To find the minimum value of the expression 5cosβ‘(2)+5sinβ‘(2)5cos(2x)+5sin(2x), we can use trigonometric identities to simplify it.
First, let's express the given terms in a single trigonometric function using the Pythagorean identity (sinβ‘2()+cosβ‘2()=1sin2(x)+cos2(x)=1):
5cosβ‘(2)+5sinβ‘(2)=5(cosβ‘(2)+sinβ‘(2))5cos(2x)+5sin(2x)=5(cos(2x)+sin(2x))
Now, we'll use a trigonometric identity sin(x+4Οβ)=2β1β(cos(x)+sin(x))) to combine the terms inside the parentheses:
5(cos(2x)+sin(2x))=52βsin(2x+4Οβ)
Since the sine function has values ranging from -1 to 1, the minimum value of βsin(2x+4Οβ) is -5252β.
Therefore, the minimum value of the given expression5cos(2x)+5sin(2x) is β52β52β.
To find the minimum value of the expression 5cos(2x) + 5sin(2x), we can use the fact that the range of both the cosine and sine functions is between -1 and 1.
Since 5cos(2x) and 5sin(2x) are both positive for any value of x, the minimum value of the expression occurs when both terms equal their minimum value of 1.
Therefore, the minimum value of 5cos(2x) + 5sin(2x) is 1 + 1 = 2.