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- Trig-Identities
- Min-Max table
- Ratta-fication formulas
- The AM GM Logic
- Find minimum value of 4 tan2θ + 9 cot2θ
- Extra facts
- SSC CGL 2012 Tier II Question
- The least value of 2 sin
^{2}θ + 3 cos^{2}θ (CGL2012T1) - The maximum value of Sin x + cos x is
- The maximum value of 3 Sin x – 4 Cos x is
- Min Max values of sin 4x + 5 are
- Minimum and maximum value of Sin Sin x is

Today we’ll see how to find the maximum value (greatest value ) or the minimum value (least value) of a trigonometric function without using differentiation. Take a pen and note-book, keep doing the steps while reading this article.

First Remember following identities:

# Trig-Identities

1. sin^{2 }θ + cos^{2} θ = 1

2. 1+ cot^{2} θ = cosec^{2 }θ

3. 1+ tan^{2 }θ = sec^{2 }θ

how did we get these formulas? Already explained, click me

# Min-Max table

Min value | Max value | Can be written as | |

sin θ, sin 2θ, sin 9θ …. sin nθ | -1 | +1 | -1 ≤ Sin nθ ≤ 1 |

cos θ, cos 4θ , cos 7θ … cos nθ | -1 ≤ Cos nθ ≤ 1 |
||

sin^{2 }θ , sin^{2} 4θ , sin^{2} 9θ …sin^{2} nθ |
0 | +1 | Can be written as0 ≤ Sin^{2} nθ ≤ 1 |

cos^{2} θ , cos^{2} 3θ , cos^{2} 8θ … cos^{2} nθ |
0 ≤ Cos^{2} nθ ≤ 1 |
||

Sin θ Cos θ | -1/2 | +1/2 | -1/2 ≤ Sin θ Cos θ ≤ ½ |

observe that in case of sin^{2}θ and cos^{2}θ, the minimum value if 0 and not (-1). Why does this happen? because (-1)^{2}=+1

## Negative Signs inside out

- Sin (- θ) = – Sin (θ)
- Cos (-θ) = Cos (θ)

# Ratta-fication formulas

- a sin θ ± b cos θ = ±√ (a
^{2}+ b^{2}) { for min. use – , for max. use + } - a sin θ ± b sin θ = ±√ (a
^{2}+ b^{2}) { for min. use – , for max. use + } - a cos θ ± b cos θ = ±√ (a
^{2}+ b^{2}) { for min. use – , for max. use + } - Min. value of (sin θ cos θ)
^{n}= (½)^{n}

# The AM GM Logic

Let A ,B are any two numbers then,

Arithmetic Mean (AM)= (A + B) / 2 and

Geometric Mean (GM) = √ (A.B)

- Hence, A.M ≥ G.M ( 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 Sweet 16 . Infact 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 min. 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

A.M ≥ G .M

Meaning, Arithmetic mean is always greater than or equal to geometric mean. For example:

# Find minimum value of 4 tan2θ + 9 cot2θ

(they’ll not ask maximum value as it is not defined. )

We know that tan2θ = 1/ cot2θ , hence applying A.M ≥ G.M logic, we get

A.M of given equation = (4 tan2θ + 9 cot2θ) / 2 …. (1)

G.M of given equation = √ (4 tan2θ . 9 cot2θ )

= √ 4 * 9 # ( tan2θ and cot2θ inverse of each other, so tan x cot =1)

= √ 36 = 6 …. (2)

Now, we know that A.M ≥ G. M

From equations (1) and (2) above we get,

=> (4 tan2 θ + 9 cot2θ) / 2 ≥ 6

Multiplying both sides by 2

=> 4 tan2 θ + 9 cot2 θ ≥ 12 ( minimum value of tan2 θ + cot2 θ is 12 )

## Deriving a common conclusion:

- Consider equation a cos2 θ + b sec2 θ ( find minimum value)
- 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.

## Summary:

While using A.M ≥ G.M logic :

- Term should be like a T1 + b T2 ; where T1 = 1 / T2
- Positive sign in between terms is mandatory. (otherwise how would you calculate mean ? )
- Directly apply 2√ab .
- Rearrange/Break terms if necessary -> priority should be given to direct use of identities -> find terms eligible for A.M ≥ G.M logic -> if any, apply -> convert remaining identities, if any, to sine and cosines -> finally put known max., min. values.

# Extra facts:

- The reciprocal of 0 is + ∞ and vice-versa.
- The reciprocal of 1 is 1 and -1 is -1.
- If a function has a maximum value its opposite has a minimum value.
- A function and its reciprocal have same sign.

Keeping these tools (not exhaustive) in mind we can easily find Maximum or Minimum values easily.

