Given that a^2+b^2=2 and that (a/b)= tan(45degee+x), find a and b in terms of sinx and cosx.
I don't know what i'm supposed to do, and i don't come to an answer! Help, thanks!
my workings:
tan(45+x)= (1+tanx)/(1-tanx)
a/b = (1+tanx)/(1-tanx)
a(1-tanx)=b(1+tanx)
i square both sides...
a^2(1-tanx)^2 = b^2(1+tanx)^2
a^2 + b^2 = 2
a^2 = 2 - b^2
substitute:
(2-b^2)(1-tanx)^2 = b^2(1+tanx)^2
I don't know if im on the right track, but i don't seem to come to an answer when i expand the whole equation!
From:
a(1-tanx)=b(1+tanx)
multiply both sides by cos(x) to get
a(cos(x)-sin(x))=b(cos(x)+sin(x))
Now square and simplify. The product terms will cancel out after you substitute b²=(2-a²) to give you a in terms of sin(x) and cos(x).
You can do similarly for b.
You're on the right track with your working. Let's continue with the steps.
Starting from where you left off:
Expand the equation (2 - b^2)(1 - tan(x))^2 = b^2(1 + tan(x))^2:
(2 - b^2)(1 - 2tan(x) + tan^2(x)) = b^2(1 + 2tan(x) + tan^2(x))
Distribute and simplify:
2 - 4tan(x) + 2tan^2(x) - b^2 + 2b^2tan(x) - b^2tan^2(x) = b^2 + 2b^2tan(x) + b^2tan^2(x)
Combine like terms:
2 - 4tan(x) + 2tan^2(x) - b^2 + 2b^2tan(x) - b^2tan^2(x) = 2b^2 + 2b^2tan(x) + b^2tan^2(x)
Rearrange the terms:
2 - 4tan(x) - b^2 = 2b^2
Move all terms to one side:
2b^2 + 4tan(x) + b^2 - 2 = 0
Combine like terms:
3b^2 + 4tan(x) - 2 = 0
Now, we have a quadratic equation in terms of b^2 with constant coefficients and a variable (tan(x)). We can solve this equation by using the quadratic formula:
b^2 = (-4tan(x) ± sqrt((4tan(x))^2 - 4 * 3 * (-2))) / (2 * 3)
Simplifying further:
b^2 = (-4tan(x) ± sqrt(16tan^2(x) + 24)) / 6
Now, we need to consider the given equation a^2 + b^2 = 2. We can substitute the expression for b^2 we found above:
a^2 + (-4tan(x) ± sqrt(16tan^2(x) + 24)) / 6 = 2
Rearrange the equation, isolating a^2:
a^2 = 2 - (-4tan(x) ± sqrt(16tan^2(x) + 24)) / 6
a^2 = 12 / 6 - (-4tan(x) ± sqrt(16tan^2(x) + 24)) / 6
a^2 = (12 + 4tan(x) ∓ sqrt(16tan^2(x) + 24)) / 6
Simplifying further:
a^2 = (2 + 2tan(x) ∓ sqrt(4tan^2(x) + 6)) / 3
Now, we have both a^2 and b^2 in terms of tan(x). To express a and b in terms of sin(x) and cos(x), we can use the trigonometric identity:
tan(x) = sin(x) / cos(x)
Substituting this into the expressions for a^2 and b^2:
a^2 = (2 + 2sin(x) / cos(x) ∓ sqrt(4(sin^2(x) / cos^2(x)) + 6)) / 3
b^2 = (-4sin(x) / cos(x) ± sqrt(16(sin^2(x) / cos^2(x)) + 24)) / 6
Simplifying further:
a = sqrt[(2 + 2sin(x) / cos(x) ∓ sqrt(4(sin^2(x) / cos^2(x)) + 6)) / 3]
b = sqrt[(-4sin(x) / cos(x) ± sqrt(16(sin^2(x) / cos^2(x)) + 24)) / 6]
Please be aware that the ± sign in the above expressions means that you will have two possible solutions for both a and b, depending on which sign you choose.