TR5 first draft #646
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TR5 first draft #646
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tdegeorge
reviewed
May 11, 2025
| <activity xml:id="intro-unit-circle"> | ||
| <introduction> | ||
| <p> | ||
| Let <m>\theta</m> be the angle shown below in standard form. Notice that the terminal side intersects with the unit circle. (Note: We will assume a circle drawn in the this context is the unit circle unless told otherwise.) We will label that point of intersection as <m>(x,y)</m>. |
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Can the theta symbol be put at the angle in the diagram? Is that an easy fix?
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@siwelwerd can you help with this? Can you adapt show_angle_value to show just
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Try setting show_angle_value="$\theta$" and report back... I think I put that feature in already but need to update the wiki
| <task> | ||
| <statement> | ||
| <p> | ||
| Solve for <m>x</m> in one of the equations you've found above to determine an expression for the <m>x</m>-value of the point <m>(x,y)</m> . |
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Suggested change
| Solve for <m>x</m> in one of the equations you've found above to determine an expression for the <m>x</m>-value of the point <m>(x,y)</m> . | |
| Solve for <m>x</m> in one of the equations you've found in part (c) to determine an expression for the <m>x</m>-value of the point <m>(x,y)</m> . |
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For this one (and the one below), I was trying to not mention which one had the x (or y) in it so they had to deduce which one helped with which coordinate. Would it be better to point it out explicitly?
| <task> | ||
| <statement> | ||
| <p> | ||
| Solve for <m>y</m> in one of the equations you've found above to determine an expression for the <m>y</m>-value of the point <m>(x,y)</m> . |
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Suggested change
| Solve for <m>y</m> in one of the equations you've found above to determine an expression for the <m>y</m>-value of the point <m>(x,y)</m> . | |
| Solve for <m>y</m> in one of the equations you've found in part (d) to determine an expression for the <m>y</m>-value of the point <m>(x,y)</m> . |
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| </activity> | ||
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| <remark> |
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I'm not sure its obvious that we've been working with the special right triangles this whole section....
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The remark to start the section mentioned it.
| <task> | ||
| <statement> | ||
| <p> | ||
| Find the exact value of the <m>y</m>-coordinate. |
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Do we want to add in a hint about using the Pythagorean Theorem?
Co-authored-by: tdegeorge <[email protected]>
Co-authored-by: tdegeorge <[email protected]>
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AbbyANoble
commented
May 12, 2025
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| <p> What are the sine and cosine of <m>\theta=150^\circ</m>? | ||
| </p> | ||
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| <p> | ||
| <ol marker= "A." cols="1"> | ||
| <li> <m>\sin 150^\circ = \dfrac{1}{2}</m> and <m>\cos 150^\circ = \dfrac{\sqrt{3}}{2}</m></li> | ||
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| <li> <m>\sin 150^\circ = \dfrac{\sqrt{3}}{2}</m> and <m>\cos 150^\circ = \dfrac{1}{2}</m></li> | ||
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| <li> <m>\sin 150^\circ = -\dfrac{\sqrt{3}}{2}</m> and <m>\cos 150^\circ = \dfrac{1}{2}</m></li> | ||
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| <li> <m>\sin 150^\circ = \dfrac{1}{2}</m> and <m>\cos 150^\circ = -\dfrac{\sqrt{3}}{2}</m></li> | ||
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| <li> <m>\sin 150^\circ = -\dfrac{1}{2}</m> and <m>\cos 150^\circ = \dfrac{\sqrt{3}}{2}</m></li> |
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| <p> What are the sine and cosine of <m>\theta=150^\circ</m>? | |
| </p> | |
| <p> | |
| <ol marker= "A." cols="1"> | |
| <li> <m>\sin 150^\circ = \dfrac{1}{2}</m> and <m>\cos 150^\circ = \dfrac{\sqrt{3}}{2}</m></li> | |
| <li> <m>\sin 150^\circ = \dfrac{\sqrt{3}}{2}</m> and <m>\cos 150^\circ = \dfrac{1}{2}</m></li> | |
| <li> <m>\sin 150^\circ = -\dfrac{\sqrt{3}}{2}</m> and <m>\cos 150^\circ = \dfrac{1}{2}</m></li> | |
| <li> <m>\sin 150^\circ = \dfrac{1}{2}</m> and <m>\cos 150^\circ = -\dfrac{\sqrt{3}}{2}</m></li> | |
| <li> <m>\sin 150^\circ = -\dfrac{1}{2}</m> and <m>\cos 150^\circ = \dfrac{\sqrt{3}}{2}</m></li> | |
| <p> What are the cosine and sine values of <m>\theta=150^\circ</m>? | |
| </p> | |
| <p> | |
