Trigonometry question. How can you write this in the form Rsin(x-α)?
If I had the equation 2cosx + sinx = 1 then I could express it as
√5 cos (x-26.6) = 1. I have no problem when it's like this :)
However, if it was 2sinx + cosx = 1, how could I do it? would it be
√5 sin (x - 63.4) = 1?
If I go further to find the values of x here, I calculate
sin^(-1) of 1/√5 which is 26.6, and then for x I get 90 and 396.8, but because it's not under 360 I take that value away and get 36.8.
So, I said x was 36.8 and 90. Whenever I put these back into the equation though, it doesn't result in 1, it results in two. Does anyone know why this is and can point out where I'm going wrong?
Thanks if you can help :)
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Verified answer
Asin(x + B) = 2sinx + cosx
expand
AsinxcosB + AcosxsinB = 2sinx + cosx
compare coefficients
AcosB = 2
AsinB = 1
divide
B = arctan(1/2)
B = 26.6
square previous equations and add
A^2cos^B = 2^2
A^2sin^2B = 1^2
A^2(cos^2B + sin^2B) = 2^2 + 1^2
A^2 = 5
A = √5
thus
√5sin(x + 26.6) = 1
solve for x
sin(x + 26.6) = 1/√5
x + 26.6 = arcsin(1/√5)
x + 26.6 = 26.6, 153.4
x = 26.6 - 26.6 = 0 deg
x = 153.4 - 26.6 = 126.9 deg