Class 11 NCERT Solutions- Chapter 3 Trigonometric Function - Exercise 3.3 | Set 1

In this article, we will be going to solve the entire Miscellaneous Exercise 3.3 of Chapter 3 of the NCERT textbook. Trigonometric functions are fundamental mathematical functions that relate the angles of a triangle to the lengths of its sides. The primary trigonometric functions are sine, cosine, and tangent, and they are defined based on the properties of right-angled triangles.

What are Trigonometric Functions?

Trigonometric Functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. These functions are crucial in the study of periodic phenomena, such as waves and oscillations, and are foundational in various fields of science and engineering.

The primary trigonometric functions are sine, cosine, and tangent, along with their reciprocals: cosecant, secant, and cotangent.

Prove That:

Question 1: sin²\frac{\pi}{6}  +cos²\frac{\pi}{3}  - tan²\frac{\pi}{4}  =-\frac{1}{2}

Solution: 

Taking LHS in consideration, we get

= sin²\frac{\pi}{6}  +cos²\frac{\pi}{3}  - tan²\frac{\pi}{4}

Substituting the values,

=(\frac{1}{2})^2 + (\frac{1}{2})^2 - (1)^2

=\frac{1}{2}  - 1

=-\frac{1}{2}

Hence, LHS = RHS

Question 2: 2sin²\frac{\pi}{6}  +cosec²\frac{7\pi}{6}  cos²\frac{\pi}{3}  =\frac{3}{2}

Solution: 

Taking LHS in consideration, we get

= 2sin²\frac{\pi}{6}  +cosec²(\pi + \frac{\pi}{6})  cos²\frac{\pi}{3}

= 2sin²\frac{\pi}{6}  + (- cosec\frac{\pi}{6})^2  cos²\frac{\pi}{3}

Substituting the values,

=2(\frac{1}{2})^2 + ((-2)^2) (\frac{1}{2})^2

= 2(\frac{1}{4})  + 4(\frac{1}{4})

=\frac{1}{2}  + 1

=\frac{3}{2}

Hence, LHS = RHS

Question 3: cot²\frac{\pi}{6}  + cosec\frac{5\pi}{6}  + 3tan²\frac{\pi}{6}  = 6

Solution: 

Taking LHS in consideration, we get

= cot²\frac{\pi}{6}  + cosec(\pi - \frac{\pi}{6})  + 3tan²\frac{\pi}{6}

= cot²\frac{\pi}{6}  + cosec\frac{\pi}{6}  + 3tan²\frac{\pi}{6}

Substituting the values,

= (√3)² + 2 + 3(\frac{1}{\sqrt{3}})^2

= 3 + 2 + 3(\frac{1}{3})

= 6

Hence, LHS = RHS

Question 4: 2sin²\frac{3\pi}{4}  + 2cos²\frac{\pi}{4}  + 2sec²\frac{\pi}{3}  = 10

Solution: 

Taking LHS in consideration, we get

= 2sin²(\pi - \frac{\pi}{4})  + 2cos²\frac{\pi}{4}  + 2sec²\frac{\pi}{3}

= 2sin²\frac{\pi}{4}  + 2cos²\frac{\pi}{4}  + 2sec²\frac{\pi}{3}

Substituting the values,

=2(\frac{1}{\sqrt{2}})^2 + 2(\frac{1}{\sqrt{2}})^2 + 2(2)^2

= 2(\frac{1}{2}) + 2(\frac{1}{2})  + 2(4)

= 1 + 1 + 8

= 10

Hence, LHS = RHS

Question 5: Find the value of:

(i) sin 75°

Solution: 

As, we don't know the angle value for 75°, so we will break into the angles which we know.

