Main content

Course: Class 11 > Unit 3

Lesson 7: Solutions to select NCERT problems

Select problems from exercise 3.3

Solutions to few problems from NCERT exercise.

In this article we will look at solutions of a few selected problems from exercise 3.3 of NCERT.

Problem 1:

Prove the following:

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

Solution:

First see that the expression on the LHS is of the form

\cos A \cos B - \sin A \sin B

. Can we recall a formula that fits this expression? Yes, remember

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

Utilising the above we can write

\begin{aligned} \cos (\frac{π}{4} - x) \cos (\frac{π}{4} - y) - \sin (\frac{π}{4} - x) \sin (\frac{π}{4} - y) \\ = \cos [(\frac{π}{4} - x) + (\frac{π}{4} - y)] \\ = \cos [\frac{π}{2} - (x + y)] \\ = \sin (x + y) ∵ \cos (\frac{π}{2} - θ) = \sin θ \end{aligned}

Problem 2:

Prove the following:

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

Solution:

First let us simplify all the terms on the LHS one-by-one.

Because trigonometric functions are periodic after an interval of

2 π

\begin{array}{r} \cos (2 π + x) = \cos x \\ \cot (2 π + x) = \cot x \end{array}

Also,

\cos (\frac{3 π}{2} + x) = \sin x

\cos (\frac{3 π}{2} + x)

, note that the argument is odd multiple of

\frac{π}{2}

plus

x

. So here

\cos

will change to

\sin

Yes. In such cases, whenever the argument is odd multiple of

\frac{π}{2}

plus

x

or minus

x

\sin

changes into

\cos

and vice versa;

\sec

changes into

cosec

and vice versa;

\tan

changes into

\cot

and vice versa.

For example,

\begin{aligned} \sin (\frac{5 π}{2} - x) & = + \cos x \\ \cos (\frac{π}{2} + x) & = - \sin x \\ \tan (\frac{3 π}{2} - x) & = + \cot x \\ \cos (\frac{7 π}{2} - x) & = - \sin x \end{aligned}

On how to determine whether it would be a

+

sign or

-

sign on the RHS, continue reading the original explanation.

Therefore,

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

. To choose between

+

and

-

, we look at the quadrant in which

\frac{3 π}{2} + x

lies (assuming

x

to be an acute angle).

\frac{3 π}{2} + x

lies in the fourth quadrant. Here

X

coordinate is positive. So,

\cos (\frac{3 π}{2} + x)

will be positive. Hence we choose the

+

sign.

\cos (\frac{3 π}{2} + x) = \sin x

try it out

Similarly,

\cot (\frac{3 π}{2} - x)

is equal to

Now we can write the expression on the LHS in the question as

\begin{aligned} \cos (\frac{3 π}{2} + x) \cos (2 π + x) [\cot (\frac{3 π}{2} - x) + \cot (2 π + x)] \\ = \sin x \cos x (\tan x + \cot x) \\ = \sin x \cos x (\tan x + \frac{1}{\tan x}) \\ = \sin x \cos x (\frac{\tan^{2} x + 1}{\tan x}) \\ = \sin x \cos x (\frac{\sec^{2} x}{\tan x}) \\ = \sin x \cos x (\frac{1}{\cos^{2} x} \frac{\cos x}{\sin x}) \\ = 1 \end{aligned}

Problem 3:

Prove the following:

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

Solution:

Let's see the expression on the left. The first thing that comes to mind is to use the

\sin A + \sin B

formula, which is

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

Here we can apply it in two different ways.

\begin{aligned} Approach 1: & (\sin 2 x + \sin 4 x) + (\sin 4 x + \sin 6 x) \\ Approach 2: & (\sin 2 x + \sin 6 x) + (2 \sin 4 x) \end{aligned}

Both the approaches will lead us to the answer. Let's follow the second approach here.

