HOW TO FIND MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC EXPRESSIONS

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Trigonometry (to find Maximum and Minimum Values of Trigonometric Expressions) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

IN BOTH ENGLISH AND HINDI VERSIONS

SHIV KISHOR

MINIMUM MAXIMUM TABLE

Observe that in case of sin² θ and cos² θ, the minimum value is 0 and not (-1). Why does this happen? because (-1)²=+1

ENGLISH VERSION

Here are some formulas, keep them memorized
1. sin² θ + cos² θ = 1
2. 1+ cot² θ = cosec² θ
3. 1+ tan² θ = sec² θ

Negative Signs inside out
Sin (- θ) = – Sin (θ)
Cos (-θ) = Cos (θ)

Ratta-fication formulas
maximum/minimum value of ‘a sin θ ± b cos θ’ = ±√(a² + b² ) —- (for min. use – , for max. use + )

maximum/minimum value of ‘a sin θ ± b sin θ’ = ±(a ± b ) —- (for min. use – , for max. use + )

maximum/minimum value of ‘a cos θ ± b cos θ’ = ±(a ± b ) —- (for min. use – , for max. use + )

maximum/minimum value of (sin θ cos θ)ⁿ = (±½)ⁿ  —- (for min. use – , for max. use + )

Now let’s talk about the logic behind AM, GM
Let A, B are any two numbers then,
Arithmetic Mean (AM)= (A + B)/2 and
Geometric Mean (GM) = √(A*B)
Hence, AM ≥ GM (We can check it by putting any values of A and B)

Consider the following statement “My age is greater than or equal to 25 years.”
What could you conclude about my age from this statement?

Answer: My age can be anywhere between 25 to infinity … means it can be 25, 50, 99, 786 or 1000 years etc… but it can not be 24 or 19 or 16. In fact it can not be less than 25, strictly. Means, We can confidently say that my age is not less 25 years. Or in other words my minimum age is 25 years. Showing numerically, if Age ≥ 25 years (minimum age = 25)

Similarly, If I say x ≥ 56 ( minimum value of x = 56 )
If, y ≥ 77 (minimum value of y = 77 )
If, x + y ≥ 133 (minimum value of x + y = 133 )
If, sin θ ≥ – 1 (minimum value of Sin θ = -1 )
If, tan θ + cot θ ≥ 2 (minimum value of tan θ + cot θ = 2 )

Sometimes, we come across a special case of trigonometric identities like to find maximum/minimum value of sin θ + cosec θ or tan θ + cot θ or cos² θ + sec² θ etc. These identities have one thing in common i.e. the first trigonometric term is opposite of the second term or vice-versa (tan θ = 1/cot θ, sin θ = 1/cosec θ, cos² θ = 1/sec² θ).

These type of problems can be easily tackled by using the concept of AM ≥ GM, meaning, Arithmetic Mean is always greater than or equal to Geometric Mean.

For example:
Find minimum value of 4tan² θ + 9cot² θ (they’ll not ask maximum value of this in the exam as it is not defined.)

We know that tan² θ = 1/cot² θ, hence applying AM ≥ GM logic, we get
AM of given equation = (4tan² θ + 9cot² θ)/2 —- (1)
GM of given equation = √ (4tan² θ*9cot² θ )
= √(4*9) (as tan² θ and cot² θ are inverse of each other, so tan*cot = 1)
= √36 = 6 —- (2)

Now, we know that A.M ≥ G. M
From equations (1) and (2) above we get,
(4tan² θ + 9cot² θ)/2 ≥ 6
Multiplying both sides by 2
4tan² θ + 9cot² θ ≥ 12 (therefore minimum value of tan² θ + cot² θ is 12 )

Deriving a common conclusion:
Consider this expression to find the minimum value ‘a cos2 θ + b sec2 θ’
As, A.M ≥ G.M
(a cos² θ + b sec² θ/2) ≥ √(a cos² θ*b sec² θ)
= a cos² θ + b sec² θ ≥ 2 √(ab)                          (minimum value 2 √ab)
So, we can use 2√ab directly in these kind of problems.

Conclusion:
While using AM ≥ GM logic, term should be like a T1 + b T2; where T1 = 1/T2

Positive sign in between terms is mandatory, otherwise how you would calculate mean!

Directly apply 2√(ab)

Rearrange/Break terms if necessary; priority should be given to direct use of identities.