75° = 30° + 45°, so lets use this and solve for sin(30° + 45°)

Using the trigonometric formula,

sin (A+B) = sin A cos B + cos A sin B

sin(30° + 45°) = sin 30° cos 45° + cos 30° sin 45°

Substituting values, we get

sin(75°) =(\frac{1}{2}) (\frac{1}{\sqrt{2}}) + (\frac{\sqrt{3}}{2}) (\frac{1}{\sqrt{2}})

sin(75°) =(\frac{1}{2\sqrt{2}}) + (\frac{\sqrt{3}}{2\sqrt{2}})

sin(75°) =(\frac{1+\sqrt{3}}{2\sqrt{2}})

(ii) tan 15°

Solution: 

As, we don't know the angle value for 15°, so we will break into the angles which we know.

15° = 60° - 45° or 45° - 30° so lets use this and solve for tan(45° - 30°)

Using the trigonometric formula,

tan (A-B) =\mathbf{\frac{tan A-tan B}{1+tan A tan B}}

tan(45° - 30°) =\frac{tan 45\degree-tan 30\degree}{1+tan 45\degree tan 30\degree}

Substituting values, we get

tan(15°) =\frac{1-\frac{1}{\sqrt{3}}}{1+(1)(\frac{1}{\sqrt{3}})}

tan(15°) =\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}

tan(15°) =\frac{\sqrt{3}-1}{\sqrt{3}+1}

Now rationalizing the denominator, multiply and divide by(\sqrt{3}-1)

tan(15°) =\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}

tan(15°) =\frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-1^2}

tan(15°) =\frac{(\sqrt{3})^2+1^2 - 2(\sqrt{3})(1)}{3-1}

tan(15°) =\frac{3+1 - 2\sqrt{3}}{2}

tan(15°) =\frac{4 - 2\sqrt{3}}{2}

tan(15°) = 2 -\sqrt{3}

Prove the following:

Question 6:cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y)  = sin (x+y)

Solution: 

Taking LHS in consideration, we get

=cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y)

As, here there is multiplication of cos cos and sin sin, we will use Defactorisation Formulae,

2 cos A cos B = cos (A+B) + cos (A-B) and, ...................(1)

2 sin A sin B = cos (A-B) - cos (A+B) ...................(2)

Multiply and divide LHS by 2, we get

=\frac{2}{2}(cos (\frac{\pi}{4}-x)cos (\frac{\pi}{4}-y)- sin (\frac{\pi}{4}-x)sin (\frac{\pi}{4}-y))

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

Subtracting (2) from (1) and using identity formulae, we get

2 cos A cos B - 2 sin A sin B = cos (A+B) + cos (A-B) - (cos (A-B) - cos (A+B))

2 cos A cos B - 2 sin A sin B = 2 cos (A+B)

Hence, using this

=\frac{1}{2}[2 cos ((\frac{\pi}{4}-x)+ (\frac{\pi}{4}-y))]

= cos(\frac{2\pi}{4}-x-y)

= cos(\frac{\pi}{2}-(x+y))

= sin (x+y) (As cos(\frac{\pi}{2}-\theta)  = sin θ)

Hence, LHS = RHS

Question 7:\frac{tan(\frac{\pi}{4}+x)}{tan(\frac{\pi}{4}-x)} = (\frac{1+tan x}{1-tan x})^2

Solution: 

Taking LHS in consideration, we get

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

As, using Factorisation Formulae of tan, we have

tan (A+B) =\mathbf{\frac{tan A + tan B}{1 - tan A tan B}}  and,

tan (A-B) =\mathbf{\frac{tan A - tan B}{1 + tan A tan B}}

Now, substituting the values

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

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

=\frac{1 + tan x}{1 - tan x} \times \frac{1 + tan x}{1 - tan x}

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

Hence, LHS = RHS

Question 8:\frac{cos(\pi + x)\hspace{0.1cm}cos(-x)}{sin (\pi-x) \hspace{0.1cm}cos(\frac{\pi}{2}+x)}  = cot²x

Solution: 

Taking LHS in consideration, we get

=\frac{cos(\pi + x)\hspace{0.1cm}cos(-x)}{sin (\pi-x) \hspace{0.1cm}cos(\frac{\pi}{2}+x)}