\begin{aligned} (\sin 2 x + \sin 6 x) + (2 \sin 4 x) \\ = 2 \sin 4 x \cos (- 2 x) + 2 \sin 4 x \\ = 2 \sin 4 x \cos 2 x + 2 \sin 4 x ∵ \cos (- x) = \cos x \\ = 2 \sin 4 x (\cos 2 x + 1) \\ = 2 \sin 4 x (2 \cos^{2} x - 1 + 1) ∵ \cos 2 x = 2 \cos^{2} x - 1 \\ = 4 \cos^{2} x \sin 4 x \end{aligned}

Problem 4:

Prove the following:

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

Solution:

How can we simplify the LHS? For the numerator, we can use the formula

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

For the denominator, we can use the formula

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

Given expression can be written as

\begin{aligned} \frac{\sin x - \sin 3 x}{\sin^{2} x - \cos^{2} x} \\ = \frac{2 \sin (- x) \cos 2 x}{- \cos 2 x} \\ = \frac{- 2 \sin x \cos 2 x}{- \cos 2 x} ∵ \sin (- x) = - \sin x \\ = 2 \sin x \end{aligned}

Problem 5:

Prove the following:

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

Solution:

Okay, to be honest, I don't remember the

\cot (x + y)

\cot (x - y)

formulae. Who memorizes $\cot$ ‍ formulae seriously?

I do however remember the

\tan (x + y)

and

\tan (x - y)

formulae. So here I'll try to convert the

\cot

expressions into

\tan

and try to solve this problem.

Let's simplify the LHS.

\begin{aligned} \cot x \cot 2 x - \cot 2 x \cot 3 x - \cot 3 x \cot x \\ = \cot x \cot 2 x - \cot 3 x (\cot 2 x + \cot x) \\ = \cot x \cot 2 x - \frac{1}{\tan 3 x} (\frac{1}{\tan 2 x} + \frac{1}{\tan x}) \\ = \cot x \cot 2 x - \frac{1}{\tan 3 x} (\frac{\tan x + \tan 2 x}{\tan x \tan 2 x}) \end{aligned}

Now we apply the formula

\begin{aligned} \tan (x + y) & = \frac{\tan x + \tan y}{1 - \tan x \tan y} \\ \tan 3 x = \tan (2 x + x) & = \frac{\tan 2 x + \tan x}{1 - \tan 2 x \tan x} \end{aligned}

Putting the above into the expression before, we get

\begin{aligned} \cot x \cot 2 x - \frac{1}{\tan 3 x} (\frac{\tan x + \tan 2 x}{\tan x \tan 2 x}) \\ = \cot x \cot 2 x - \frac{(1 - \tan 2 x \tan x)}{(\tan 2 x + \tan x)} (\frac{\tan x + \tan 2 x}{\tan x \tan 2 x}) \\ = \frac{1}{\tan x \tan 2 x} - (\frac{1 - \tan 2 x \tan x}{\tan x \tan 2 x}) \\ = \frac{1 - 1 + \tan 2 x \tan x}{\tan x \tan 2 x} \\ = 1 \\ = RHS \end{aligned}

In general if you don't remember the $\cot$ ‍ formulae, many problems can be solved by converting $\cot$ ‍ into $\tan$ ‍.

Problem 6:

Prove the following:

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

Solution:

Recall the double angle identity for

\cos

\begin{aligned} \cos (2 x) & = \cos^{2} x - \sin^{2} x \\ = 2 \cos^{2} x - 1 \\ = 1 - 2 \sin^{2} x \end{aligned}

The question is which one do we use for given problem? See the RHS of the problem. The pattern there suggests that we should apply the third identity.

\begin{aligned} \cos (4 x) & = 1 - 2 \sin^{2} 2 x \\ = 1 - 2 (2 \sin x \cos x)^{2} ∵ \sin 2 x = 2 \sin x \cos x \\ = 1 - 8 \sin^{2} x \cos^{2} x \end{aligned}

Want to join the conversation?

Sort by:

oliveratwork
Posted 8 months ago. Direct link to oliveratwork's post “The end of problem 5 conf...”
The end of problem 5 confused me. Can you explain from the 3rd to last step to the last step which gets the final answer as 1? Thanks.
Button navigates to signup pageComment on oliveratwork's post “The end of problem 5 conf...”
(0 votes)
Answer