Asked by Leash
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 ?
2.) determine the general solution to the equation root2*cosx= sin2x ?
Answers
Answered by
oobleck
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
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
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