Can you help me to solve this please? -provide some examples of two decreasing functions f(x) and g(x) on the interval (0,+infinity) such that the function f(x)g(x) is increasing on the interval (0,+infinity). 🙏
Can you help me solve this? Cos(2x)+sin(2x)=1 Why this dont work if i solve like this Cos(2x)=1-sin(2x) 1= sec(2x)-tan(2x) Subtitute Sec(2x)=square root of 1+(tan(2x))^2
It doesn’t seem to be simplifying nicely. At least I can’t see how to solve it this way. Another method is to multiple through by 1/sqrt(2) giving you: (1/sqrt(2))*cos2x + (1/sqrt(2))*sin2x=1/sqrt(2) Because cos(45)=sin(45)=1/sqrt(2) this allows you to write as: Sin2x*cos(45)+cos2x*sin(45)=1/sqrt(2) Then you can use the sum and diff formula for sin: Sin(2x +45)=1/sqrt(2) and maybe you can take it from here :)
Can you help me to solve this please?
-provide some examples of two decreasing functions f(x) and g(x) on the interval (0,+infinity) such that the function f(x)g(x) is increasing on the interval (0,+infinity). 🙏
when will you be doing a live stream?
Should be sometime this week. Check scheduled streams later in the week for the time and day :)
Can you help me solve this?
Cos(2x)+sin(2x)=1
Why this dont work if i solve like this
Cos(2x)=1-sin(2x)
1= sec(2x)-tan(2x)
Subtitute Sec(2x)=square root of 1+(tan(2x))^2
It doesn’t seem to be simplifying nicely. At least I can’t see how to solve it this way. Another method is to multiple through by 1/sqrt(2) giving you:
(1/sqrt(2))*cos2x + (1/sqrt(2))*sin2x=1/sqrt(2)
Because cos(45)=sin(45)=1/sqrt(2) this allows you to write as:
Sin2x*cos(45)+cos2x*sin(45)=1/sqrt(2)
Then you can use the sum and diff formula for sin:
Sin(2x +45)=1/sqrt(2) and maybe you can take it from here :)