As, we know these standard values

cos(-x) = cos x

cos(\pi + x)  = - cos x

sin(\pi-x)  = sin x

cos(\frac{\pi}{2} + x)  = - sin x

Substituting these values, we have

=\frac{(-cos x)\hspace{0.1cm}cos x}{sin x \hspace{0.1cm}(- sin x)}

=\frac{cos^2 x}{sin^2 x}

= cot² x

Hence, LHS = RHS

Question 9:cos (\frac{3\pi}{2}+x)\hspace{0.1cm}cos (2\pi+x)[\hspace{0.1cm}cot (\frac{3\pi}{2}-x)+cot(2\pi+x)]  = 1

Solution: 

Taking LHS in consideration, we get

=cos (\frac{3\pi}{2}+x)\hspace{0.1cm}cos (2\pi+x)[\hspace{0.1cm}cot (\frac{3\pi}{2}-x)+cot(2\pi+x)]

As, we know these standard values

cos(\frac{3\pi}{2}+x)  = sin x

cos(2\pi + x)  = cos x

cot(2\pi + x)  = cot x

cot(\frac{3\pi}{2}-x)  = tan x

Substituting the values, we have

=sin x\hspace{0.1cm}cos x[\hspace{0.1cm}tan x + cot x]

=sin x\hspace{0.1cm}cos x[\frac{sin x}{cos x} + \frac{cos x}{sin x}]

=sin x\hspace{0.1cm}cos x[\frac{sin^2 x + cos^2 x}{sin x \hspace{0.1cm}cos x}]

As sin² x + cos² x = 1

= 1

Hence, LHS = RHS

Question 10: sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x

Solution: 

Taking LHS in consideration, we get

= sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x

As, here there is multiplication of cos cos and sin sin, we will use Defactorisation Formulae,

2 cos A cos B = cos (A+B) + cos (A-B) and, ...................(1)

2 sin A sin B = cos (A-B) - cos (A+B) ...................(2)

Multiply and divide LHS by 2, we get

=\frac{2}{2}  (sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x)

=\frac{1}{2}  (2 sin(n + 1)x sin(n + 2)x + 2 cos(n + 1)x cos(n + 2)x)

Adding (1) and (2) and using identity formulae, we get

2 cos A cos B + 2 sin A sin B = cos (A+B) + cos (A-B) + cos (A-B) - cos (A+B)

2 cos A cos B + 2 sin A sin B = 2 cos (A-B)

Hence, using this

=\frac{1}{2}  (2 cos((n + 1)x - (n + 2)x))

= cos((n + 1)x - (n + 2)x)

= cos (x-2x)

= cos (- x)

= cos x (As, cos(-x) = cos x)

Hence, LHS = RHS

Question 11:cos (\frac{3\pi}{4}+x)-cos (\frac{3\pi}{4}-x)  = -\sqrt{2}  sin x

Solution: 

Taking LHS in consideration, we get

=cos (\frac{3\pi}{4}+x)-cos (\frac{3\pi}{4}-x)

Using the identity,

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

Substituting the values,

=2 sin (\frac{(\frac{3\pi}{4}-x)+(\frac{3\pi}{4}+x)}{2}) sin (\frac{(\frac{3\pi}{4}-x)-(\frac{3\pi}{4}+x)}{2})

= 2 sin(\frac{6\pi}{4\times 2})  sin(\frac{(-x-x)}{2})

= 2 sin(\frac{3\pi}{4})  sin (-x)

= 2 sin(\pi - \frac{\pi}{4})  sin (-x)

= 2 ( sin(\frac{\pi}{4})  ) sin (-x)

= 2(\frac{1}{\sqrt{2}})  (- sin x)

