Math Assignment Class XI Ch-3 | Trigonometric Functions

Math Assignment | Class XI | Chapter 3

Trigonometric Functions

Extra questions of chapter 3 class 11 Trigonometric Functions with answer and hints to the difficult questions. Important and useful math. assignment for the students of class 11

For better results

Students should learn all the basic points of Trigonometry up to 11th standard
Student should revise NCERT book thoroughly with examples.
Now revise this assignment. This assignment integrate the knowledge of the students.

ASSIGNMENT FOR XI STANDARD TRIGONOMETRY

STRICTLY ACCORDING TO THE CBSE AND DAV BOARD

Question 1

Find the degree measure for the following radian measure

$i)\:\; \; \frac{2\pi }{15}Radian\; \; \; \; \; ............. \; \; [Ans\; \; 24^{o}]$

$ii)\:\; \; \frac{11 }{16}Radian\; \; \; \; \; ............. \; \; [Ans\; \; 39^{o}\: 22'\: 30'']$

$iii)\:\; \; (-2)Radian\; \; \; \; \; ............. \; \; [Ans\; \; -(114^{o}\: 32'\: 43'')]$

Question 2

Find the radian measure for the following degree measure

$i)\:\; \; -37^{o}30'\; \; \; \; \; ............. \; \; [Ans\; \; -\left ( \frac{5\pi }{24} \right )]$

$ii)\:\; \; 40^{o}\: 20'\; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{121\pi }{540} \right )]$

$iii)\:\; \; 5^{o}\: 37'\:30'' \; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{\pi }{32} \right )]$

Question 3

Find the magnitude, radian and degree, of the interior angles of a regular

$i)\:\; \; Pentagon \; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{3\pi }{5}\; :\; 108^{o} \right )]$

$ii)\:\; \; Octagon \; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{3\pi }{4}\; :\; 135^{o} \right )]$

$iii)\; \; \; Heptagon \; \; \; \; \; ............. \; \; Ans\; \; \left ( \frac{5\pi }{7}\; :\; 128^{o}\; 34'\; 17'' \right )$

Solution Hint:

Note: Each interior angle of a regular polygon is given by :

$\frac{(n-2)}{n}\times 180$ Where n is the number of sides.

Question 4

If sin θ =12/13 and θ lie in the second quadrant, then find the value of sec θ + tan θ. Ans. [- 5]

Question 5

Prove the followings

i) cos24^o + cos 55^o + cos 125^o + cos 204^o + cos 300^o = 1/2

ii) sin 600^o tan(-690^o) + sec 840^o cot(-945^o) = 3/2

iii) $sin^{2}\left \{ \frac{\pi }{8} \right \} +sin^{2}\left \{ \frac{3\pi }{8} \right \}+sin^{2}\left \{ \frac{5\pi }{8} \right \}+sin^{2}\left \{ \frac{7\pi }{8} \right \}=2$

Question 6: Simplify

$tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right )tan\left ( \frac{5\pi }{20} \right )tan\left ( \frac{7\pi }{20} \right )tan\left ( \frac{9\pi }{20} \right )$

Ans: 1

Solution Hint:

$tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right )tan\left ( \frac{\pi }{4} \right )tan\left ( \frac{10\pi-3\pi }{20} \right )tan\left ( \frac{10\pi-\pi }{20} \right )$

$=tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right ).1.tan\left ( \frac{\pi}{2}-\frac{3\pi}{20} \right )tan\left ( \frac{\pi}{2}-\frac{\pi}{20} \right )$

$=tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right )cot\left ( \frac{3\pi}{20} \right )cot\left (\frac{\pi}{20} \right ) = 1$

Question 7

Evaluate the following :

$i) \; \; tan105^{o}\; \; ......\; \; Ans \left [ \frac{\sqrt{3}+1}{1-\sqrt{3}} \right ]$

$ii) \; \; cos105^{o}+sin105^{o}\; \; ......\; \; Ans \left [ \frac{1}{\sqrt{2}} \right ]$

Question 8

Prove that : tan70^o = tan20^o + 2tan50^o

Solution Hint:

$70^{o}=50^{o}+20^{o}\;$

$tan\: 70^{o}=tan(50^{o}+20^{o})$

$tan\: 70^{o}=\frac{tan50^{o}+tan20^{o}}{1-tan50^{o}\; tan20^{o}}$

Now cross-multiply and simplify the above fraction we get the required result.

