ - \sin 2x}}{{\cos 10x}} = RHS\]

Question (17)

\[\frac{{\sin 5x + \sin 3x}}{{\cos 5x + \cos 3x}} = \tan 4x\]

Solution

\[LHS = \frac{{\sin 5x + \sin 3x}}{{\cos 5x + \cos 3x}}\]
\[\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\] and
\[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[ = \frac{{2\sin \left( {\frac{{5x + 3x}}{2}} \right) \cdot \cos \left( {\frac{{5x - 3x}}{2}} \right)}}{{2\cos \left( {\frac{{5x + 3x}}{2}} \right) \cdot \cos \left( {\frac{{5x - 3x}}{2}} \right)}}\] \[ = \frac{{2\sin 4x \cdot \cos x}}{{2\cos 4x \cdot \cos x}}\] \[ = \frac{{\sin 4x}}{{\cos 4x}}\] \[ = \tan 4x = RHS\]

Question (18)

\[\frac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \frac{{x - y}}{2}\]

Solution

\[LHS = \frac{{\sin x - \sin y}}{{\cos x + \cos y}}\]
\[\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\] and
\[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[ = \frac{{2\cos \left( {\frac{{x + y}}{2}} \right) \cdot \sin \left( {\frac{{x - y}}{2}} \right)}}{{2\cos \left( {\frac{{x + y}}{2}} \right) \cdot \cos \left( {\frac{{x - y}}{2}} \right)}}\] \[ = \frac{{\sin \left( {\frac{{x - y}}{2}} \right)}}{{\cos \left( {\frac{{x - y}}{2}} \right)}}\] \[ = \tan \left( {\frac{{x - y}}{2}} \right) = RHS\]

Question (19)

\[\frac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} = \tan 2x\]

Solution

\[LHS = \frac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}}\]
\[\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\] and
\[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[ = \frac{{2\sin \left( {\frac{{x + 3x}}{2}} \right)\cos \left( {\frac{{x - 3x}}{2}} \right)}}{{2\cos \left( {\frac{{x + 3x}}{2}} \right)\cos \left( {\frac{{x - 3x}}{2}} \right)}}\] \[ = \frac{{\sin 2x}}{{\cos 2x}}\] \[ = \tan 2x = RHS\]

Question (20)

\[\frac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x\]

Solution

\[LHS = \frac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}}\]
\[\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\] and
\[{\cos ^2}A - {\sin ^2}A = \cos 2A\]
\[ = \frac{{2\cos \left( {\frac{{x + 3x}}{2}} \right)\sin \left( {\frac{{x - 3x}}{2}} \right)}}{{ - \cos 2x}}\] \[ = \frac{{2\cos 2x\sin \left( { - x} \right)}}{{ - \cos 2x}}\] \[ = 2 \times \left( { - \sin x} \right)\] \[ = 2\sin x = RHS\]

Question (21)

\[\frac{{\cos 4x + \cos 3x + \cos 2x}}{{\sin 4x + \sin 3x + \sin 2x}} = \cot 3x\]

Solution

\[LHS = \frac{{\cos 4x + \cos 3x + \cos 2x}}{{\sin 4x + \sin 3x + \sin 2x}}\] \[ = \frac{{\left( {\cos 4x + \cos 2x} \right) + \cos 3x}}{{\left( {\sin 4x + \sin 2x} \right) + \sin 3x}}\]
\[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\] and
\[\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
\[ = \frac{{2\cos \left( {\frac{{4x + 2x}}{2}} \right)\cos \left( {\frac{{4x - 2x}}{2}} \right) + \cos 3x}}{{2\sin \left( {\frac{{4x + 2x}}{2}} \right)\cos \left( {\frac{{4x - 2x}}{2}} \right) + \sin 3x}}\] \[ = \frac{{2\cos 3x\cos x + \cos 3x}}{{2\sin 3x\cos x + \sin 3x}}\] \[ = \frac{{\cos 3x\left( {2\cos x + 1} \right)}}{{\sin 3x\left( {2\cos x + 1} \right)}}\] \[ = \cot 3x = RHS\]

Question (22)

\[\cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1\]

Solution

\[3x = 2x + x\] \[\cot 3x = \cot \left( {2x + x} \right)\] \[\cot 3x = \frac{{\cot 2x\cot x - 1}}{{\cot 2x + \cot x}}\] \[\cot 3x\left( {\cot 2x + \cot x} \right) = \cot 2x\cot x - 1\] \[\cot 3x\cot 2x + \cot 3x\cot x = \cot 2x\cot x - 1\] \[ \therefore \cot 2x\cot x - \cot 3x\cot 2x - \cot 3x\cot x = 1\]

Question (23)

\[\tan 4x = \frac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}\]

Solution

\[LHS = \tan 4x = \tan 2\left( {2x} \right)\]
Use formula
\[\tan 2A = \frac{{2\tan A}}{{1 - {{\tan }^2}A}}\]
\[\tan 4x = \frac{{2\tan 2x}}{{1 - {{\tan }^2}\left( {2x} \right)}}\] \[\ = \frac{{2\left( {\frac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}}{{1 - {{\left( {\frac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}^2}}}\] \[\ = \frac{{\left( {\frac{{4\tan x}}{{1 - {{\tan }^2}x}}} \right)}}{{\left[ {1 - \frac{{4{{\tan }^2}x}}{{{{\left( {1 - {{\tan }^2}x} \right)}^2}}}} \right]}}\] \[\ = \frac{{\left( {\frac{{4\tan x}}{{1 - {{\tan }^2}x}}} \right)}}{{\left[ {\frac{{\left( {1 - {{\tan }^2}x} \right)^2 - 4{{\tan }^2}x}}{{{{\left( {1 - {{\tan }^2}x} \right)}^2}}}} \right]}}\] \[\tan 4x = \frac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{{{\left( {1 - {{\tan }^2}x} \right)}^2} - 4{{\tan }^2}x}}\] \[ = \frac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}\] \[ = RHS\]

Question (24)

\[\cos 4x = 1 - 8{\sin ^2}x{\cos ^2}x\]

Solution

\[LHS = \cos 4x\] \[ = \cos 2\left( {2x} \right)\]
Use following formula
\[\cos 2A = 1 - 2{\sin ^2}x\] \[\sin 2A = 2\sin A\cos A\]
\[ = 1 - 2{\sin ^2}2x\] \[ = 1 - 2{\left( {2\sin x\cos x} \right)^2}\] \[ = 1 - 8{\sin ^2}x{\cos ^2}x\] \[ = RHS\]

Question (25)

\[\cos 6x = 32co{s^6}x - 48{\cos ^4}x - 1\]

Solution

\[LHS = \cos 6x\] \[ = \cos 2\left( {3x} \right)\] \[ = 2{\cos ^2}3x - 1\] \[ = 2{\left[ {4{{\cos }^3}x - 3\cos x} \right]^2} - 1\] \[ = 2\left[ {16{{\cos }^6}x - 24{{\cos }^4} + 9{{\cos }^2}x} \right] - 1\] \[ = 32{\cos ^6}x - 48{\cos ^4}x + 18{\cos ^2}x - 1\] \[ = RHS\]
Exercise 3.2 ⇐
⇒ Exercise 3.4