Question

how would you find the value of sin pi/12 using a half angle formula?

Answer 1

eli do you know what's \(sin^2x\)=?

Answer 2

2sin(x)cos(x)

Answer 3

no sin^2x= 1-cos^2x

Answer 4

would the answer be 1/2 srq of 2- srq of 3?

Answer 5

now let me guide you
we have to find \(sin\pi/12\)
let's find \(cos \pi/12\) then find \(sin\pi/12\) using = \(\sqrt{1-cos^2 \pi/12}\)
we know \[cos 2x= 2cos^2 x-1\]
here x = pi/12
so
\[cos 2\pi/12= 2cos^2 \pi/12-1\]
cos 2pi/12= cos pi/6
eli what's cos pi/6

Answer 6

the half-angle formula for sin is
\[sin(a/2)= \sqrt{\frac{1}{2}(1-cos(a))} \]

Answer 7

cos pi/6= \(\frac{\sqrt 3}{2}\)
so we get
\[ \frac{\sqrt 3}{2}=2 cos^2 \pi/12-1\]
so \[ cos^2 \pi/12= \frac{1}{2}(\frac{\sqrt 3}{2}+1)\]
now
\[sin^2 \pi/12= 1- cos^2 \pi/12\]
so we get
\[sin^2 \pi/12= 1- \frac{1}{2}(\frac{\sqrt 3}{2}+1)\]
simplifying we get
\[sin^2 \pi/12= \frac{1}{2} - \frac{\sqrt 3}{4}\]
so
we get
\[sin \frac{\pi}{12}= \sqrt{\frac{1}{2} - \frac{\sqrt 3}{4}}\]