same trig topic
- Thread starter she18
- Start date
Hello, she18!
This one looked horrible . . . until I baby-talked my way through it.
We'll need this identity: /\(\displaystyle \sin^2\left(\frac{\theta}{2}\right)\:=\:\frac{1\,-\,\cos\theta}{2}\;\;\Rightarrow\;\;\sin\left(\frac{\theta}{2}\right)\:=\:\pm\sqrt{\frac{1\,-\,\cos\theta}{2}}\)
\(\displaystyle \arccos\left(\frac{4}{5}\right)\) is just some angle, \(\displaystyle \theta\), such that: \(\displaystyle \cos(\theta)\,=\,\frac{4}{5}\) . [1]
. . And we are asked to find: .\(\displaystyle \sin\left(\frac{\theta}{2}\right)\)
From the identity: .\(\displaystyle \sin\left(\frac{\theta}{2}\right)\;=\;\pm\sqrt{\frac{1\,-\,\cos\theta}{2}}\)
and from [1], we know that: \(\displaystyle \cos(\theta)\,=\,\frac{4}{5}\)
Therefore: .\(\displaystyle \sin\left(\frac{\theta}{2}\right)\;=\;\pm\sqrt{\frac{1-\frac{4}{5}}{2}}\;=\;\pm\frac{1}{\sqrt{10}}\;=\;\pm\frac{\sqrt{10}}{10}\)
This one looked horrible . . . until I baby-talked my way through it.
We'll need this identity: /\(\displaystyle \sin^2\left(\frac{\theta}{2}\right)\:=\:\frac{1\,-\,\cos\theta}{2}\;\;\Rightarrow\;\;\sin\left(\frac{\theta}{2}\right)\:=\:\pm\sqrt{\frac{1\,-\,\cos\theta}{2}}\)
Here's the baby-talk . . .
\(\displaystyle \arccos\left(\frac{4}{5}\right)\) is just some angle, \(\displaystyle \theta\), such that: \(\displaystyle \cos(\theta)\,=\,\frac{4}{5}\) . [1]
. . And we are asked to find: .\(\displaystyle \sin\left(\frac{\theta}{2}\right)\)
From the identity: .\(\displaystyle \sin\left(\frac{\theta}{2}\right)\;=\;\pm\sqrt{\frac{1\,-\,\cos\theta}{2}}\)
and from [1], we know that: \(\displaystyle \cos(\theta)\,=\,\frac{4}{5}\)
Therefore: .\(\displaystyle \sin\left(\frac{\theta}{2}\right)\;=\;\pm\sqrt{\frac{1-\frac{4}{5}}{2}}\;=\;\pm\frac{1}{\sqrt{10}}\;=\;\pm\frac{\sqrt{10}}{10}\)
