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 + cos2 θ = 1 2. 1+ cot2 θ = cosec2 θ 3. 1+ tan2 θ = sec2 θ how did we get these formulas? Already explained, click me Min-Max table Can be written as -1 ≤ Sin nθ ≤ 1 -1 ≤ Cos nθ ≤ 1 0 +1 Can be written as 0 ≤ Sin2 nθ ≤ 1 cos2 θ , cos2 3θ , cos2 8θ … cos2 nθ 0 ≤ Cos2 nθ ≤ 1 Sin θ Cos θ -1/2 +1/2 -1/2 ≤ Sin θ Cos θ ≤ ½ observe that in case of sin2θ and cos2θ, the minimum value if 0 and not (-1). Why does this happen? because (-1)2=+1 sin θ, sin 2θ, sin 9θ …. sin nθ cos θ, cos 4θ , cos 7θ … cos nθ sin2 θ , sin2 4θ , sin2 9θ …sin2 nθ Min value -1 Max value +1 Negative Signs inside out Sin (- θ) = - Sin (θ) Cos (-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 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 . These type of problems can be easily tackled by using the concept of A. strictly.M of given equation = (4 tan2θ + 9 cot2θ) / 2 …. For example: Find minimum value of 4 tan2θ + 9 cot2θ (they’ll not ask maximum value as it is not defined. If I say x ≥ 56 ( minimum value of x = 56 ) If. the first trigonometric term is opposite of the second term or vice-versa ( tan θ = 1/ cot θ . ” 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 . 9 cot2θ ) = √ 4 * 9 # ( tan2θ and cot2θ inverse of each other. Infact it can not be less than 25. Showing numerically. if Age ≥ 25 years ( minimum age = 25 ) Similarly.The AM GM Logic Let A .M Meaning.M ≥ G . 786 or 1000 years etc… but it can not be 24 or 19 or Sweet 16 .e. x + y ≥ 133 ( minimum value of x + y = 133 ) If.M ≥ G. Means. 50 . A. . so tan x cot =1) . value of sin θ + cosec θ or tan θ + cot θ or cos2 θ + sec2 θ etc. tan θ + cot θ ≥ 2 (minimum value of tan θ + cot θ = 2 ) ]] Sometimes.99. sin θ = 1/ cosec θ . These identities have one thing in common i. hence applying A. cos2 θ = 1/ sec2 θ ). ) We know that tan2θ = 1/ cot2θ . Arithmetic Mean (AM)= (A + B) / 2 and Geometric Mean (GM) = √ (A. y ≥ 77 ( minimum value of y = 77 ) If. We can confidently say that my age is not less 25 years.. we come across a special case of trigonometric identities like to find min.B) Hence. Or in other words my minimum age is 25 years. sin θ ≥ .B are any two numbers then. Arithmetic mean is always greater than or equal to geometric mean.1 ( minimum value of Sin θ = -1 ) If. (1) G. we get A.M logic.M of given equation = √ (4 tan2θ . we can use 2 √ab directly in these kind of problems.M ≥ G. if any.M ≥ G. (2) Now. Summary: While using A. A function and its reciprocal have same sign. values.M (a cos2 θ + b sec2 θ / 2 ) ≥ √ (a cos2 θ . M From equations (1) and (2) above we get. 1 B. The reciprocal of 1 is 1 and -1 is -1. min.M ≥ G. apply -> convert remaining identities. A. => (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. SSC CGL 2012 Tier II Question What is The minimum value of sin2 θ + cos2 θ + sec2 θ + cosec2 θ + tan2 θ + cot2 θ A. (otherwise how would you calculate mean ? ) Directly apply 2√ab . If a function has a maximum value its opposite has a minimum value.. where T1 = 1 / T2 Positive sign in between terms is mandatory.