| <ol marker= "A." cols="1"> | |
| <li> <m>\cos 150^\circ = \dfrac{\sqrt{3}}{2}</m> and <m>\sin 150^\circ = \dfrac{1}{2}</m> </li> | |
| <li> <m>\cos 150^\circ = \dfrac{1}{2}</m> and <m>\sin 150^\circ = \dfrac{\sqrt{3}}{2}</m> </li> | |
| <li> <m>\cos 150^\circ = \dfrac{1}{2}</m> and <m>\sin 150^\circ = -\dfrac{\sqrt{3}}{2}</m></li> | |
| <li> <m>\cos 150^\circ = -\dfrac{\sqrt{3}}{2}</m> and <m>\sin 150^\circ = \dfrac{1}{2}</m> </li> | |
| <li> <m>\cos 150^\circ = \dfrac{\sqrt{3}}{2}</m> and <m>\sin 150^\circ = -\dfrac{1}{2}</m> </li> |
Co-authored-by: tdegeorge <[email protected]>
AbbyANoble
commented
May 12, 2025
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| <statement> | ||
| <p> What are the sine and cosine of <m>\theta=\dfrac{4\pi}{3}</m>? | ||
| </p> | ||
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| <p> | ||
| <ol marker= "A." cols="1"> | ||
| <li> <m>\sin \dfrac{4\pi}{3} = \dfrac{1}{2}</m> and <m>\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}</m></li> | ||
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| <li> <m>\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m> and <m>\cos \dfrac{4\pi}{3} = -\dfrac{1}{2}</m></li> | ||
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| <li> <m>\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m> and <m>\cos \dfrac{4\pi}{3} = \dfrac{1}{2}</m></li> | ||
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| <li> <m>\sin \dfrac{4\pi}{3} = \dfrac{1}{2}</m> and <m>\cos \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m></li> | ||
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| <li> <m>\sin \dfrac{4\pi}{3} = -\dfrac{1}{2}</m> and <m>\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}</m></li> | ||
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Suggested change
| <statement> | |
| <p> What are the sine and cosine of <m>\theta=\dfrac{4\pi}{3}</m>? | |
| </p> | |
| <p> | |
| <ol marker= "A." cols="1"> | |
| <li> <m>\sin \dfrac{4\pi}{3} = \dfrac{1}{2}</m> and <m>\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}</m></li> | |
| <li> <m>\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m> and <m>\cos \dfrac{4\pi}{3} = -\dfrac{1}{2}</m></li> | |
| <li> <m>\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m> and <m>\cos \dfrac{4\pi}{3} = \dfrac{1}{2}</m></li> | |
| <li> <m>\sin \dfrac{4\pi}{3} = \dfrac{1}{2}</m> and <m>\cos \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m></li> | |
| <li> <m>\sin \dfrac{4\pi}{3} = -\dfrac{1}{2}</m> and <m>\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}</m></li> | |
| <statement> | |
| <p> What are the cosine and sine values of <m>\theta=\dfrac{4\pi}{3}</m>? | |
| </p> | |
| <p> | |
| <ol marker= "A." cols="1"> | |
| <li> <m>\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}</m> and <m>\sin \dfrac{4\pi}{3} = \dfrac{1}{2}</m> </li> | |
| <li> <m>\cos \dfrac{4\pi}{3} = -\dfrac{1}{2}</m> and <m>\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m></li> | |
| <li> <m>\cos \dfrac{4\pi}{3} = \dfrac{1}{2}</m> and <m>\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m></li> | |
| <li> <m>\cos \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}</m> and <m>\sin \dfrac{4\pi}{3} = \dfrac{1}{2}</m></li> | |
| <li> <m>\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}</m> and <m>\sin \dfrac{4\pi}{3} = -\dfrac{1}{2}</m> </li> | |
Co-authored-by: tdegeorge <[email protected]>
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Co-authored-by: tdegeorge <[email protected]>
Co-authored-by: tdegeorge <[email protected]>
Co-authored-by: tdegeorge <[email protected]>
Co-authored-by: tdegeorge <[email protected]>
Co-authored-by: tdegeorge <[email protected]>
Co-authored-by: tdegeorge <[email protected]>
AbbyANoble
commented
May 12, 2025
| </p> | ||
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| </statement> | ||
| <answer> |
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Suggested change
| <answer> | |
| <hint> | |
| <p> | |
| Use the Pythagorean Theorem to help. | |
| </p> | |
| </hint> | |
| <answer> |
siwelwerd
reviewed
May 12, 2025
siwelwerd
reviewed
May 12, 2025
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Co-authored-by: Drew Lewis <[email protected]>
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@siwelwerd show_angle_value="$\theta$" gives this: 
Whomp whomp. |
Oh yeah, sorry, |
AbbyANoble
commented
May 12, 2025
| <image width="50%"> | ||
| <sageplot> | ||
| <xi:include parse="text" href="../../../common/sagemath/library.sage"/> | ||
| p=TBIL.plot_angle(pi/4,show_unit_circle=True,show_angle_value="$\theta$") |
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Suggested change
| p=TBIL.plot_angle(pi/4,show_unit_circle=True,show_angle_value="$\theta$") | |
| p=TBIL.plot_angle(pi/4,show_unit_circle=True,show_angle_value=r"$\theta$") |
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