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Maths C3 - Trigonometry... Help??

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Original post by Philip-flop
Thanks for making realise that! I'm still a little stuck on this question though :frown:

If tanθ=34 tan \theta = \frac{3}{4} how do I find tanθ4 tan \frac{\theta}{4} ? :frown:

Sorry if I'm sounding pretty silly, but I've never come across a question like this :frown:


Forgot to mention this, but a more straightforward method of doing this is to do the following mapping: 2θ↦θ⇒tan⁡(θ)=2tan⁡(θ2)1−tan⁡2(θ2)\displaystyle 2\theta \mapsto \theta \Rightarrow \tan(\theta)=\frac{2 \tan(\frac{\theta}{2})}{1-\tan^2(\frac{\theta}{2})} and you know the value of tan⁡(θ)\tan(\theta) so just solve for tan⁡(θ2)\tan(\frac{\theta}{2}) by forming a quadratic. Your final angle should satisfy the bound of it.
Original post by RDKGames
I would simply find what θ\theta is (as an exact value) from the given condition since we know it is bounded π<θ<32π\pi<\theta<\frac{3}{2}\pi. Then substitute it into the formula.

Ok so solving tanθ=34 tan \theta = \frac{3}{4}

I get... θ=0.6435011088 \theta = 0.6435011088

then I have to substitute theta into... tanθ2=2tanθ41−tan2θ4 tan \frac{\theta}{2} = \frac{2 tan \frac{\theta}{4}}{1-tan^2 \frac{\theta}{4}} ??

the only problem is, I don't think I even know how to type something like tan2() tan^2() into my calculator as obviously calculators wont allow you to put 'squared' after tan just before the angle inside the brackets () :frown:
Original post by Philip-flop
Ok so solving tanθ=34 tan \theta = \frac{3}{4}

I get... θ=0.6435011088 \theta = 0.6435011088

then I have to substitute theta into... tanθ2=2tanθ41−tan2θ4 tan \frac{\theta}{2} = \frac{2 tan \frac{\theta}{4}}{1-tan^2 \frac{\theta}{4}} ??

the only problem is, I don't think I even know how to type something like tan2() tan^2() into my calculator as obviously calculators wont allow you to put 'squared' after tan just before the angle inside the brackets () :frown:

Think about what tan⁡2θ\tan^2\theta means : it means "tan⁡θ\tan\theta squared" i.e. (tan⁡θ)2\left(\tan\theta\right)^2

So to work out something like tan⁡21.2\tan^2 1.2 on your calculator you would type in (tan⁡1.2)2(\tan 1.2)^2.
Original post by notnek
Think about what tan⁡2θ\tan^2\theta means : it means "tan⁡θ\tan\theta squared" i.e. (tan⁡θ)2\left(\tan\theta\right)^2


So to work out something like tan⁡21.2\tan^2 1.2 on your calculator you would type in (tan⁡1.2)2(\tan 1.2)^2.

Thank you it turns out I knew how to type it into my calculator but my workings were letting me down!
But I managed to get there in the end!!

θ=0.6435011088,3.785093762 \theta = 0.6435011088, 3.785093762

But since... π<θ<3π2 \pi < \theta < \frac{3 \pi}{2}

then...
θ≠0.6435011088[br]θ=3.785093762 \theta \not= 0.6435011088 [br]\theta = 3.785093762

Then sub theta into...tanθ2=2tanθ41−tan2θ4 tan \frac{\theta}{2} = \frac{2 tan \frac{\theta}{4}}{1-tan^2 \frac{\theta}{4}}

So this gives me... tanθ2=2tan(3.785..4)1−tan2(3.785..4) tan \frac{\theta}{2} = \frac{2 tan (\frac{3.785..}{4})}{1-tan^2 (\frac{3.785..}{4})}

tanθ2=−3 tan \frac{\theta}{2} = -3


Thanks a lot for your help!