Question 9

$Prove \; that: \frac{cos9^{o}+sin9^{o}}{cos9^{o}-sin9^{o}}=tan54^{o}$

Question 10

If tan A = k tan B, then show that: $sin(A+B)=\left (\frac{k+1}{k-1} \right )sin(A-B)$

Solution Hint: tan A = k tan B

$\frac{sinA}{cosA}=k\; \frac{sinB}{cosB}$

$\Rightarrow \frac{sinA\: cosB}{cosA\: sinB}=\frac{k}{1}$

Now apply componendo and dividendo and using trigonometric formulas.

Question 11

If tan(A + B) = p, tan(A – B) = q, then show that : $tan\: 2A = \frac{p+q}{1-pq}$

Solution Hint:

Start from RHS and then putting the value of p and q then simplify and get the result.

Question 12

An angle α is divided into two parts such that the ratio of the tangents of the two parts = k and difference of two parts = x then show that:

$sin\: x=\frac{k-1}{k+1}\: sin\alpha$

Solution hint:

Let two parts of α are p and q, Then

ATQ : p + q = α, p – q = x, $\frac{tanp}{tanq}=\frac{k}{1}$

Now applying componendo and dividendo and simplify we get the required result.

Question 13

Prove that : cos20^o cos40^o cos60^o cos80^o = 1/16

Question 14

Prove that : sin20^o sin40^o sin60^o sin80^o = 3/16

Question 15

Prove That : $tan20^{o}\; tan40^{o}\; tan80^{o}=tan60^{o}$

Solution Hint :

Use tan θ = sin θ/cos θ, then using AB and CD formulas

Question 16

If cos(θ + 2 α) = n cos θ , then prove that : $cot\; \alpha =\frac{1+n}{1-n}\; tan(\theta +\alpha )$

Solution Hint: $\frac{cos\: \theta }{cos(\theta +2\alpha) }=\frac{1}{n}$

Applying componendo and dividendo then applying CD formulas then simplify the fractions we get the required result.

Question 17

Prove that: $sec\left ( \frac{\pi }{4}+\theta \right )sec\left ( \frac{\pi }{4}-\theta \right )=2\: sec\: 2\theta$

Solution Hint

Taking LHS and convert these into cosine functions.

Multiply and divide by 2 and the apply AB formulas

Question 18

Prove that: $\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}=tan8A$

Solution Hint:

Multiply and divide by 2 and then apply AB formulas

Question 19

Find the value of sin18^o

Ans: $\left [sin18^{o}=\frac{\sqrt{5}-1}{4} \right ]$

Solution:

θ = 18^o ⇒ 5 θ = 90^o ⇒ 2 θ + 3 θ = 90^o ⇒ 2 θ = 90^o - 3 θ

Now taking sin on both side we get

Sin (2 θ) = Sin (90^o - 3 θ) Sin (2 θ) = Cos (3 θ)

2Sin θ cos θ = 4Cos³ θ – 3cos θ 2Sin θ = 4Cos² θ – 3

2Sin θ = 4(1-sin² θ) – 3

4 sin² θ + 2sin θ – 1 = 0

Now using quadratic formula here and find the value of sin θ

Question 20

Prove that : $\frac{sec8A-1}{sec4A-1}=\frac{tan8A}{tan2A}$

Solution Hint:

$\frac{sec8A-1}{sec4A-1}=\frac{1-cos8A}{cos8A}\times \frac{cos4A}{1-cos4A}$

$=\frac{2sin^{2}4A.\: cos4A}{cos8A.\: 2sin^{2}2A}=\frac{(2sin4A\: cos4A)sin4A}{cos8A.\: 2sin^{2}2A}\$

