# Double angleWatch

This discussion is closed.
#1
sin 2x = cos x

---

2sinx cosx = cos x

2 sinx = 1

sinx = 1/2 .... 30º and 150º

But the book also adds 90º and 270º

Jorge
0
14 years ago
#2
(Original post by Jorge)
sin 2x = cos x

---

2sinx cosx = cos x

2 sinx = 1

sinx = 1/2 .... 30º and 150º

But the book also adds 90º and 270º

Jorge
Your mistake was in dividing by cosx hence losing two more solutions.

cosx = 0 => x = 90, 270 ...

Euclid.
0
#3
Thanks Euclid

But how do I get cosx to equal 0 ?

Jorge
0
14 years ago
#4
(Original post by Jorge)
Thanks Euclid

But how do I get cosx to equal 0 ?

Jorge
Because you have,

cosx(2sinx)=cosx

Either cosx = 0 or 2sinx = 1 for the equality to hold.

The solutions then follow from there.

Euclid.
0
#5
Got it. Thanks.

But I'll never pass the exam!

Jorge
0
14 years ago
#6
(Original post by Jorge)
Got it. Thanks.

But I'll never pass the exam!

Jorge
have faith, practice makes perfect
0
#7
(Original post by Euclid)
Because you have,

cosx(2sinx)=cosx

Either cosx = 0 or 2sinx = 1 for the equality to hold.

The solutions then follow from there.

Euclid.
troubles, troubles, troubles...

Take this one:

sin^2x + sin2x = 0

---
I got the answers but I'm sure there is a quicker way.

sin^2x + 2sinx cosx = 0

sinx sinx = -2sinx cosx

sinx = -2cosx

tanx = -2 ... 296.6º and 116.6º

but there are more answers and I started all over again~:

sin^2x + 2sinx cosx = 0

sinx (sinx + 2cosx) = 0

so sinx could be 0 or something else (-2 is a good guess)

sinx = 0 ... 0º and 360º

but the book still has 180º

help...

jorge
0
14 years ago
#8
sin^2x + sin2x = 0
sinx(sinx+2cosx)=0

sinx = 0
x = 0, 180, 360

tanx = -2
x = 116.6, 296.6
0
#9
(Original post by kikzen)
sin^2x + sin2x = 0
sinx(sinx+2cosx)=0

sinx = 0
x = 0, 180, 360

tanx = -2
x = 116.6, 296.6
Right...

But you still have to make two calculations like I did? One for tan ,one for sin?

jorge
0
14 years ago
#10
yep! you just missed out the 180 in sin
0
#11
Sorry for going on and on...

But how do I know (or rather decide) that having worked out my solution I still need to go looking for a further one?

jorge
0
14 years ago
#12
(Original post by Jorge)
Sorry for going on and on...

But how do I know (or rather decide) that having worked out my solution I still need to go looking for a further one?

jorge
do you know about the quadrants where sin, cos or tan are positive or negative?
0
#13
Yes, I do.

I think I did not explain myself well.

In my last problem I solved it by finding that tanx = -2

sin^2x + 2sinx cosx = 0

sinx sinx = -2sinx cosx

sinx = -2cosx

tanx = -2 ... 296.6º and 116.6º

But there were other solutions.

How should I know? because having got that far I assumed there was nothing further to look for.

Jorge
0
14 years ago
#14
what?

the question will give a range to look for solutions in.

keep looking for solutions until you exceed that range.

another set of solutions come from the sinx=0
0
14 years ago
#15
You can also sometimes tell how many solutions you need from marks available from the question
0
#16
Thanks Kitzen

That's exactly what I meant:

I know about the different solutions withinn the range.

What I could not have guesses is that having arrived at my tan solution, there was still a sin solution lurking in the dark...

How do YOU go about it?

jorge
0
14 years ago
#17
(Original post by Jorge)
Thanks Kitzen

That's exactly what I meant:

I know about the different solutions withinn the range.

What I could not have guesses is that having arrived at my tan solution, there was still a sin solution lurking in the dark...

How do YOU go about it?

jorge
Always keep track of what you cancel out, like where you divided through by sinx or cosx.

Remember when cancelling out you may be loosing one/two more solutions.

Euclid.
0
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