...And thanks @RDKGames too! You really helped me out on this one!
So I've found myself hitting a brick wall (yet again) with another question...
C3 EXE7B Q10.png

I've drawn the following...
Attachment not found


I realise that I can't use what I've learnt from C2 and apply the 'cosine rule' or 'sine rule'. I can't seem to progress anywhere with this question. All I can work out from the diagram that I've drawn is that angleBAC = 180−3θ 180-3 \theta :frown:

Unless I have to think of the line AC = 5cm as the Hypotenuse and use the fact that H2=O2+A2H^2 = O^2 + A^2 in order to find the opposite side of the triangle? But this only works with right angled triangles :frown:
(edited 7 years ago)
Diagram for C3 EXE7B Q10.png2.3KB
Original post by Philip-flop
So I've found myself hitting a brick wall (yet again) with another question...
C3 EXE7B Q10.png

I've drawn the following...
Attachment not found


I realise that I can't use what I've learnt from C2 and apply the 'cosine rule' or 'sine rule'. I can't seem to progress anywhere with this question. All I can work out from the diagram that I've drawn is that angleBAC = 180−3θ 180-3 \theta :frown:


Remember sin(A)a=sin(B)b\frac{sin(A)}{a} =\frac{sin(B)}{b} and sin2A=2sin(A)cos(A)sin2A = 2sin(A)cos(A)
Original post by NotNotBatman
Remember sin(A)a=sin(B)b\frac{sin(A)}{a} =\frac{sin(B)}{b} and sin2A=2sin(A)cos(A)sin2A = 2sin(A)cos(A)


Oh right. So I was wrong, I do actually have to use the 'sine rule'. I thought I could only do that if I have two side lengths (which there are) and at least one of their opposite angles (in degrees or radians) to find the missing angle, but both angles are missing, so what do I do? :frown:
Original post by Philip-flop
Oh right. So I was wrong, I do actually have to use the 'sine rule'. I thought I could only do that if I have two side lengths (which there are) and at least one of their opposite angles (in degrees or radians) to find the missing angle, but both angles are missing, so what do I do? :frown:


In this case you can use the sine rule in terms of theta rearrange and use double angle identities, then solve.
Original post by NotNotBatman
In this case you can use the sine rule in terms of theta rearrange and use double angle identities, then solve.


Ok, so using the sine rule I would do...

sin2θ5=sinθ4 \frac{sin2 \theta}{5} = \frac{sin \theta}{4}

sin2θ=5sinθ4 sin2 \theta = \frac{5sin \theta}{4}

Apply double angle formula for sin...
2sinθcosθ=5sinθ4 2sin \theta cos \theta = \frac{5sin \theta}{4}

not sure I actually know what I'm doing here/next :frown:
Original post by Philip-flop
Ok, so using the sine rule I would do...

sin2θ5=sinθ4 \frac{sin2 \theta}{5} = \frac{sin \theta}{4}

sin2θ=5sinθ4 sin2 \theta = \frac{5sin \theta}{4}

Apply double angle formula for sin...
2sinθcosθ=5sinθ4 2sin \theta cos \theta = \frac{5sin \theta}{4}

not sure I actually know what I'm doing here/next :frown:


rearrange so you have 0 on the RHS, factorise and solve each factor equaling 0. Then reject the obvious wrong one.
Original post by NotNotBatman
rearrange so you have 0 on the RHS, factorise and solve each factor equaling 0. Then reject the obvious wrong one.

Thanks a lot! Managed to work out the answer as θ=51.31781255\theta = 51.31781255
(edited 7 years ago)
Can someone try to help understand this question...
C3 EXE7B Q12.png

I've drawn out a little diagram of what I think this question is describing...
Attachment not found


I understand that tan is +ve in the 1st & 3rd quadrant which shows for y=34x y= \frac{3}{4}x ... But how do I find out the value of tanθ tan \theta ? Does it have something to do with the gradient of line L : y=34x y = \frac{3}{4}x ??
(edited 7 years ago)
Photo 15-10-2016, 14 54 15.jpg113.7KB
Original post by Philip-flop
Can someone try to help understand this question...
C3 EXE7B Q12.png

I've drawn out a little diagram of what I think this question is describing...


I understand that tan is +ve in the 1st & 3rd quadrant which shows for y=34x y= \frac{3}{4}x ... But how do I find out the value of tanθ tan \theta ? Does it have something to do with the gradient of line L : y=34x y = \frac{3}{4}x ??


tan⁡(θ)\tan(\theta) is the gradient. It is easy to see if you refer to your unit circle; as sine is the vertical distance, and cosine is the horizontal distance. Vertical over horizontal gives the gradient, or tan in this case.
Original post by RDKGames
tan⁡(θ)\tan(\theta) is the gradient. It is easy to see if you refer to your unit circle; as sine is the vertical distance, and cosine is the horizontal distance. Vertical over horizontal gives the gradient, or tan in this case.