$=\left (\frac{sin8A}{cos8A} \right ).\frac{2sin2Acos2A}{2sin^{2}A}=\frac{tan8A}{tan4A}$

Question 21

i) Evaluate: $sin\left ( 22\frac{1}{2} \right )^{o}\: \: ...\: \: Ans\left [ \frac{\sqrt{2-\sqrt{2}}}{2} \right ]$

Solution Hint: Using: $Sin\theta =\sqrt{\frac{1-cos2\theta }{2}},\; \; where\; \theta =\frac{45}{2}$

ii) Evaluate: $tan\left ( 22\frac{1}{2} \right )^{o}\: \: ...\: \: Ans\left [ \sqrt{2}-1 \right ]$

Question 22

Prove that: $\frac{cos8Acos5A-cos12Acos9A}{sin8Acos5A+cos12Asin9A}=tan4A$

Solution Hint:

Multiply numerator and denominator by 2

Applying AB formulas we get

$\frac{cos13A-cos21A}{sin13A+sin21A}$

Now applying CD formulas we get the required result.

Question 23

Find $sin\left ( \frac{x}{2} \right )$ , $cos\left ( \frac{x}{2} \right )$ and $tan\left ( \frac{x}{2} \right )$ , when tan x = $-\frac{4}{3}$ , and x lie in II quadrant

Ans:

$sin\left ( \frac{x}{2} \right )=\frac{2}{\sqrt{5}}$

$cos\left ( \frac{x}{2} \right )=\frac{1}{\sqrt{5}}$ , $tan\left ( \frac{x}{2} \right )=2$

Question 24

Prove that: $4sinxsin\left ( x+\frac{\pi }{3} \right )sin\left ( x+\frac{2\pi }{3} \right )=sin3x$

Solution Hint: Apply AB formulas

Question 25

If tan35^o = α , then find the value of $\frac{tan145^{o}-tan125^{o}}{1+tan145^{o}tan125^{o}}$ in terms of α

Ans: $\frac{1-\alpha ^{2}}{2\alpha }$

Question 26

Prove that : $\frac{sin5x-2sin3x+sinx}{cos5x-cosx}=tanx$

Question: 27

Prove that : $sin^{2}x+sin^{2}\left ( x+\frac{\pi }{3} \right )+sin^{2}\left ( x-\frac{\pi }{3} \right )=\frac{3}{2}$

Solution Hint:

Use formula: $sin^{2}x=\frac{1-cos2x}{2}$

Now simplify and then apply CD formula we get the required result

Question 28

Prove that: $tan4x=\frac{4tanx(1-tan^{2}x)}{1-6tan^{2}x+tan^{4}x}$

Question 29

If tan x + tan y + tan x tan y = 1, find (x + y).

Ans: x + y = 45^o

Solution Hint:

tan x + tan y = 1-tan x tan y

Dividing on both side by 1-tanx tany we get

$\frac{tanx+tany}{1-tanxtany}=1=tan45^{o}$

tan(x + y) = tan 45^o ⇒ x + y = 45^o

Question: 30

Prove that: Cos6θ = 32cos⁶ θ - 48cos⁴ θ + 18cos² θ - 1

Solution Hint: Use cos6θ = cos3(2θ) = - 3cos2θ + 4cos³ 2θ and then proceed.

Question: 31

Prove that: cos6x = 1 - 18sin²x + 48sin⁴x - 32sin⁶x

Solution Hint: Use cos6θ = cos3(2θ) = -3cos2θ + 4cos³ 2θ .

Now putting cos2θ = 1 - 2sin²θ we get

cos6θ = -3(1 - 2sin²θ) + 4(1 - 2sin²θ)³

Solving this we get the required result

Question 32

Prove that: $\frac{tan5\theta +tan3\theta }{tan5\theta -tan3\theta}=4cos2\theta cos4\theta$

Solution Hint

Using tanθ = sinθ/cosθ in numerator and in denominator

Taking LCM then using the formula sin(A+B) and sin (A-B) in numerator and in denominator respectively.

