Another question on identities and trig etc

Hello AGAIN! :o:

So heres another I can't get my head around, if anyone can start me off in the right direction I'll be grateful.

Given that 2 sin (2x - pi/4) = cos (2x - pi/4), show that tan 2x=3

I can do stuff with it, I just cannot get to a point where I have sin2x/cos2x and it's doing my head in. Any help appreciated.

Reply 1

Swayum

I would literally just expand both sides of the equation, noting that sin(pi/4) = cos(pi/4) = (root2)/2.

Post some working along these lines.

Reply 2

HaNzY

Swayum

I would literally just expand both sides of the equation, noting that sin(pi/4) = cos(pi/4) = (root2)/2.

Post some working along these lines.

Thats what I did, I have-

2 sin 2x ((root 2)/2) - cos 2x (-(root 2)/2) = cos 2x ((root 2)/2) + sin 2x (- (root 2)/2)

= root 2 sin 2x + root 2 cos 2x = (root 2)/2 cos 2x - (root 2)/2 sin 2x

= 3 (root 2)/2 sin 2x + (root 2)/2 cos 2x = 0

and then I probably just got carried away and went-

(root 2)/2 (3 sin 2x + cos 2x) = 0

and I have noooo idea where to get a tan from. I tried shoving some cos's in but it didn't seem to work, however I cannot see any other way to do it apart from multiplying it by cos 2x/cos 2x or something similar......

Reply 3

Pheylan

Alternatively...

2sin(2x - \frac {\pi}{4}) = cos(2x - \frac {\pi}{4})

\frac {sin(2x - \frac {\pi}{4})}{cos(2x - \frac {\pi}{4})} = \frac {1}{2}

tan(2x - \frac {\pi}{4}) = \frac {1}{2}

\frac {tan2x - tan\frac {\pi}{4}}{1 + tan2xtan\frac {\pi}{4}} = \frac {1}{2}

tan\frac {\pi}{4} \equiv 1

\frac {tan2x - 1}{tan2x + 1} = \frac {1}{2}

2(tan2x - 1) = tan2x + 1

2tan2x - 2 = tan2x + 1

tan2x = 3

Reply 4

EM(Lin)

umm I can get it down to -3=tan2x...

3(root2)/2sin2x=-(root2)/2cos2x

multiply by 2sin2x: 3(root2)=-(root2)2sin2x/2cos2x
divide by (root2): -3=2sin2x/2cos2x
cancel the 2s: -3=sin2x/cos2x=tanx

I haven't checked your working, so you could have made a mistake with a sign somewhere along the way. Or I could just be wrong. Sorry I can't be more help :s-smilie:

Reply 5

EM(Lin)

Oh, ignore my post- Pheylan's got there first. I guess that's what comes of actually bothering to look at the problem from the beginning :p:

Reply 6

HaNzY

Pheylan

Alternatively...

2sin(2x - \frac {\pi}{4}) = cos(2x - \frac {\pi}{4})

\frac {sin(2x - \frac {\pi}{4})}{cos(2x - \frac {\pi}{4})} = \frac {1}{2}

Omg, I totally didn't see that you could put all that sinx bit over 1/2 and then swippy swappy the cosx for it :frown:

oh jeez, how do you spot these things?? How do I learn to spot that? I would never have seen that on my own :frown:

Thank you as well :smile:

Reply 7

Swayum

HaNzY

Thats what I did, I have-

2 sin 2x ((root 2)/2) - cos 2x (-(root 2)/2) = cos 2x ((root 2)/2) + sin 2x (- (root 2)/2)

It falls apart at the start:

cos(A - B) = cosAcosB + sinAsinB

So RHS = cos(2x)cos(pi/4) + sin(2x)sin(pi/4)

= cos2x * (root2)/2 + sin(2x) * (root2)/2

You've got a minus sign from somewhere.

Reply 8

Swayum

HaNzY

Omg, I totally didn't see that you could put all that sinx bit over 1/2 and then swippy swappy the cosx for it :frown:

oh jeez, how do you spot these things?? How do I learn to spot that? I would never have seen that on my own :frown:

Thank you as well :smile:

He was aiming to get a tan term from somewhere. He noticed that the equation is of the form 2sin(u) = cos(u) (where u = x - pi/4).

It's easy to see that 2sinu = cosu can be transformed to 2tanu = 1, which you can see will go to tanu = 1/2.

Reply 9

Pheylan

HaNzY

How do I learn to spot that?

practice

your method would work too

2sin(2x - \frac {\pi}{4}) = 2(sin2xcos\frac {\pi}{4} - sin\frac {\pi}{4}cos2x) =

cos(2x - \frac {\pi}{4}) = cos2xcos\frac {\pi}{4} + sin2xsin\frac {\pi}{4} =

go from there and see if you get it :h:

Reply 10

HaNzY

Swayum

It falls apart at the start:

cos(A - B) = cosAcosB + sinAsinB

So RHS = cos(2x)cos(pi/4) + sin(2x)sin(pi/4)

= cos2x * (root2)/2 + sin(2x) * (root2)/2

You've got a minus sign from somewhere.

Oh so is B just pi/4 and not -pi/4?? Thats where I got it from, I was using -pi/4 as B.

Reply 11

EM(Lin)

Swayum

It falls apart at the start:

cos(A - B) = cosAcosB + sinAsinB

So RHS = cos(2x)cos(pi/4) + sin(2x)sin(pi/4)

= cos2x * (root2)/2 + sin(2x) * (root2)/2

You've got a minus sign from somewhere.

Which could be where the error in my explanation came from? If that's the case, OP you could have kept on going with your original method and still got the answer (albeit in a slightly more roundabout way :p:

) without having to spot the other method.

Reply 12

Swayum

HaNzY

Oh so is B just pi/4 and not -pi/4?? Thats where I got it from, I was using -pi/4 as B.

You can do that, in which case you'd have this sort of situation:

cos(A + B) = cosAcosB - sinAsinB

But sinB = sin(-pi/4) = -(root2)/2

So -sinAsinB = -(-(root2)/2) = (root2)/2

So we're still getting a positive term.