=\frac{-2 sin x}{\sqrt{2}}

Rationalizing the denominator, by multiplying and dividing by\sqrt{2}

=\frac{-2 sin x}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}

=\frac{-2 \sqrt{2} \hspace{0.1cm}sin x}{(\sqrt{2})^2}

=\frac{-2 \sqrt{2}\hspace{0.1cm} sin x}{2}

=-\sqrt{2}  sin x

Hence, LHS = RHS

Question 12: sin² 6x – sin² 4x = sin 2x sin 10x

Solution: 

Taking LHS in consideration, we get

= sin² 6x – sin² 4x

= sin 6x sin 6x – sin 4x sin 4x

As, here there is multiplication of sin sin, we will use Defactorisation Formulae,

2 sin A sin B = cos (A-B) - cos (A+B)

Multiply and divide LHS by 2, we get

=\frac{2}{2}  (sin 6x sin 6x – sin 4x sin 4x)

=\frac{1}{2}  (2 sin 6x sin 6x – 2 sin 4x sin 4x)

Using the identity, we can simplify

=\frac{1}{2}  [(cos(6x-6x) - cos(6x+6x)) – (cos(4x-4x) - cos(4x+4x))]

=\frac{1}{2}  [(cos(0) - cos(12x)) – (cos(0) - cos(8x))]

=\frac{1}{2}  [1 - cos(12x) – 1 + cos(8x)] (As, cos 0 = 1)

=\frac{1}{2}  [cos(8x) - cos (12x)]

Now, using the identity

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

Substituting the values, we have

=\frac{1}{2}[2 \hspace{0.1cm}sin (\frac{8x+12x}{2}) \hspace{0.1cm}sin (\frac{12x-8x}{2})]

= sin(\frac{20x}{2})  sin(\frac{4x}{2})

= sin (10 x) sin (2x)

Hence, LHS = RHS

Question 13: cos² 2x – cos² 6x = sin 4x sin 8x

Solution: 

Taking LHS in consideration, we get

= cos² 2x – cos² 6x

= cos 2x cos 2x – cos 6x cos 6x

As, here there is multiplication of cos cos, we will use Defactorisation Formulae,

2 cos A cos B = cos (A+B) + cos (A-B)

Multiply and divide LHS by 2, we get

=\frac{2}{2}  (cos 2x cos 2x – cos 6x cos 6x)

=\frac{1}{2}  (2 cos 2x cos 2x – 2 cos 6x cos 6x)

Using the identity, we can simplify

=\frac{1}{2}  [(cos(2x+2x) + cos(2x-2x)) – (cos(6x+6x) + cos(6x-6x))]

=\frac{1}{2}  [(cos(2x+2x) + cos(0)) – (cos(6x+6x) + cos(0))]

=\frac{1}{2}  [(cos(4x) + 1 – cos(12x) - 1)] (As, cos 0 = 1)

=\frac{1}{2}  [cos(4x) – cos(12x)]

Now, using the identity

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

Substituting the values, we have

=\frac{1}{2}[2 \hspace{0.1cm}sin (\frac{4x+12x}{2}) \hspace{0.1cm}sin (\frac{12x-4x}{2})]

= sin(\frac{16x}{2})  sin(\frac{8x}{2})

= sin (8x) sin (4x)

Hence, LHS = RHS

Question 14: sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x

Solution: 

Taking LHS in consideration, we get

sin 2x + 2 sin 4x + sin 6x

After rearranging, we have

= (sin 2x + sin 6x) + 2 sin 4x

Using the identity, we can simplify

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

= 2 sin(\frac{6x+2x}{2})  cos(\frac{6x-2x}{2})  + 2 sin 4x

= 2 sin(\frac{8x}{2})  cos(\frac{4x}{2})  + 2 sin 4x

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

Taking (2 sin 4x), we have

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

= 2 sin (4x) (2 cos² x - 1 + 1) (As, cos 2θ = 2cos² θ - 1)

= 2 sin (4x) (2 cos² x)

= 4 sin (4x) cos² x

Hence, LHS = RHS