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My answer for this question was $2\sin(θ - 30)$, but the answer in the textbook says $-2\sin(θ - 30)$. I tried to use the concept of odd functions so using the idea that $-\sin(θ)= \sin(-θ)$.

So $-2\sin(θ-30) = 2\sin(30-θ)$ but that still is not the same as my answer and I was wondering why?

In addition, the question also asks to give the maximum and minimum values of the function and the values between $0$ and $360$ at which they occur.

I get:

max value $y = 2$ max $x$ at $θ = 120$

min $y$ value = $ -2$ min $x$ at $θ = 300$

But the textbook answers say:

max value $y = 2$ at $θ = 300$ min value $y = -2$ at $θ = 106.3$

I am quite confused as to why.

Thank you.

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3 Answers 3

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Hint: Assume that the expression can be rewritten as

$$A\sin(\theta+\phi)=A\sin\theta\cos\phi + A\cos\theta \sin \phi$$

I used the formula for the sum of angles for the sine function. Now, compare your expression to this expression to obtain

$$-\sqrt{3}=A\cos \phi$$ $$1=A\sin \phi.$$

Square both equations and add them to obtain $$A^2\left[\sin^2\phi + \cos^2\phi \right]=1^2+(-\sqrt{3})^2$$ $$\implies A^2 = 1+3$$ $$\implies A^2=4.$$ I used the theorem of Pythagoras. Now, divide both equations to obtain $$\dfrac{\sin \phi}{\cos \phi}= \dfrac{1}{-\sqrt{3}}.$$ $$\implies \tan \phi = -1/\sqrt{3}.$$ Can you determine $A$ and $\phi$ from that? You can use $\phi=-\alpha$ to obtain your value for alpha.

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  • $\begingroup$ Ah! I thought that 1 becomes -1, which is where i made my error. Why is it that you do 1 and why is it that you use this 𝐴sin(𝜃+𝜙) expression instead of the one with minus. The question requries me to use 𝐴sin(𝜃-𝜙). $\endgroup$
    – user639649
    Commented Apr 27, 2019 at 14:43
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    $\begingroup$ You can replace $\phi = -\alpha$ to obtain your form. I don't like to put a minus in there when it is not necessary. What do you mean by "Why is it that you do $1$"? $\endgroup$ Commented Apr 27, 2019 at 14:46
  • $\begingroup$ The question for why is it that you do 1, was answer by telling me to replace 𝜙=−𝛼, so thank you for that! I was also wondering why the minus value for theta is 106.3? Do you think it was an error on the textbook? $\endgroup$
    – user639649
    Commented Apr 27, 2019 at 15:17
  • $\begingroup$ $\tan \phi=-1/\sqrt{3}$ has two solutions in the unit circle $\phi_1 = -\pi/6$ and $\phi_2 = 5/6\pi$. By looking at the signs of $\cos \phi=-\sqrt{3}/2$ and $\sin \phi = 1/2$ we see that $\phi_2$ is the correct solution. Then $2\sin(x-5/6\pi)$ can be rewritten to the form in your textbook. $\endgroup$ Commented Apr 27, 2019 at 15:31
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\begin{align}\cos\theta-\sqrt3\sin\theta&=2\left(\frac12\cos\theta-\frac{\sqrt3}2\sin\theta\right)\\&=-2\left(\sin(-30^\circ)\cos\theta+\cos(-30^\circ)\sin\theta\right)\\&=-2\sin(\theta-30^\circ).\end{align}Therefore, your textbook is right.

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  • $\begingroup$ Sir in the RHS of the 1st equation, the coeff. of sin(theta) is sqrt(3)/2, please change it. $\endgroup$
    – 19aksh
    Commented Apr 27, 2019 at 14:38
  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ Commented Apr 27, 2019 at 14:39
  • $\begingroup$ Do you know why i was wrong for the max/min x values. Because Im not too sure where theta = 106.3 comes from? $\endgroup$
    – user639649
    Commented Apr 27, 2019 at 14:42
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Hint: Write your function in the form $$f(\theta)=4\left(\frac{1}{4}\cos(x)-\frac{\sqrt{3}}{4}\sin(x)\right)$$

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