= √ 36 = 6 …. b sec2 θ) a cos2 θ + b sec2 θ ≥ 2 √ (ab) ( minimum value 2 √ab ) So. we know that A. to sine and cosines -> finally put known max. Keeping these tools (not exhaustive) in mind we can easily find Maximum or Minimum values easily. Extra facts: The reciprocal of 0 is + ∞ and vice-versa. 3 .M logic : Term should be like a T1 + b T2 . Rearrange/Break terms if necessary -> priority should be given to direct use of identities -> find terms eligible for A.M logic -> if any.M ≥ G. = 1 + 2 + sec2 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.M = 1 + 2 + (sin2 θ + cos2 θ)/( sin2 θ . cos2 θ)….C.) = 1 + 2 + (1/ cos2 θ) + (1/ sin2 θ) solving taking L. and we get = 1 + 2 + (sin2 θ + cos2 θ)/( sin2 θ . (sin2 θ + cos2 θ) + sec2 θ + cosec2 θ + tan2 θ + cot2 θ = (1) + sec2 θ + cosec2 θ + tan2 θ + cot2 θ Using A. B.M logic for tan2 θ + cot2 θ we get . value of (sin θ cos θ)n = (½)n (Ratta-fication formula #4) Apply them into eq1. C. 1 2 3 5 We can solve this question via two approaches . 7 Solution: We know that sin2 θ + cos2 θ = 1 (identitiy#1) Therefore.eq1 but we already know two things sin2 θ + cos2 θ=1 (trig identity #1) Min.C. 5 D. cos2 θ) = 1 + 2 + (1/1/4) = 1+2+4 = 7 (correct answer D) The least value of 2 sin2 θ + 3 cos2 θ (CGL2012T1) A.M ≥ G. D.. value of sin2 θ in order to get minimum value . we’ve to find the max value of Sin x + cos x = + √ (12+ 12 ) = √2 ( correct answer A ) .either sin or cos first convert it into a sin equation : = 2 sin2 θ + 3 (1. use + } in the given question. for max.Approach #1 Break the equation and use identity no. (but sin2 θ + cos2 θ=1) = 2 + cos2 θ .(but as per min-max table. value of sin2 θ is 0 …confusing ???? ) As sin2 θ is preceded by a negative sign therefore we have to take max. use . 1 = 2 sin2 θ + 2 cos2 θ + cos2 θ =2(sin2 θ + cos2 θ) + cos2 θ .sin2 θ) = 2 sin2 θ + 3 – 3 sin2 θ = 3 .. the minimum value of cos2 θ=0) = 2 + 0 = 2 (correct answer B) Approach #2 convert equation into one identity .sin2 θ) . C.sin2 θ = 3 – ( 1) = 2 (but Min.1 a sin θ ± b cos θ = ±√ (a2 + b2 ) { for min.(because sin2 θ + cos2 θ=1=>cos2 θ=1. √2 1/ √2 1 2 Applying Ratta-fication formulae No.cos2 θ) + 3 cos2 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. D. B. Converting into a cos equation : = 2 sin2 θ + 3 cos2 θ = 2 (1. -5 (D) 4. use + } in the given question. D. we’ve to find the max value of 3 Sin x – 4 Cos x = + √ (32+ 42 ) = √25 = 5 ( correct answer B ) Min Max values of sin 4x + 5 are (A) 2.. use . Sin 1 .Sin 1 . Sin +1 . C. C. Do not exist -1. the minimum value is 4 and maximum value is 6 ( correct answer D ) Minimum and maximum value of Sin Sin x is A. for max. 6 (B) 4. B. D. -1 ≤ Sin nx ≤ 1 = -1 ≤ Sin 4x ≤ 1 Adding 5 throughout. B.The maximum value of 3 Sin x – 4 Cos x is A.1 a sin θ ± b cos θ = ±√ (a2 + b2 ) { for min. 4 ≤ Sin 4x +5 ≤ 6 Therefore. 1 Sin -1 . 5 (C) -4. -1 5 7 9 Solution: Applying Ratta-fication formulae No. 6 Solution: We know that. visit mrunal. For more articles on trigonometry and aptitude. 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. [Sin(-θ) is same as – Sin θ ] Therefore.org/aptitude .We know that.Sin 1 ≤ Sin x ≤ Sin 1 . -1 ≤ Sin nx ≤ 1 = Sin (-1) ≤ Sin x ≤ Sin (1) = .