- #1
subzero0137
- 91
- 4
Use de Moivre's identity to find real values of a and b in the equation below such that the equation is valid.
[itex]cos^6(x)+sin^6(x)+a(cos^4(x)+sin^4(x))+b=0[/itex]
Hint: Write [itex]cos(x)[/itex] & [itex]sin(x)[/itex] in terms of [itex]e^{ix}[/itex] & [itex]e^{-ix}[/itex].
Check your values of [itex]a[/itex] and [itex]b[/itex] are valid by substituting in a value of [itex]x[/itex]. State, with explanation, two values of [itex]x[/itex]which would not have been sufficient checks on your values of [itex]a[/itex] and [itex]b[/itex].
I've managed to obtain the following expression:
[itex]\frac{3}{8}cos(4x)+\frac{a}{4}cos(4x)+\frac{3a}{4}+\frac{5}{8}+b=0[/itex], and I checked the model solution, and this is the expression they've got too. But then they simply state [itex]a=\frac{-3}{2}[/itex], and work out [itex]b[/itex] from there. But I don't understand how they got that value for [itex]a[/itex]. Can someone explain to me what I'm missing?
[itex]cos^6(x)+sin^6(x)+a(cos^4(x)+sin^4(x))+b=0[/itex]
Hint: Write [itex]cos(x)[/itex] & [itex]sin(x)[/itex] in terms of [itex]e^{ix}[/itex] & [itex]e^{-ix}[/itex].
Check your values of [itex]a[/itex] and [itex]b[/itex] are valid by substituting in a value of [itex]x[/itex]. State, with explanation, two values of [itex]x[/itex]which would not have been sufficient checks on your values of [itex]a[/itex] and [itex]b[/itex].
I've managed to obtain the following expression:
[itex]\frac{3}{8}cos(4x)+\frac{a}{4}cos(4x)+\frac{3a}{4}+\frac{5}{8}+b=0[/itex], and I checked the model solution, and this is the expression they've got too. But then they simply state [itex]a=\frac{-3}{2}[/itex], and work out [itex]b[/itex] from there. But I don't understand how they got that value for [itex]a[/itex]. Can someone explain to me what I'm missing?
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