Simplify into Double Angle Formula

KatieJM

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Aug 14, 2019
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Hi there,

I need some help with this answer ASAP, please and thank you!
Simplify the expression by using a double-angle formula:
Screen Shot 2019-08-13 at 10.28.16 PM.png

(I am attaching a screen shot of the formula because I can't find how to type the symbols).

Thanks!
Katie
 

Subhotosh Khan

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Jun 18, 2007
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19,713
Hi there,

I need some help with this answer ASAP, please and thank you!
Simplify the expression by using a double-angle formula:
View attachment 13240

(I am attaching a screen shot of the formula because I can't find how to type the symbols).

Thanks!
Katie
Can you expand

cos(2x)

in terms of cos(x)?
 

KatieJM

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Joined
Aug 14, 2019
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Thank you, does this look correct?
Screen Shot 2019-08-14 at 7.44.25 AM.png
 

ksdhart2

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Mar 25, 2016
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In the future, if you're ever unsure of an answer, you can always check it yourself. In this case, evaluating by hand would be next to impossible, so let's punch both the left-hand side and right-hand side into a calculator and see what we find:

Left-hand side: \(\cos(16^{\circ}) \approx 0.96126169 \implies 2 \cos^2(16^{\circ}) - 1 \approx 2 \cdot 0.96136169^2 - 1 \approx 0.84804807\)

Right-hand side: \(\cos(32^{\circ}) \approx 0.84804809\)

What say you? You may notice there's a very tiny discrepancy - a difference in the eighth decimal place. Why did this difference appear? What does it mean?
 

KatieJM

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Aug 14, 2019
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Thank you for that explanation!
I am stumped as to why the 8th decimal is different. What am I missing?

Thank you,
Katie
 

pka

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Thank you for that explanation!
I am stumped as to why the 8th decimal is different. What am I missing?
It is a difference in rounding, that is all.
 

ksdhart2

Senior Member
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Mar 25, 2016
Messages
1,297
Consider a different equation altogether that's much simpler to solve: \(\displaystyle \frac{1}{3} \cdot \frac{1}{6}\). What would happen if we used approximations instead of the real value?

\(\displaystyle \frac{1}{3} \approx 0.3333\) and \(\displaystyle \frac{1}{6} \approx 0.1666\)

\(\displaystyle 0.3333 \cdot 0.1666 = 0.0555{\color{red}2778}\)

\(\displaystyle \frac{1}{18} = 0.0555{\color{red}5555}\)

Could we make the approximation tighter? What happens then?

\(\displaystyle \frac{1}{3} \approx 0.33333333\) and \(\displaystyle \frac{1}{6} \approx 0.16666666\)

\(\displaystyle 0.33333333 \cdot 0.16666666 = 0.05555555{\color{red}27777778}\)

\(\displaystyle \frac{1}{18} = 0.05555555{\color{red}55555555}\)

🤔 What happens if we use the real, exact value?

\(\displaystyle \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18}\)
 

KatieJM

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Joined
Aug 14, 2019
Messages
5
Consider a different equation altogether that's much simpler to solve: \(\displaystyle \frac{1}{3} \cdot \frac{1}{6}\). What would happen if we used approximations instead of the real value?

\(\displaystyle \frac{1}{3} \approx 0.3333\) and \(\displaystyle \frac{1}{6} \approx 0.1666\)

\(\displaystyle 0.3333 \cdot 0.1666 = 0.0555{\color{red}2778}\)

\(\displaystyle \frac{1}{18} = 0.0555{\color{red}5555}\)

Could we make the approximation tighter? What happens then?

\(\displaystyle \frac{1}{3} \approx 0.33333333\) and \(\displaystyle \frac{1}{6} \approx 0.16666666\)

\(\displaystyle 0.33333333 \cdot 0.16666666 = 0.05555555{\color{red}27777778}\)

\(\displaystyle \frac{1}{18} = 0.05555555{\color{red}55555555}\)

🤔 What happens if we use the real, exact value?

\(\displaystyle \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18}\)
Got it, I was overthinking it, thank you! 🙏🏼
 
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