Now using sin2θ = 2sinθcosθ in the numerator for two times.

Question: 33

Prove that: sin3x cos³x + cos3xsin³x = $\frac{3}{4}$ sin4x

Solution Hint: Expand sin3x and cos3x

Now multiply and divide by 2

Applying 2sinxcosx = sin2x for two times we get the required result.

Question 34

Prove that : $cos\theta\; cos\frac{\theta }{2}-cos3\theta\; cos\frac{9\theta }{2} = sin4\theta \; sin\frac{7\theta}{2}$

Solution : $cos\theta.\; cos\frac{\theta }{2}-cos3\theta.\; cos\frac{9\theta }{2}$

$=\frac{1}{2}\left [2cos\theta.\; cos\frac{\theta }{2}-2cos3\theta.\; cos\frac{9\theta }{2} \right ]$

$=\frac{1}{2}\left [ cos\left ( \theta +\frac{\theta }{2} \right )+cos\left ( \theta -\frac{\theta }{2} \right ) \right ]- \frac{1}{2}\left [ cos\left ( 3\theta +\frac{9\theta }{2} \right )+cos\left ( 3\theta -\frac{9\theta }{2} \right ) \right ]\$

$=\frac{1}{2}\left [ cos\left (\frac{3\theta }{2} \right )+cos\left (\frac{\theta }{2} \right )-cos\left (\frac{15\theta }{2} \right )-cos\left (\frac{3\theta }{2} \right ) \right ]\$

$=\frac{1}{2}\left [ cos\left (\frac{\theta }{2} \right )-cos\left (\frac{15\theta }{2} \right ) \right ]$

$=\frac{1}{2}\left [ -2sin\frac{1}{2}\left ( \frac{\theta +15\theta }{2} \right ).\; sin\frac{1}{2}\left ( \frac{\theta -15\theta }{2} \right ) \right ]$

$=\frac{1}{2}\left [ -2sin4\theta .\; sin\left ( \frac{-7\theta }{2} \right ) \right ]$

$=sin4\theta .\; sin\left ( \frac{7\theta }{2} \right )$

Question 35

Evaluate: $If\; cos(\alpha +\beta )=\frac{4}{5}\; and\; sin(\alpha -\beta )=\frac{5}{13}$ and

$0<\alpha <\frac{\pi }{4}, \; \; Find \; \; tan2\alpha$

Solution:

$cos(\alpha +\beta )=\frac{4}{5}\Rightarrow sin(\alpha +\beta )=\frac{3}{5}$

$sin(\alpha -\beta )=\frac{5}{13}\Rightarrow cos(\alpha -\beta )=\frac{12}{13}$

$cos[(\alpha +\beta )+(\alpha -\beta )]=cos(\alpha +\beta)cos(\alpha -\beta)-sin(\alpha +\beta)sin(\alpha -\beta)$

$cos(2\alpha )=\frac{4}{5}\times \frac{12}{13}-\frac{3}{5}\times \frac{5}{13}=\frac{33}{65}$

$sin[(\alpha +\beta )+(\alpha -\beta )]=sin(\alpha +\beta)cos(\alpha -\beta)+cos(\alpha +\beta)sin(\alpha -\beta)$

$sin(2\alpha )=\frac{3}{5}\times \frac{12}{13}+\frac{4}{5}\times \frac{5}{13}=\frac{56}{65}$

$tan(2\alpha )=\frac{sin(2\alpha )}{cos(2\alpha )}=\frac{56/65}{33/65}=\frac{56}{33}$

Question: 36

Prove that: sin²α + sin²(α - β) - 2sinα cosβ sin(α - β) = sin²β

Solution Hint:

LHS = sin²α + sin²(α - β) - [sin(α + β) + sin(α - β)]sin(α - β)

