If the question is express 4 cos x - 3 sin x in the form R cos(x + α)]
everyone knows that R = 5
but regarding the alpha(&)
tan & = B/A = -3/4 = -36.9
R forumla is 5cos (x + (-36.9))
What i was tot is that if it is cos, the sign in the (x -&) is opposite that of the orginal function. So if it is "4 cos x - 3 sin x in the form R cos(x + α)" teh sign inside would be +. This is opposite for sin. For sin, the sign follows. if it is + you write x+&
But firstly, ans is x+36.9
also, is my method correct?
Secondly, could someone explain why
max value of R sin(x+&) = R where x+a = 90
min value of R sin(x+&) = -R where x+a 270
max for R cos(x+&) = R when x+ & = 0, 360 ... and so on....
Originally posted by davidche:Secondly, could someone explain why
max value of R sin(x+&) = R where x+a = 90
min value of R sin(x+&) = -R where x+a 270
max for R cos(x+&) = R when x+ & = 0, 360 ... and so on....
max value of sin (x+&) is 1
So max value of R sin(x+&) = R
when sin (x+&) = 1, x+&=90 degrees
Same for the rest... think about it :D
Originally posted by davidche:If the question is express 4 cos x - 3 sin x in the form R cos(x + α)]
everyone knows that R = 5
but regarding the alpha(&)
tan & = B/A = -3/4 = -36.9
R forumla is 5cos (x + (-36.9))
What i was tot is that if it is cos, the sign in the (x -&) is opposite that of the orginal function. So if it is "4 cos x - 3 sin x in the form R cos(x + α)" teh sign inside would be +. This is opposite for sin. For sin, the sign follows. if it is + you write x+&
But firstly, ans is x+36.9
also, is my method correct?
because that is the short cut.
I don't advocate short cuts and pure memorizing of formulas like that (same for physics) because of instances like this... could easily make you blur... stupid O levels
R cos(x + α) = R cos x cos α - R sin x sin α
So, comparing coefficients,
R cos α = 4 ---(1)
R sin α = 3 --- (2)
(1)² + (2)² = R² (cos² α+ sin² α) = R² ===> that's why you get R
(2) / (1) = tan α
tan α = 3/4
And hence you get positive 36.9
my teacher wrote
tan & = A/B if R sin(x+&)
and tan & = B/A if R cos(x+&)
Why is it that when it is sine it is A/B?
Originally posted by davidche:my teacher wrote
tan & = A/B if R sin(x+&)
and tan & = B/A if R cos(x+&)
Why is it that when it is sine it is A/B?
Actually if you understand the compare coefficients method
A sin x + B cos x = R sin(x + α)
R sin(x + α) = R sin x cos α + R cos x sin α
compare coefficients,
R cos α = A --- (1)
R sin α = B --- (2)
(2) / (1): tan α = B/A