Extra facts:
The reciprocal of 0 is + ∞ and vice-versa.
The reciprocal of 1 is 1 and -1 is -1

SSC CGL 2012 Tier II Question

1. What is The minimum value of sin² θ + cos² θ + sec² θ + cosec² θ + tan² θ + cot² θ

A) 1
B) 3
C) 5
D) 7

Solution:
We know that sin² θ + cos² θ = 1
Therefore,
(sin² θ + cos² θ) + sec² θ + cosec² θ + tan² θ + cot² θ
= 1 + sec² θ + cosec² θ + tan² θ + cot² θ

Using AM ≥ GM logic for tan² θ + cot² θ we get
= 1 + 2 + sec² θ + cosec² θ

Changing into sin and cos values
= 1 + 2 + (1/cos² θ) + (1/sin² θ)
solving taking L.C.M
= 1 + 2 + (sin² θ + cos² θ)/( sin² θ . cos² θ) —- (1)

but we already know two things
sin² θ + cos² θ=1
Min. value of (sin θ cos θ)n = (½)n

Apply them into equation (1) we get
= 1 + 2 + (sin² θ + cos² θ)/( sin² θ . cos² θ)
= 1 + 2 + (1/1/4) = 1+2+4
= 7 (option ‘D’)

2. The least value of 2 sin² θ + 3 cos² θ

A) 1
B) 2
C) 3
D) 5

We can solve this question via two approaches

Approach #1
Break the equation and use identity no. 1
= 2 sin² θ + 2 cos² θ + cos² θ
= 2 (sin² θ + cos² θ) + cos² θ; (but sin² θ + cos² θ = 1)
= 2 + cos² θ; (but as per min-max table, the minimum value of cos² θ = 0)
= 2 + 0 = 2 (option ‘B’)

Approach #2
convert equation into one identity, either sin or cos

first convert it into a sin equation :
= 2 sin² θ + 3 (1- sin² θ); (because sin² θ + cos² θ=1 => cos² θ = 1- sin² θ)
= 2 sin² θ + 3 – 3 sin² θ
= 3 – sin² θ
= 3 – 1 = 2 (but minimum value of sin² θ is 0 —-confusing ???? )
As sin² θ is preceded by a negative sign therefore we have to take maximum value of sin² θ in order to get minimum value.

Converting into a cos equation:
= 2sin² θ + 3cos² θ
= 2(1- cos² θ) + 3cos² θ
= 2 – 2cos² θ + 3cos² θ
= 2 + cos² θ
= 2 + 0 = 2 (option ‘B’)

3. The maximum value of sin x + cos x is

A) √2
B) 1/√2
C) 1
D) 2

Solution:
Applying Ratta-fication formulae No. 1
a sin θ ± b cos θ = ±√(a² + b² ) —- (for min. use – , for max. use + )
in the given question, we’ve to find the max value of
Sin x + cos x
= +√(1² + 1² )
= √2 (option ‘A’)

4. The maximum value of 3 sin x – 4 Cos x

A) -1
B) 5
C) 7
D) 9

Solution:
Applying Ratta-fication formulae No. 1
a sin θ ± b cos θ = ±√(a² + b²) —- (for min. use – , for max. use + )

In the given question, we’ve to find the max value of
3sin x – 4cos x
= +√(3²+ 4²)
= √25
= 5 (option ‘B’)

5. Min Max values of sin 4x + 5 are

A) 2, 6
B) 4, 5
C) -4, -5
D) 4, 6

Solution:
We know that, -1 ≤ sin nx ≤ 1
= -1 ≤ sin 4x ≤ 1
Adding 5 throughout, 4 ≤ sin 4x +5 ≤ 6
Therefore, the minimum value is 4 and maximum value is 6 (option ‘D’)

6. Minimum and maximum value of Sin Sin x is

A) do not exist
B) -1, 1
C) sin -1, sin +1
D) -sin 1, sin 1

Solution:
We know that, -1 ≤ sin nx ≤ 1
= sin (-1) ≤ sin x ≤ sin (1)
= – sin 1 ≤ sin x ≤ sin 1;                         [sin(-θ) is same as – sin θ ]
Therefore, Minimum value is –sin 1 and maximum is sin 1 (option ‘D’)

HINDI VERSION

अब एक दो ये फोर्मुले हैं, इन्हें अवश्य याद रखें।
1. sin^2 θ + cos^2 θ = 1
2. 1+ cot^2 θ = cosec^2 θ
3. 1+ tan^2 θ = sec^2

साथ ही ये भी याद रखें।
sin (- θ) = – sin (θ)
cos (-θ) = cos (θ)