# SSC CGL 2012 Tier II Question

What is The minimum value of sin^{2} θ + cos^{2} θ + sec^{2} θ + cosec^{2} θ + tan^{2} θ + cot^{2} θ

- 1
- 3
- 5
- 7

## Solution:

We know that sin^{2 }θ + cos^{2} θ = 1 (identitiy#1)

Therefore,

** click (sin ^{2} θ + cos^{2} θ)** + sec

^{2}θ + cosec

^{2}θ + tan

^{2}θ + cot

^{2}θ

= (1) + sec^{2} θ + cosec^{2} θ + tan^{2} θ + cot^{2} θ

Using A.M ≥ G.M logic for tan^{2} θ + cot^{2} θ we get ,

= 1 + 2 + sec^{2} θ + cosec^{2} θ

changing into sin and cos values

( Because we know maximum and minimum values of Sin θ, Cos θ :P and by using simple identities we can convert all trigonometric functions into equation with Sine and Cosine.)

= 1 + 2 + (1/ cos^{2} θ) + (1/ sin^{2} θ)

solving taking L.C.M

= 1 + 2 + (sin^{2} θ + cos^{2} θ)/( sin^{2} θ . cos^{2} θ)…..eq1

but we already know two things

** acquistare viagra generico 100 mg a Venezia sin ^{2} θ + cos^{2} θ=1 (trig identity #1)**

** http://cinziamazzamakeup.com/?x=vardenafil-generico-durata-effetto Min. value of (sin θ cos θ) ^{n} = (½)^{n }(Ratta-fication formula #4)**

Apply them into eq1, and we get

= 1 + 2 + (sin^{2} θ + cos^{2} θ)/( sin^{2} θ . cos^{2} θ)

= 1 + 2 + (1/1/4) = 1+2+4

= 7 (correct answer D)

# The least value of 2 sin^{2} θ + 3 cos^{2} θ (CGL2012T1)

- 1
- 2
- 3
- 5

We can solve this question via two approaches

## Approach #1

Break the equation and use identity no. 1

= 2 sin^{2} θ + 2 cos^{2} θ + cos^{2} θ

=2(sin^{2} θ + cos^{2} θ) + cos^{2} θ ; (but sin^{2} θ + cos^{2} θ=1)

= 2 + cos^{2} θ ;(but as per min-max table, the minimum value of cos^{2} θ=0)

= 2 + 0 = 2 (correct answer B)

## Approach #2

convert equation into one identity ,either sin or cos

** http://cinziamazzamakeup.com/?x=viagra-generico-100-mg-pagamento-online-a-Firenze first convert it into a sin equation :**

= 2 sin^{2} θ + 3 (1- sin^{2} θ) ;(because sin^{2} θ + cos^{2} θ=1=>cos^{2} θ=1- sin^{2} θ)

= 2 sin^{2} θ + 3 – 3 sin^{2} θ

= 3 – sin^{2} θ

= 3 – ( 1) = 2 (but Min. value of sin^{2} θ is 0 …confusing ???? )

As sin^{2} θ is preceded by a negative sign therefore we have to take max. value of sin^{2} θ in order to get minimum value .

** source Converting into a cos equation :**

= 2 sin^{2} θ + 3 cos^{2} θ

= 2 (1- cos^{2} θ) + 3 cos^{2} θ

= 2 – 2 cos^{2} θ + 3 cos^{2} θ

= 2 + cos^{2} θ

= 2 + 0 = 2 ( correct answer B )

# The maximum value of Sin x + cos x is

- √2
- 1/ √2
- 1
- 2

Applying Ratta-fication formulae No.1

a sin θ ± b cos θ = ±√ (a^{2} + b^{2} ) { for min. use – , for max. use + }

in the given question, we’ve to find the max value of

Sin x + cos x

= + √ (1^{2}+ 1^{2} )

= √2 ( correct answer A )

# The maximum value of 3 Sin x – 4 Cos x is

- -1
- 5
- 7
- 9

Solution:

Applying Ratta-fication formulae No.1

a sin θ ± b cos θ = ±√ (a^{2} + b^{2} ) { for min. use – , for max. use + }

in the given question, we’ve to find the max value of

3 Sin x – 4 Cos x

= + √ (3^{2}+ 4^{2} )

= √25

= 5 ( correct answer B )

# Min Max values of sin 4x + 5 are

- 2, 6
- 4, 5
- -4, -5
- 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 ( correct answer D )

# Minimum and maximum value of Sin Sin x is

- Do not exist
- -1, 1
- Sin -1 , Sin +1
- – Sin 1 , Sin 1

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 ( correct answer D)

The key to success is Practice! Practice! Practice!

Drop your problems in the comment box.

For more articles on trigonometry and aptitude, visit mrunal.org/aptitude

## 301 Comments on “[Trigonometry] Finding Minimum Maximum Values for SSC CGL Made Easy without differentiation”

What is the extreme values of cos 2x+cos square x

cos(2x) = cos^2 (x) – sin ^2 (x)

so if we convert cos(2x) and simplify , it will come as 3cos^2(x) -1

and min value of cos^2 (x) is 0,

so

min value is “-1″

max value is ” 2″

please correct me if i am wrong.

This answer was really helpful for me.. thank you sir

8sin^2x-cos^2x find min. and max. Value

Please give me all your new.update information about jee main

excellent description !!

Maximum and minimum value of Cos(cosx)

Its cos1 both min and max…