Finding the value of Sin (pi/12)

[CWB1234]

Hi all,

i am asked to find the value of sin (pi/12) in 2 different ways

at the top of the image i use the exact values of sine and cosine of both (pi/4) and (pi/6)
i believe this is correct

however i use the half angle identity (HAI in the image) and i think i have messed up, but i don't know where
italso asks me to use the exact value of cos (pi/6) but not sure how to do that


when i square both these methods it should show that both expressions are equivalent, but there not

any help would be appreciated, hopefully you can read the image easily

 

[Deleted member 4993]

Hi all,

i am asked to find the value of sin (pi/12) in 2 different ways

at the top of the image i use the exact values of sine and cosine of both (pi/4) and (pi/6)
i believe this is correct

however i use the half angle identity (HAI in the image) and i think i have messed up, but i don't know where
italso asks me to use the exact value of cos (pi/6) but not sure how to do that


when i square both these methods it should show that both expressions are equivalent, but there not

any help would be appreciated, hopefully you can read the image easily

For part (b)

$\displaystyle \displaystyle{\cos(\frac{\pi}{6}) \ = \ \dfrac{\sqrt{3}}{2}}$

 

[Prove It]

Hi all,

i am asked to find the value of sin (pi/12) in 2 different ways

at the top of the image i use the exact values of sine and cosine of both (pi/4) and (pi/6)
i believe this is correct

however i use the half angle identity (HAI in the image) and i think i have messed up, but i don't know where
italso asks me to use the exact value of cos (pi/6) but not sure how to do that


when i square both these methods it should show that both expressions are equivalent, but there not

any help would be appreciated, hopefully you can read the image easily

Since \(\displaystyle \displaystyle \begin{align*} \cos{ \left( 2\,\theta \right) } \equiv 1 - 2\sin^2{ \left( \theta \right) } \end{align*}\) we can write

\(\displaystyle \displaystyle \begin{align*} \sin^2{ \left( \theta \right) } &\equiv \frac{1}{2} \, \left[ 1 - \cos{ \left( 2 \, \theta \right) } \right] \\ \\ \sin^2{ \left( \frac{\pi}{12} \right) } &= \frac{1}{2} \, \left[ 1 - \cos{ \left( 2 \cdot \frac{\pi}{12} \right) } \right] \\ &= \frac{1}{2} \,\left[ 1 - \cos{ \left( \frac{\pi}{6} \right) } \right] \\ &= \frac{1}{2} \, \left( 1 - \frac{\sqrt{3}}{2} \right) \\ &= \frac{1}{2} \, \left( \frac{2 - \sqrt{3}}{2} \right) \\ &= \frac{2 - \sqrt{3}}{4} \\ \\ \sin{ \left( \frac{\pi}{12} \right) } &= \sqrt{ \frac{2 - \sqrt{3}}{4} } \\ &= \frac{\sqrt{2 - \sqrt{3}}}{2} \end{align*}\)

Notice we only have to worry about the positive square root as \(\displaystyle \displaystyle \begin{align*} \frac{\pi}{12} \end{align*}\) is in the first quadrant, so we know that its sine must be positive.