कुछ फॉर्मुले
maximum/minimum value of ‘a sin θ ± b cos θ’ = ±√(a² + b² ) [min. के लिए – , तथा max. के लिए + प्रयुक्त करें]

maximum/minimum value of ‘a sin θ ± b sin θ’ = ±(a ± b) [min. के लिए – , तथा max. के लिए + प्रयुक्त करें]

maximum/minimum value of ‘a cos θ ± b cos θ’ = ±(a ± b ) [min. के लिए – , तथा max. के लिए + प्रयुक्त करें]

maximum/minimum value of (sin θ cos θ)ⁿ = (± ½)ⁿ  [min. के लिए – , तथा max. के लिए + प्रयुक्त करें]

Arithmetic Mean (AM), Geometric Mean (GM) Logic के बारे में भी बात कर लेते हैं।
माना A, B कोई दो नम्बर हैं तब,
समांतर माध्य (AM)= (A + B)/2 तथा
गुणोत्तर माध्य (GM) = √(A*B)
इसलिए AM ≥ GM

अब इस कथन पे ध्यान दो “रमेश की आयु 20 साल के बराबर या उससे ज्यादा है।”
अब इस कथन से आप रमेश की आयु के बारे में क्या राय देंगे?

उत्तर : रमेश की आयु 20 साल से लेकर अनंत तक कुछ भी अर्थात 25, 30, 50 , 99, 112 या 1000 साल इत्यादि हो सकती है लेकिन यह 19 अथवा 16 नही हो सकती।
मतलब की रमेश की आयु 20 से कम नही हो सकती अर्थात रमेश की न्यूनतम आयु (minimum age) 20 साल है। गणितीय भाषा में इसे ऐसे भी लिख देते हैं। आयु ≥ 20 साल।

इसी तरह से, यदि x ≥ 56 (तब x की minimum value = 56)
यदि, y ≥ 72 (y की minimum value = 72)
यदि, x + y ≥ 133 (तब x+y की minimum value = 133)
यदि, sin θ ≥ – 1 (तब Sin θ का न्यूनतम मान = -1)

कभी कभी हमे विशेष त्रिकोणमितिक फलनों जैसे कि sin θ + cosec θ या tan θ + cot θ या फिर cos² θ + sec² θ इत्यादि का अधिकतम/न्यूनतम मान निकालना पड़ता है। इनमे आपने देखा की एक चीज़ तो कॉमन है कि पहला पद दुसरे का व्युत्क्रम है (अर्थात tan θ = 1/cot θ, sin θ = 1/cosec θ, cos² θ = 1/sec² θ)

ऐसी सभी समस्याएं AM/GM लॉजिक से आसानी से हल की जा सकती हैं।

We know that AM ≥ GM
मान लो हमे 4tan² θ + 9cot² θ का न्यूनतम मान निकालना है।
(यहाँ परीक्षा में इसकी max value नही पूछी जायेगी क्यूंकि वह not defined है)
क्यूंकि tan² θ = 1/cot² θ, इसलिए AM ≥ GM logic से
दिए हुए फंक्शन का A.M = (4tan² θ + 9cot² θ)/2 —-(1)

इसी प्रकार G.M = √(4tan² θ * 9cot² θ )
= √(4*9)
= √36 = 6 —-(2)
अतः समीकरण (1) व (2) से,
(4tan² θ + 9cot² θ)/2 ≥ 6
=> 4tan² θ + 9cot² θ ≥ 12 (इस प्रकार 4tan² θ + 9cot² θ का न्यूनतम मान 12 है )

अब एक कॉमन अनालिसिस पर आते हैं:
मान लो एक फलन a cos² θ + b sec² θ है जिसकी हमे minimum value निकालनी है।
क्यूंकि AM ≥ GM

Therefore (a cos² θ + b sec² θ)/2 ≥ √(a cos² θ * b sec² θ)
=> a cos² θ + b sec² θ ≥ 2√(ab)
अतः 2√(ab) को सीधे ही हम प्रयोग कर सकते हैं।

परिणाम क्या रहा?
A.M ≥ G.M logic का प्रयोग हम कर सकते हैं यदि:
यदि दिया हुआ फलन इस प्रकार है a T1 + b T2; जबकि T1 = 1/T2
दोनों के मध्य Positive sign ही होना चाहिए
अतः आप सीधे 2√(ab) को अप्लाई कर सकते हैं।
ध्यान दें कि कभी कभी सीधे सीधे कोई सवाल ऐसा नही दिया जाता है लेकिन उसके पदों को rearrange करने पर उन्हें ऐसा बनाया जा सकता है ।

अतिरिक्त तथ्य:
0 का व्युत्क्रम + ∞ है तथा इसका विपरीत भी सत्य है।
1 तथा -1 तो स्वयं के ही व्युत्क्रम होते हैं।

SOME MORE QUESTIONS ON FINDING OF MAXIMUM/MINIMUM VALUES

QUERY 1

The least value of 2cos² θ + 3sin² θ ?

