# Minimum and Maximum values of Trigonometric Functions

Here we are providing you with the tricks to find the minimum and maximum values of Trigonometric Identities which are useful in SSC, Railways and other exams.

Type I: a sinɸ ± b cosɸ,       a sinɸ ± b sinɸ,       a cosɸ ± b cosɸ
Maximum value = √ (a2 + b2)
Minimum value  = – √ (a2 + b2)

Example: Find the minimum and maximum value of 3 sinɸ + 4 sinɸ
Minimum value  = – √ (32 + 42) = -5
Maximum value = √ (32 + 42) = 5

Type 2: (sinɸ cosɸ)n
Minimum value = (1/2)n
The maximum value can go up to infinity.

Example: Find the minimum value of sin4ɸ cos4ɸ
Minimum value = (1/2)4 = 1/16

Type 3: a sin2ɸ + b cos2ɸ
If a > b, Maximum value = a and Minimum value = b
If a < b, Maximum value = b and Minimum value = a

Example: Find the minimum and maximum values of 3 sin2ɸ + 5 cos2ɸ
First check, here a < b
So Maximum value = 5 and Minimum value = 3

*Note: You do not have to learn this formula, just observe here that if the equation is of type a sin2ɸ + b cos2ɸ, no matter what, the maximum value is the larger of values (a, b) and minimum value is smaller of values(a, b).

Type 4: a sin2ɸ + b cosec2ɸ,       a cos2ɸ + b sec2ɸ,       a tan2ɸ + b cot2ɸ
Minimum value = 2√(ab)
The maximum value can go up to infinity.

Example: Find the minimum value of 4 cos2ɸ + 9 sec2ɸ
Observe the case, so Minimum value = 2√(4*9) = 12

*Note: In these formulae, reciprocal of one another is there.

Now these above given formulae can be used to deduce minimum and maximum values of other trigonometric functions also. But remember that first we have to deduce teh equation up to the point we can.

We are providing some examples:

Example 1: Minimum value of a sec2ɸ + b cosec2ɸ ?
We do not have a formula for this, lets continue
The given function can be written as
a (1 + tan2ɸ) + b(1 + cot2ɸ)
= a + b + a tan2ɸ + b cot2ɸ
= (a+b) + (a tan2ɸ + b cot2ɸ)
Now observe that we have the formula for finding min value of a tan2ɸ + b cot2ɸ
So here minimum value is
(a+b) + 2√(ab)

Now find the minimum value of a question asked in SSC CGL 2012 Exam:
Example 2: sin2ɸ + cos2ɸ + sec2ɸ + cosec2ɸ + tan2ɸ + cot2ɸ
First try it yourself and then see the solution
sin2ɸ + cos2ɸ + sec2ɸ + cosec2ɸ + tan2ɸ + cot2ɸ
= 1 + sec2ɸ + cosec2ɸ + tan2ɸ + cot2ɸ…………………….(i)
Now 1 + tan2ɸ = sec2ɸ
So (i) = 2sec2ɸ + cosec2ɸ + cot2ɸ…………………………………(ii)
Now cot2ɸ = cosec2ɸ – 1
So (ii) = 2sec2ɸ + cosec2ɸ + cosec2ɸ – 1
= 2sec2ɸ + 2cosec2ɸ – 1
Now see we had find the min value of 2sec2ɸ + 2cosec2ɸ type above
So now min value = 2 + 2 + 2√(2*2) – 1 = 4 + 4 – 1 = 7