# Thread: Trig without using a calculator help

1. ## Trig without using a calculator help

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

2. Originally Posted by jamm89
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
I think you need to double check that you've written the question right. ATM it doesn't make any sense.

3. Originally Posted by jamm89
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)?

4. Originally Posted by sinewave85
Ok, the basic principle here is that tanΘ = sinΘ/cosΘ, so sinΘ = 8 and cosΘ = 9. That said, the sin2 is sin2, 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)?
Sorry guys it is sin(2Θ)

That makes more sense now lol

5. Hello, jamm89!

We need this identity: . $\sin2\theta \:=\:2\sin\theta\cos\theta$

Given that $\theta$ is an acute angle and that $\tan\theta = \tfrac{8}{9}$
find, without using a calculator the value of: $\sin 2\theta$

We are given: . $\tan\theta \:=\:\frac{8}{9} \:=\:\frac{opp}{adj}$

So $\theta$ is an acute angle in a right triangle with: . $opp = 8,\:adj = 9$

Using Pythagorus, we find that: . $hyp \:=\:\sqrt{145}$

. . Then: . $\begin{array}{c}\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{8}{\sqrt{145}} \\ \\[-3mm]\cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{9}{\sqrt{145}} \end{array}$

Therefore: . $\sin2\theta \;=\;2\sin\theta\cos\theta \;=\;2\left(\frac{8}{\sqrt{145}}\right)\left(\frac {9}{\sqrt{145}}\right) \;=\;\boxed{\frac{144}{145}}$