Really struggling with this one.
Given that Θ is an acute angle and that tanΘ=8/9 find, without using a calculator the value of sin 2
Can anyone shed some light on how it is done
Many thanks
Ok, the basic principle here is that tanΘ = sinΘ/cosΘ, so sinΘ = 8/h and cosΘ = 9/h. That said, the value of the sin 2 is the value of the sin 2, which is not dependent on what you introduced the problem with. Are you sure you typed it out exactly as is written in the book? Is it not sin(2Θ)?
Hello, jamm89!
We need this identity: .$\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta$
Given that $\displaystyle \theta$ is an acute angle and that $\displaystyle \tan\theta = \tfrac{8}{9}$
find, without using a calculator the value of: $\displaystyle \sin 2\theta$
We are given: .$\displaystyle \tan\theta \:=\:\frac{8}{9} \:=\:\frac{opp}{adj}$
So $\displaystyle \theta$ is an acute angle in a right triangle with: .$\displaystyle opp = 8,\:adj = 9$
Using Pythagorus, we find that: .$\displaystyle hyp \:=\:\sqrt{145}$
. . Then: .$\displaystyle \begin{array}{c}\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{8}{\sqrt{145}} \\ \\[-3mm]\cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{9}{\sqrt{145}} \end{array} $
Therefore: .$\displaystyle \sin2\theta \;=\;2\sin\theta\cos\theta \;=\;2\left(\frac{8}{\sqrt{145}}\right)\left(\frac {9}{\sqrt{145}}\right) \;=\;\boxed{\frac{144}{145}}$