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If v = 2i -3j, then find $\sin \theta$ and $ \cos \theta$?

Solution manual says:

Can someone expound on what solution manual said, I am having hard time understanding it.

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3 Answers

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This is the definition of $\sin{t}$ and $\cos{t}$ that you should be thinking of:

This is on the unit circle, so you want to find a vector $\mathbf{w} = (\cos{t},\sin{t}) = \cos{t}\mathbf{i} + \sin{t}\mathbf{j}$ that points in the same direction as $\mathbf{v}$ (so that it preserves the angle $t$) and has length one (so that it lies on the unit circle). This is the normalized vector

$$ \mathbf{w} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{2\mathbf{i} - 3\mathbf{j}}{\sqrt{2^2 + (-3)^3}} = \frac{2}{\sqrt{13}}\mathbf{i} - \frac{3}{\sqrt{13}}\mathbf{j} $$

Therefore,

$$ \cos{t} = \frac{2}{\sqrt{13}} \\ \sin{t} = - \frac{3}{\sqrt{13}} $$

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At first, calculate $r=\sqrt{2^2+(-3)^2}=\sqrt{13}.$ If you write $$v=r(\cos\theta \,\mathbf{i}+\sin\theta\,\mathbf{j}),$$ then $\mathbf{v}=2\mathbf{i}-3\mathbf{j}$ yields $\cos\theta=\frac{2}{\sqrt{13}}$ and $\sin \theta=-\frac{3}{\sqrt{13}}.$

Is this the answer you was searching for?

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define a circle $A$ by the collection $A=\{(x,y) \in \mathbb{R} \mid x^2+y^2=1\}$.

Then clearly, since $\cos^2t+\sin^2t=1$ for all $t \in \mathbb{R}$. Similarly, $-1 \leq x,y \leq 1$ and $g,h:\mathbb{R} \to [-1,1]$ where $g(t)=\cos(t)$ and $h(t)=\sin(t)$ are both surjective. So every point $(x,y) \in A$ can be described by $(x,y)=(\cos(t),\sin(t))$.

Then let $f(x,y)=(\cos(t),\sin(t))$ for $ 0 \leq t \leq 2\pi$.

note: for $t=0$, we have that $f(x,y)=(0,0$. This means that we are starting on the x-axis at the right, initially, $\sin(t)$ is increasing and $\cos(t)$ is decreasing, so $t$ is an angle rotating counter-clockwise!

Example $(3/5)^2+(4/5)^2=1$, therefore $3/5,4/5$ are on the circle. Then $\cos(t)=3/5$ and $t=\cos^{-1}(3/5)$. Similarly, $\sin(t)=4/5$, so $t=\sin^{-1}(4/5)$ [both $t$ are the same]

Now, you want to generalize a bit to get $x^2+y^2=r^2$. You know that the magnitude of $\vec{v}$ is $\sqrt(2^2+3^2)$. Can you take it from here?

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