#Trigo

77 messages · Page 1 of 1 (latest)

near saffron
rapid birchBOT
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near saffron
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Hey there can someone teach me this question?

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Here's my work so far .... Am i in the right way ? Bcz the ans given for the min value at 63.4° and 243.4°

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But i got different

sacred goblet
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Try not to use degrees when dealing with trigonometric functions.

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Use radians, and if the answer requires degrees, convert to them at the very end.

near saffron
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Well the ans given exactly is degree

sacred goblet
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The result in degress is a transcendental number. It decimal expression doesn't terminate.

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That's why it's better to find the exact angle in radians first.

near saffron
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Okay

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So what's next ?

sacred goblet
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Well, first of all, what would α be?

near saffron
sacred goblet
near saffron
sacred goblet
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Radians.

near saffron
sacred goblet
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Um...

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As I said, what is α exactly?

near saffron
sacred goblet
near saffron
sacred goblet
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That's C, yes.

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What about φ0?

near saffron
sacred goblet
near saffron
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Or i should +180 the angle ?

sacred goblet
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We do have tan(φ0) = B/A. However, there are two solutions in a unit circle to that equation.
To pick the one we need, note that sin(φ0) has the same sign as B and cos(φ0) has the same sign as A. That allows you to determine the quarter that φ0 lies in. And then you solve tan(φ0) = B/A in that quarter.

near saffron
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So 233.13 also my alpha?

sacred goblet
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Forget about degrees for now.

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Let's apply this to our case.

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In our case A = 3, B = -4. So, which quarter does φ0 lie in our case?

near saffron
sacred goblet
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Why II?

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Those signs determine it uniquely.

near saffron
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no ?

sacred goblet
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So, if cosine is positive and sign is negative, what quarter will φ0 be in?

near saffron
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whats that mean

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III ?

sacred goblet
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No. They are both negative there.

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Draw a unit circle if you're having trouble.

near saffron
sacred goblet
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Yeah, nice!

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So, now solve tan(φ0) = B/A so that φ0 is in the fourth quadrant.

near saffron
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φ0 = 53.13 and 306.87

sacred goblet
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No, don't use degrees (espeically if you don't identify them as such). Find an exact value in radians for now.

near saffron
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φ0 = 0.927 and 1.705

sacred goblet
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It appears you are not listening to me...

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First, what's the general solution of tan(x) = -4/3?

near saffron
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im sorry

sacred goblet
near saffron
sacred goblet
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Huh. That's weird...

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Well, that's the general solution of tan(x) = a.

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So, as in our case we have tan(φ0) = -4/3, the general solution is:
φ0 = -arctan(4/3) + πn, n ∈ ℤ
So, now we need to determine which one lies in the fourth quarter.

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The fourth quarter is from 3π/2 to 2π. So, we have:
3π/2 < -arctan(4/3) + πn < 2π
So, now try solving for n. Note that it must be an integer.

near saffron
sacred goblet
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Yeah, nice! So, we get φ0 = 2π - arctan(4/3).

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Thus, we get:
3cos(2θ) - 4sin(2θ) = 5cos(2θ + 2π - arctan(4/3))

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We can also simplify that to just 5cos(2θ - arctan(4/3)), of course.

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So, now we determine the values of θ that give minimum and maximum values.

near saffron
sacred goblet
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So, first, let's solve 5cos(2θ - arctan(4/3)) = 5.

near saffron
sacred goblet
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Let's solve it as usual first.
cos(2θ - arctan(4/3)) = 1
2θ - arctan(4/3) = 2πn, n ∈ ℤ
θ = (1/2)arctan(4/3) + πn, n ∈ ℤ
And now we do as before.
0 ≤ (1/2)arctan(4/3) + πn ≤ 2π

near saffron
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whats n stand for ?