A) 1
B) 2
C) 3
D) 4

Atul Roy
2cos² θ + 3sin² θ

= 2(1-sin² θ) + 3sin² θ

= 2 + sin² θ

Its value will be least for the minimum value of sin² θ. As sin² θ can’t be negative so minimum value of sin² θ = 0

So the least value of the given expression = 2 + 0 = 2

And its value will be maximum for the greatest value of sin² θ. The maximum value of sin² θ = 1

So the maximum value of the given expression = 2 + 1 = 3 (option ‘3’)

QUERY 2

What is the minimum value of 16tan² A + 25cot² A

A) 20
B) 40
C) 41
D) 82

JAYANTCHARAN CHARAN
Make it a perfect square. It can be done in the following two ways

i) (4tan A + 5cot A)² – 2 x 4tan A x 5cot A

ii) (4tan A – 5cot A)² + 2 x 4tan A x 5cot A

The minimum value of a square number is always zero
Therefore i) is 0 – 40 = -40 & ii) is 0 + 40 = 40

But as the question itself is an addition of squares; so it has to be positive.

Therefore 40 (option ‘B’) is the answer

QUERY 3

What is the minimum value of cos² @

A) 5
B) 1
C) 4
D) 0

YOGESH DAHIYA
Method1
Square of a number or identity is always greater than or equal to 0.

so given identity has minimum value 0 because cos @ is 0 at 90°. (option ‘D’)

Method 2
-1<cos @<1

so 0<cos² @<1

hence minimum value is 0. (option ‘D’)

Method 3
cos² @ = (1 – sin2 @)/2
Given will be minimum when sin² @ is maximum; and maximum value of sin² @ is 1

so given expression =
(1 – 1)/2 = 0 (option ‘D’)

Method 4
Use maxima/minima approach.

Method 5 (fastest method) –
Make graph of given expression using approaches used in 12th class.

QUERY 4

What is the minimum value of 2sin² @ + 3cos² @

A) 0
B) 1
C) 2
D) 5

RONNIE BANSAL
sin² x + cos² x = 1

=> 2sin² x = 2 – 2cos² x

So, 2sin² x + 3cos² x
= 2 – 2cos² x + 3cos² x

= cos² x + 2

A number squared can’t be negative so the minimum value cos² x can take is 0,
The minimum value of cos² x + 2 is 2, and so is the minimum value of 2sin² x + 3cos² x

Answer: 2 (option ‘C’)

QUERY 5

The maximum value of sin⁸ A + cos¹⁴ A for all real values of A is

A) 1/√2
B) 0
C) 1
D) √2

Yogesh Dahiya
Value of given expression is equal to 1 at angle 0 or 90; hence either option C or D

but it can’t be √2 because powers of sin and cos are 8 and 14
respectively i.e. large even powers. So by the method of elimination of options the answer to the given question is 1. (option ‘C’)

NOTE: This question can also be solved by maxima /minima technique but it would be lengthy to solve that way.

QUERY 6

Find the minimum value of 4tan² X + 9cot² X

A) 10
B) 5
C) 7
D) 12

SHIV KISHOR
Suppose tan X= A
Then given expression is 4A² + 9/A²
And we have to evaluate its minimum value

It can be expressed as [2A- (3/A)]² + 12
Since minimum value of [2A – (3/A)]² is only zero.
Thus minimum value of given expression is 0 + 12 = 12 (option ‘D’)

QUERY 7

Maximum value of cos @ – sin @

A) √5
B) 5
C) 2
D) √2

MAHA GUPTA
Ratta-fication formula
Maximum/minimum value of ‘a sin θ ± b cos θ’ = ±√(a² + b² ) —- (for min. use – , for max. use + )

Here we see that a = 1, and b = -1

Therefore according to the above formula maximum value of cos@ – sin@ = +√(a² + b²)
= +√[1² + (-1)²)]
= √2 (option ‘D’)