# Simplify into Double Angle Formula

#### KatieJM

##### New member
Hi there,

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

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

Thanks!
Katie

#### Subhotosh Khan

##### Super Moderator
Staff member
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

##### New member
Thank you, does this look correct?

#### ksdhart2

##### Senior Member
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

##### New member
Thank you for that explanation!
I am stumped as to why the 8th decimal is different. What am I missing?

Thank you,
Katie

#### pka

##### Elite Member
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
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

##### New member
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!