Transcript
This is a guest article by Mr.Dipak Singh.
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. sin2 θ + cos 2 θ = 1 2. 1+ cot 2 θ = cosec2 θ 3. 1+ tan 2 θ = sec2 θ how did we get these formulas? Already explained, click me
Min-Max table Min value -1
Max value +1
Can be written as sin θ, sin 2θ, sin 9θ …. sin nθ -1 ≤ Sin nθ ≤ 1 cos θ, cos 4θ , cos 7θ … cos nθ -1 ≤ Cos nθ ≤ 1 2 2 2 2 sin θ , sin 4θ , sin 9θ …sin nθ 0 +1 Can be written as 0 ≤ Sin2 nθ ≤ 1 cos2 θ , cos2 3θ , 3θ , cos2 8θ … cos2 nθ 0 ≤ Cos2 nθ ≤ 1 Sin θ Cos θ -1/2 +1/2 -1/2 ≤ Sin θ Cos θ ≤ ½ 2 2 observe that in case of sin θ and cos θ, the minimum value if 0 a nd not (-1). ( -1). Why does this happen?
because (-1)2=+1
Negative Signs inside out
Sin (- θ) = - Sin (θ)
Cos (-θ) (-θ) = Cos (θ)
Ratta-fication formulas 1.
a sin θ ± b cos θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }
2.
a sin θ ± b sin θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }
3.
a cos θ ± b cos θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }
4.
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 m y 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 = 1 33 )
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 + 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 (identitiy#1) Therefore, (sin2 θ + cos 2 θ) + sec2 θ + cosec2 θ + tan2 θ + cot2 θ
= (1) + sec 2 θ + cosec2 θ + tan2 θ + cot2 θ Using A.M ≥ G.M logic for tan 2 θ + cot2 θ we get , = 1 + 2 + sec 2 θ + cosec2 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/ cos2 θ) + (1/ sin 2 θ) solving taking L.C.M = 1 + 2 + (sin 2 θ + cos2 θ)/( sin2 θ . cos2 θ)…..eq1 but we already know two things sin2 θ + cos 2 θ=1 (trig identity #1) 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 + cos2 θ)/( sin2 θ . cos2 θ) = 1 + 2 + (1/1/4) = 1+2+4 = 7 (correct answer D)
The least value of 2 sin 2 θ + 3 cos2 θ (CGL2012T1) 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 cos 2 θ + cos2 θ =2(sin2 θ + cos2 θ) + cos 2 θ ; (but sin 2 θ + cos2 θ=1) = 2 + cos2 θ ;(but as per min-max table, the minimum value of cos2 θ=0) = 2 + 0 = 2 (correct answer 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 Min. value of sin 2 θ is 0 …confusing ???? ) As sin2 θ is preceded by a negative sign therefore we have to take max. value of sin 2 θ in order to get minimum value . Converting into a cos equation :
= 2 sin2 θ + 3 cos 2 θ = 2 (1- cos 2 θ) + 3 cos 2 θ = 2 – 2 cos2 θ + 3 cos2 θ = 2 + cos2 θ = 2 + 0 = 2 ( correct answer B )
The maximum value of Sin x + cos x is A. √2 B. 1/ √2 C.
1
D. 2 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 = + √ (1 2+ 12 ) = √2 ( correct answer A )
The maximum value of 3 Sin x – 4 Cos x is 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 3 Sin x – 4 Cos x = + √ (3 2+ 42 ) = √25 = 5 ( correct answer B )
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 ( correct answer D )
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
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
