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# C2: Trig identities Watch

1. Find all values of θ in the range 0° ≤ θ ≤ 360° satisfying
7/cosθ - 10/sinθ = 0

I recognise that you use the identity tanθ = sinθ/cosθ
but how is it tanθ= 10/7???

7/cosθ=10/sinθ then what do you do? I need the steps please! It's just that I am revising Core 2 stuff for my resit exam and I've completely forgotten what to do!
2. Start by cross multipling your 7/cosθ=10/sinθ (Multiply both sides by cosθsinθ) and simplify this.

Now, try looking at your identity of tanθ = sinθ/cosθ.

Can you rearrange this to get it in terms of sinθ = ?

Then, try replacing what you get for sinθ into your cross multiplied equation and rearrange it so you get tanθ = 10/7.

Do you wanna try this for me and see if you can get it?
3. (Original post by backtraced)
Start by cross multipling your 7/cosθ=10/sinθ (Multiply both sides by cosθsinθ) and simplify this.

Now, try looking at your identity of tanθ = sinθ/cosθ.

Can you rearrange this to get it in terms of sinθ = ?

Then, try replacing what you get for sinθ into your cross multiplied equation and rearrange it so you get tanθ = 10/7.

Do you wanna try this for me and see if you can get it?
7sinθ=10cosθ
sinθ=10cosθ/7

tanθ=10cosθ/7/cosθ

I'm sorry, I just can't figure this out for some reason.
4. tanθ= 10cosθ/7 * 1/cosθ and then the cosθ's get cancelled out?? 10cosθ/7cosθ 10/7
5. (Original post by BubbleLover98)
tanθ= 10cosθ/7 * 1/cosθ and then the cosθ's get cancelled out?? 10cosθ/7cosθ 10/7
easy multiply everything by cosθ and sinθ
move cosθ onto the other side
divide both sides by 7 and cosθ at the same time
you have
6. (Original post by BubbleLover98)
tanθ= 10cosθ/7 * 1/cosθ and then the cosθ's get cancelled out?? 10cosθ/7cosθ 10/7
Yeah, that's it!

A bit of an easier way to do it (or the longer way that I did it) was like this:
7sinθ=10cosθ

tanθ = sinθ/cosθ
So, sinθ = tanθ*cosθ

Plug this in to 7sinθ=10cosθ
7tanθ*cosθ=10cosθ

Divide both sides by cosθ
7tanθ=10

Divide through by 7
tanθ=10/7

I find it a bit easier to go through step by step, hopefully that helps a little!
7. (Original post by BubbleLover98)
tanθ= 10cosθ/7 * 1/cosθ and then the cosθ's get cancelled out?? 10cosθ/7cosθ 10/7
(Original post by backtraced)
Yeah, that's it!

A bit of an easier way to do it (or the longer way that I did it) was like this:
7sinθ=10cosθ

tanθ = sinθ/cosθ
So, sinθ = tanθ*cosθ

Plug this in to 7sinθ=10cosθ
7tanθ*cosθ=10cosθ

Divide both sides by cosθ
7tanθ=10

Divide through by 7
tanθ=10/7

I find it a bit easier to go through step by step, hopefully that helps a little!
imma latex this, and get some practise..

done
8. Let's tackle it in another way.
Let's use a property of proportions which goes like this:
9. Looking at it, my way was much more complicated than it needed to be - i'm sorry!

Try looking at thefatone and depymak's work for an easier way to do it. Sorry again for being longwinded!
10. (Original post by depymak)
Let's tackle it in another way.
Let's use a property of proportions which goes like this:
put a d in front of your \frac it makes them bigger like so ^^
11. or

(Original post by depymak)
Let's tackle it in another way.
Let's use a property of proportions which goes like this:
12. (Original post by depymak)
Let's tackle it in another way.
Let's use a property of proportions which goes like this:
The arrow you're using isn't the one you want to be, that's just incorrect notation. Logical connectives are notated as double-barred arrows, so (or ) or .
13. You are right. I tried \Leftrightarrow but something didn't go well. But after your answer, I have just found out why. Thank you.
(Original post by Zacken)
The arrow you're using isn't the one you want to be, that's just incorrect notation. Logical connectives are notated as double-barred arrows, so (or ) or .
14. (Original post by thefatone)
put a d in front of your \frac it makes them bigger like so ^^
Or just use \displaystyle...
15. (Original post by Zacken)
Or just use \displaystyle...
that's the one
16. (Original post by thefatone)
that's the one

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