= sin²α + sin²(α - β) - sin(α + β)sin(α - β) - sin²(α - β)

= sin²α - [sin²α - sin²β] ..... [Using sin(A+B)sin(A-B) = sin²A - sin²B ]

= sin²β

Question 37

If $xcos\theta =ycos\left ( \theta +\frac{2\pi }{3} \right )=zcos\left ( \theta +\frac{4\pi }{3} \right )$ , then find the value of xy + yz + zx

Answer 0

Solution Hint:

Let all equations = k

Now find the value of x, y, z in terms of k

Now find the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ , it should be '0'

Putting this value in xy + yz + zx = $\frac{xyz}{z}+\frac{xyz}{x}+\frac{xyz}{y}$

= $xyz\left [ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right ]$

= xyz ×0 = 0

Question 38

Find the value of : $\frac{tan^{2}15^{o}-1}{tan^{2}15^{o}+1}$

Ans: $-\frac{\sqrt{3}}{2}$

Solution Hint:

Taking "-" common from numerator

Using the formula: $cos2\theta =\frac{1-tan^{2}\theta }{1+tan^{2}\theta }$

General Solutions of Trigonometric Functions

These questions are not in syllabus for the session 2022-23

Question 39

Find the general solution of the equation: cos x + cos 2x + cos 3x = 0.

Question 40

Solve: $cos\theta + cos2\theta + cos3\theta =0$

$Ans\left [ \theta =(2n+1)\frac{\pi }{4}, \; \; or \; \; 2n\pi \pm \frac{2\pi }{3} \right ]$

Question 41

Solve the equation: sin 3x + sin 5x + sin 7x = 0, $\frac{\pi }{2}<x<\pi$

Question 42

Solve: $cos3\theta +8cos^{3}\theta =0$

$Ans\left [ \theta =(2n+1)\frac{\pi }{2} \; or\; \; n\pi \pm \frac{\pi }{3}\right ]$

Question 43

Solve: $\sqrt{2}sec\; \theta +tan\; \theta =1$

Solution Hint: $\sqrt{2}sec\; \theta +tan\; \theta =1$

$\sqrt{2}\; \frac{1}{cos\theta } +\frac{sin\theta }{cos\theta } =1\Rightarrow \sqrt{2}+sin\theta =cos\theta$

$cos\theta -sin\theta =\sqrt{2}$

$\frac{1}{\sqrt{2}}cos\theta-\frac{1}{\sqrt{2}}sin\theta=1$

$cos\frac{\pi }{4}\; cos\theta-sin\frac{\pi }{4}\; sin\theta=1$

$cos\left ( \theta +\frac{\pi }{4} \right )=1=cos\: 0$

$\theta +\frac{\pi }{4}=2n\pi \pm 0\; \Rightarrow \; \theta =2n\pi -\frac{\pi }{4}$

Question 44

Solve: $\sqrt{3}\; cosx-sinx=1$

$Ans\left [ x=2n\pi +\frac{\pi }{6}\; or\; 2n\pi -\frac{\pi }{2} \right ]$

Solution Hint:

$\sqrt{3}\; cosx-sinx=1$

Dividing on both side by

$\sqrt{(\sqrt{3})^{2}+(-1)^{2}}$ i.e. by 2

$\frac{\sqrt{3}}{2}\; cosx-\frac{1}{2}\; sinx=\frac{1}{2}$
$cos\frac{\pi }{6}\; cosx-sin\frac{\pi }{6}\; sinx=\frac{1}{2}$
Now proceed this question as previous one.

Question 45

Find the general solution of: 2 cos²x + 3 sin x = 0.

Question 46

Find general solution of cos²x cosec x + 3sinx + 3 = 0

Ans: $x=n\pi +(-1)^{n}\frac{3\pi }{2}$ , $x=n\pi +(-1)^{n}\frac{7\pi }{2}$

Solution Hint:

Convert all terms into sin x , we get a quadratic equation.

Solve the quadratic equations.

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