1.) How would I simplify: sin4x*cos3x -cos3x*sin3x into a single trigonometric function?
2.) determine the general solution to the equation root2*cosx= sin2x ?
sin(A-B) = sinAcosB - cosAsinB
I suspect a typo, and you meant sin4x*cos3x -cos4x*sin3x
so you have just sin(4x-3x) = sin(x)
If not, then you have cos3x(sin4x-sin3x) and that does not simplify to much.
Now, for
√2 cosx = sin2x
√2 cosx - 2sinx cosx = 0
cosx (√2 - 2sinx) = 0
cosx = 0
sinx = 1/√2
So x = π/2 + kπ or π/2±π/4 + 2kπ
To simplify the expression sin(4x) * cos(3x) - cos(3x) * sin(3x) into a single trigonometric function, you can use the trigonometric identity for the product of sines and cosines.
The identity is: sin(A - B) = sin(A) * cos(B) - cos(A) * sin(B)
Let's apply this identity to the given expression:
sin(4x) * cos(3x) - cos(3x) * sin(3x)
= sin(4x - 3x)
= sin(x)
Therefore, the simplified expression is sin(x).
Now, let's move on to the second question.
To determine the general solution to the equation √2 * cos(x) = sin(2x), we can rewrite it using trigonometric identities.
The double-angle identity for sine is: sin(2x) = 2 * sin(x) * cos(x)
Let's substitute that into the equation:
√2 * cos(x) = 2 * sin(x) * cos(x)
Now, we can rearrange the equation:
√2 = 2 * sin(x)
Divide both sides by 2:
sin(x) = √2/2
This means that the angle x must be equal to π/4 plus any multiple of π.
So, the general solution to the equation is:
x = π/4 + nπ, where n is an integer.