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Find the maximum value and minimum period of a trigonometry function

Question:

Given function f(x) = sin (2x + pi/6) - cos2x, find the maximum value and minimum period of the trigonometry function.

Solution:

Using the sine sum and difference formula: sin (a + b) = sin a cos b + cos a sin b.

Using the cosine power reducing formula: cos2 = (1/2) (1 + cos 2x)

f(x) = sin (2x + pi/6) - cos2
= sin 2x cos pi/6 + cos 2x sin pi/6 - (1/2) (1 + cos 2x)
note: cos pi/6 = cos 30o = sin 60o = (square roor of 3)/2
note: sin pi/6 = sin 30o = 1/2
f(x) = [(square root of 3)/2] sin 2x + (1/2) cos 2x - 1/2 - (1/2) cos 2x
= [(square root of 3)/2] sin 2x - 1/2

The minimum period of the function f(x) = [(square root of 3)/2] sin 2x is: T = 2pi/2 = pi.

Now we will find the value of x in which f(x) = [(square root of 3)/2] sin 2x - 1/2 reachs the maximum value.

For the graph of y = sin t, its minimum period is 2pi. When t = pi/2 + 2pi × k (k is an integer), the value of y = sin t reachs the maximum malue.

So, for the function f(x) = [(square root of 3)/2] sin 2x - 1/2, when 2x = pi/2 + 2pi × k, f(x) reachs the maximum value

2x = pi/2 + 2pi × k, (k is an integer)
divided by 2 in both sides of the equation
x = pi/4 + pi × k

when x = pi/4 + pi × k, (k is an integer) f(x) reaches the maximum value

when k = 0, x = pi/4, the maximum value of the function f(x) is (square root of 3)/2 - 1/2 = [(square root of 3) -1]/2

when k = 1, x = pi/4 + pi = 5pi/4, the maximum value of the function f(x) is [(square root of 3) -1]/2

find the maximum value and the minimum period of y = sin (2x + pi/6) - cos x^2

Therefore, the minimum period of the function f(x) = sin (2x + pi/6) - cos2x is pi. When x = pi/4 + pi × k (k is an integer), f(x) reaches the maximum value that is [(square root of 3) -1]/2. For more details, please watch the video.