If the equation sinx+cosx=m processes solution then range of the range of values of m is given by?
-(2)^(1/2)<m<2^(1/2)
since 2^(1/2)*sin(x+(pie/4))=m
& -1<sin y<1
where do all you guys learn to spell "pi" as "pie"?
Does no one ever teach about the Greek alphabet any more?
To find the range of values for m that satisfy the equation sin(x) + cos(x) = m, we can re-arrange the equation to solve for m.
sin(x) + cos(x) = m
We can manipulate the equation by squaring both sides to eliminate the square root:
(sin(x) + cos(x))^2 = m^2
Expanding the square:
sin^2(x) + 2sin(x)cos(x) + cos^2(x) = m^2
Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can simplify further:
1 + 2sin(x)cos(x) = m^2
Now, let's analyze the expression 2sin(x)cos(x) in the equation. Since the range of the sine function and the cosine function is between -1 and 1 (inclusive), the product of sin(x) and cos(x) can range from -1 to 1 inclusive, resulting in a range for 2sin(x)cos(x) from -2 to 2 inclusive.
Substituting this range into the equation:
1 + (-2) ≤ m^2 ≤ 1 + 2
-1 ≤ m^2 ≤ 3
Taking the square root of both sides of the equation, we need to consider both the positive and negative square root values:
-√3 ≤ m ≤ √3
Therefore, the range of values for m that satisfy the equation sin(x) + cos(x) = m is -√3 ≤ m ≤ √3.