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secondary 4 | A Maths
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How do you do (ii)? This looks like r formulae but I’m not sure how to go about with finding the constants
Date Posted: 3 years ago
Views: 244
Girl, we bring the sin x over to the other side.
4 cos x - 3 sin x = 2
For the LHS,
R
= sqrt (4^2 + 3^2)
= 5
Phase shift alpha
= taninv (3/4)
= 0.6435011088 rad
~ 0.644 rad
So, the LHS becomes 5 cos (x + 0.644).
The equation becomes 5 cos (x + 0.644) = 2.
4 cos x - 3 sin x = 2
For the LHS,
R
= sqrt (4^2 + 3^2)
= 5
Phase shift alpha
= taninv (3/4)
= 0.6435011088 rad
~ 0.644 rad
So, the LHS becomes 5 cos (x + 0.644).
The equation becomes 5 cos (x + 0.644) = 2.
Oh I get it now, I just didn’t divide 2/5 to get 0.4. Thank you!!
The equation solving requires the division of 5 first so that the LHS will be a cosine-only expression without the coefficient 5. Then, you should be able to solve the equation from there. Let me know if you also need help for the equation solving partz
Okay I’ll try it out after I eat
yeaaah I got it
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