##### VMC VIQ Scholarship Test

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Edited By Sumit Saini | Updated on Jul 29, 2022 03:19 PM IST

Questions based on different concepts are asked in Exercise 3.3 Class 11 Maths which includes proof related questions of trigonometric functions. NCERT Solutions for class 11 maths chapter 3 exercise 3.3 is relatively difficult as compared to previous exercises. These questions are asked only in subjective papers like School final examinations. Exercise 3.3 Class 11 Maths questions might take more time to solve but once done, one can expect a good score in the examination. NCERT Solutions for class 11 maths chapter 3 exercise 3.3 is a must to do exercise. Also for the other exercises of NCERT one can find the below.

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

**Suggested: ****JEE Main: high scoring chapters | ****Past 10 year's papers**

**Scholarship Test:** **Vidyamandir Intellect Quest (VIQ)**

- Trigonometric Functions NCERT Exercise 3.1
- Trigonometric Functions NCERT Exercise 3.2
- Trigonometric Functions NCERT Exercise 3.4
- Trigonometric Functions NCERT Miscellaneous Exercise

Question:1 Prove that

Answer:

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:

= R.H.S.

Question:3 Prove that

Answer:

We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)

R.H.S.

Question:5(i) Find the value of

Answer:

We know that

(sin(x+y)=sinxcosy + cosxsiny)

Using this idendity

Question:6 Prove the following:

Answer:

Multiply and divide by 2 both cos and sin functions

We get,

Now, we know that

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

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

We use these two identities

In our question A =

B =

So,

As we know that

By using this

R.H.S

Question:8 Prove the following

Answer:

As we know that,

, ,

and

By using these our equation simplify to

R.H.S.

Question:9 Prove the following

Answer:

We know that

So, by using these our equation simplifies to

R.H.S.

Question:10 Prove the following

Answer:

Multiply and divide by 2

Now by using identities

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

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

R.H.S.

Question:11 Prove the following

Answer:

We know that

[ cos(A+B) - cos (A-B) = -2sinAsinB ]

By using this identity

R.H.S.

Question:12 Prove the following

Answer:

We know that

So,

Now, we know that

By using these identities

sin6x + sin4x = 2sin5x cosx

sin6x - sin4x = 2cos5x sinx

Now,

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

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

by using these identities

2cos5x sin5x = sin10x - 0

2sinx cosx = sin2x + 0

hence

Question:13 Prove the following

Answer:

As we know that

Now

By using these identities

cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( sin(-x) = -sin x

cos(-x) = cosx)

cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes

R.H.S.

Question:14 Prove the following

Answer:

We know that

We are using this identity

sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x

sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( cos(-x) = cos x)

So, our equation becomes

sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x

Now, take the 2sin4x common

sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( )

=2sin4x( +1 )

=2sin4x( )

=

R.H.S.

Question:15 Prove the following

Answer:

We know that

By using this , we get

sin5x + sin3x = 2sin4xcosx

now nultiply and divide by sin x

Now we know that

By using this our equation becomes

R.H.S.

Question:20 Prove the following

Answer:

We know that

We use these identities

R.H.S.

Question:22 prove the following

Answer:

cot x cot2x - cot3x(cot2x - cotx)

Now we can write cot3x = cot(2x + x)

and we know that

So,

= cotx cot2x - (cot2xcotx -1)

= cotx cot2x - cot2xcotx +1

= 1 = R.H.S.

Question:23 Prove that

Answer:

We know that

and we can write tan 4x = tan 2(2x)

So, =

=

=

= = R.H.S.

Question:24 Prove the following

Answer:

We know that

We use this in our problem

cos 4x = cos 2(2x)

=

=

= = R.H.S.

Question:25 Prove the following

Answer:

We know that

cos 3x = 4 - 3cos x

we use this in our problem

we can write cos 6x as cos 3(2x)

cos 3(2x) = 4 - 3 cos 2x

= -

=

= 32 - 4 - 48 + 24 -

= 32 - 48 + 18 - 1 = R.H.S.

The NCERT syllabus class 11 maths chapter Trigonometric functions mainly deals with the application part as the theoretical part is discussed in the earlier classes. Exercise 3.3 Class 11 Maths has most of the questions which needs analytical approach to solve. Although basic concepts will be required but more than that, approach will matter as these require stop b step solutions in the right direction. NCERT Solutions for class 11 maths chapter 3 exercise 3.3 has some questions which are very good for the examination point of view.

The NCERT book class 11th maths chapter 3 exercise once completed properly, can help in other subjects also.

Exercise 3.3 Class 11 Maths needs an analytical approach which can be developed by practice only.

Class 11 maths chapter 3 exercise 3.3 solutions are prepared by referring various good sources. Hence students can rely on these without any doubt.

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Download EBook1. Is it very tough to solve some of the questions from this exercise?

Yes, in that case one can refer to the solutions provided here.

2. Can there be different methods to solve questions ?

Yes, in proof related questions, more than one method can be used.

3. Can I get some marks if I am not able to complete the solutions but have done a few steps in the CBSE exam ?

Yes, there are marks for the steps in the CBSE exams

4. In how much time can one master exercise 3.3 maths class 11 ?

This exercise can take 5 to 6 hours if done properly in step by step manner.

5. How many questions are there in exercise 3.3 class 11 maths ?

There are 25 questions in this exercise, mostly proof related questions.

6. Can I skip some of the questions from this chapter for the CBSE exam ?

Not recommended. Can do if there is paucity of time .

7. Can one score 100 percent marks in this chapter questions ?

Yes, as most of the questions are simple and Maths is a very scoring subject.

Sep 06, 2024

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