Trig help please!
Holaa!
I'm probably being dense again, but I've re-written this out a thousand times now and can't see what I'm supposed to do, so any help would be great, thanks!
Solve the equation: (3cos2x)/(cos2x+1)=2(tanx+1) ; range 0-360
I'm assuming you expand each cos2x, but then every time I try and rearrange what I have left, I end up with disgustingly long expressions and I'm sure it must be simpler than this!
Thank you!
I'm probably being dense again, but I've re-written this out a thousand times now and can't see what I'm supposed to do, so any help would be great, thanks!
Solve the equation: (3cos2x)/(cos2x+1)=2(tanx+1) ; range 0-360
I'm assuming you expand each cos2x, but then every time I try and rearrange what I have left, I end up with disgustingly long expressions and I'm sure it must be simpler than this!
Thank you!
Original post by Lainathiel
Holaa!
I'm probably being dense again, but I've re-written this out a thousand times now and can't see what I'm supposed to do, so any help would be great, thanks!
Solve the equation: (3cos2x)/(cos2x+1)=2(tanx+1) ; range 0-360
I'm assuming you expand each cos2x, but then every time I try and rearrange what I have left, I end up with disgustingly long expressions and I'm sure it must be simpler than this!
Thank you!
I'm probably being dense again, but I've re-written this out a thousand times now and can't see what I'm supposed to do, so any help would be great, thanks!
Solve the equation: (3cos2x)/(cos2x+1)=2(tanx+1) ; range 0-360
I'm assuming you expand each cos2x, but then every time I try and rearrange what I have left, I end up with disgustingly long expressions and I'm sure it must be simpler than this!
Thank you!
Cos2x + 1 should trigger which double angle formulae for cos you should be using (the one that cancels out nicely!).
And then thinking about what that leaves you on the denominator for the left hand side, think about which one you should use for the top cos2x.
Spoiler
Original post by SeanFM
Cos2x + 1 should trigger which double angle formulae for cos you should be using (the one that cancels out nicely!).
And then thinking about what that leaves you on the denominator for the left hand side, think about which one you should use for the top cos2x.
And then thinking about what that leaves you on the denominator for the left hand side, think about which one you should use for the top cos2x.
Spoiler
That's what I did before and still got confused; I'm at (3-6sin^2x)/(2cos^2x) which can then be written as 3/(2cos^2x)-3tan^2x, but then you still have the sec in there, which I don't know how I'm supposed to get rid of it!
Posted from TSR Mobile
Original post by Lainathiel
That's what I did before and still got confused; I'm at (3-6sin^2x)/(2cos^2x) which can then be written as 3/(2cos^2x)-3tan^2x, but then you still have the sec in there, which I don't know how I'm supposed to get rid of it!
Posted from TSR Mobile
Posted from TSR Mobile
Your substitution for the denominator is correct.
I'm not sure if you can work from the substitution you've done for the numerator, but instead of cos2x = 1 - 2sin^2(x), I would suggest using cos2x = (cos^2(x) - sin^2(x)).
Original post by SeanFM
Your substitution for the denominator is correct.
I'm not sure if you can work from the substitution you've done for the numerator, but instead of cos2x = 1 - 2sin^2(x), I would suggest using cos2x = (cos^2(x) - sin^2(x)).
I'm not sure if you can work from the substitution you've done for the numerator, but instead of cos2x = 1 - 2sin^2(x), I would suggest using cos2x = (cos^2(x) - sin^2(x)).
So, (3cos^2x)/(2cos^2x)-(3/2*tanx)=4tanx-4 - but I rearranged that and the quadratic is unsolvable?!
I'm sorry for being a pain - it's been a long week!
Original post by Lainathiel
So, (3cos^2x)/(2cos^2x)-(3/2*tanx)=4tanx-4 - but I rearranged that and the quadratic is unsolvable?!
I'm sorry for being a pain - it's been a long week!
I'm sorry for being a pain - it's been a long week!
It's okay
I think you've half-multiplied the equation by 2 (it should be what you've got on the left hand side = 2tanx + 2 rather than 4tanx - 4).
Original post by SeanFM
It's okay
I think you've half-multiplied the equation by 2 (it should be what you've got on the left hand side = 2tanx + 2 rather than 4tanx - 4).
I think you've half-multiplied the equation by 2 (it should be what you've got on the left hand side = 2tanx + 2 rather than 4tanx - 4).
Damn it - the reason it was unsolvable was because I had a quadratic of 3x^2+4x+7, not +1. *facepalm* I told you I was being dense!! Okay, I FINALLY have it. :')
Thank you so much!
Ok so step 1:
3(cos^2(X)-sin^2(X))/cos^2(X)-sin^2(X)+1
Step 2:
You need to use sin^2(X)+cos^2(X)=1
This can rearrange to sin^2(X)=1-cos^2(X)
Sub this in on the bottem half of the fraction to get
3(cos^2(X)-sin^2(X))/cos^2(X)-(1-cos^2(X))+1
step 3:
Simplify the bottem
3(cos^2(X)-sin^2(X))/2cos^2(X)
Step 4:
Split it all up to make it easier
3(cos^2(X)/2cos^2(X))(-sin^2(X))/2cos^2(X))
And simplify
3(1/2)(-tan^2(X)/2)
-3/2tan^2(X)
Step 5:
We know it equals 2(tanx+1)
So set them equal
-3/2tan^2(X)=2(tanx+1)
Expand the right side
-3/2tan^2(X)=2tanx+2
This makes a quadratic
3/2tan^2(X)+2tanx+2 = 0
Step 6:
Let tanx = a
So 3/2a^2+2a+2=0
And solve it like a normal quadratic equation (you might need to use the formula)
I will let you do this bit
Step 7:
When you know what a equals you know what tanx equals
Just do X=tan^-1(a) (inverse tan of what ever a was)
Then find all the corisponding values within you set range
Hope that helps!
Let me know if you are still stuck
3(cos^2(X)-sin^2(X))/cos^2(X)-sin^2(X)+1
Step 2:
You need to use sin^2(X)+cos^2(X)=1
This can rearrange to sin^2(X)=1-cos^2(X)
Sub this in on the bottem half of the fraction to get
3(cos^2(X)-sin^2(X))/cos^2(X)-(1-cos^2(X))+1
step 3:
Simplify the bottem
3(cos^2(X)-sin^2(X))/2cos^2(X)
Step 4:
Split it all up to make it easier
3(cos^2(X)/2cos^2(X))(-sin^2(X))/2cos^2(X))
And simplify
3(1/2)(-tan^2(X)/2)
-3/2tan^2(X)
Step 5:
We know it equals 2(tanx+1)
So set them equal
-3/2tan^2(X)=2(tanx+1)
Expand the right side
-3/2tan^2(X)=2tanx+2
This makes a quadratic
3/2tan^2(X)+2tanx+2 = 0
Step 6:
Let tanx = a
So 3/2a^2+2a+2=0
And solve it like a normal quadratic equation (you might need to use the formula)
I will let you do this bit
Step 7:
When you know what a equals you know what tanx equals
Just do X=tan^-1(a) (inverse tan of what ever a was)
Then find all the corisponding values within you set range
Hope that helps!
Let me know if you are still stuck
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