Oh yeah because of "SOH" and "CAH"...
sinθ1=sinθ\frac{sin \theta}{1} = sin \theta which gives the vertical distance and...
cosθ1=sinθ\frac{cos \theta}{1} = sin \theta which gives the horizontal distance.

This means coordinates of where y=34xy= \frac{3}{4}x meets the unit circle is...
(cosθ,sinθ) (cos \theta , sin \theta)
(4,3)(4,3)

because... tanθ=sinθcosθ=34 tan \theta = \frac{sin \theta}{cos \theta} = \frac{3}{4}


Thank you soooooo much :smile: :smile: :smile:
(edited 7 years ago)
Came across this question last night (part (i)), and have come home from work to attempt it again but I still find myself stuck :frown:
C3 EXE7C Q1.png

For part (i) I've managed to get up to...
1−tanx1+tanx \frac{1-tan x}{1+tan x}

I've used the identity tanθ=sinθcosθ tan \theta = \frac{sin \theta}{cos\theta} and have done a lot of re-arranging of the equation only to end up with...
cos3x−sinxcos3x+sinx \frac{cos^3x - sinx}{cos^3x + sinx} which I have a feeling isn't right and is stopping me from progressing any further :/
Original post by Philip-flop
Came across this question last night (part (i)), and have come home from work to attempt it again but I still find myself stuck :frown:
C3 EXE7C Q1.png

For part (i) I've managed to get up to...
1−tanx1+tanx \frac{1-tan x}{1+tan x}

I've used the identity tanθ=sinθcosθ tan \theta = \frac{sin \theta}{cos\theta} and have done a lot of re-arranging of the equation only to end up with...
cos3x−sinxcos3x+sinx \frac{cos^3x - sinx}{cos^3x + sinx} which I have a feeling isn't right and is stopping me from progressing any further :/

Your working is correct until you said

cos3x−sinxcos3x+sinx \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}

It should be this instead

cosx−sinxcosx+sinx \displaystyle \frac{cos x - sin x}{cos x + sin x}

From here I would recommend multiplying top and bottom by

cosx−sinx \displaystyle cos x - sin x


Have another go and post your working if you still get

cos3x−sinxcos3x+sinx \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}
Original post by notnek
Your working is correct until you said

cos3x−sinxcos3x+sinx \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}

It should be this instead

cosx−sinxcosx+sinx \displaystyle \frac{cos x - sin x}{cos x + sin x}

From here I would recommend multiplying top and bottom by

cosx−sinx \displaystyle cos x - sin x


Have another go and post your working if you still get

cos3x−sinxcos3x+sinx \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}

But how do I get from...
1−tanx1+tanx \frac{1-tan x}{1+tan x}

to..
cosx−sinxcosx+sinx\frac{cos x - sin x}{cos x + sinx}??
Original post by Philip-flop
But how do I get from...
1−tanx1+tanx \frac{1-tan x}{1+tan x}

to..
cosx−sinxcosx+sinx\frac{cos x - sin x}{cos x + sinx}??


Multiply top and bottom by cosine.
Original post by Philip-flop
But how do I get from...
1−tanx1+tanx \frac{1-tan x}{1+tan x}

to..
cosx−sinxcosx+sinx\frac{cos x - sin x}{cos x + sinx}??

1−tanx1+tanx=1−sin⁡xcos⁡x1+sin⁡xcos⁡x \displaystyle \frac{1-tan x}{1+tan x} = \frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}

The next step is to multiply top and bottom of the fraction by cos⁡x\cos x.
Original post by RDKGames
Multiply top and bottom by cosine.

Thank you!!! How did you know to multiply top and bottom by cos(x)? :frown:


Original post by notnek
1−tanx1+tanx=1−sin⁡xcos⁡x1+sin⁡xcos⁡x \displaystyle \frac{1-tan x}{1+tan x} = \frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}

The next step is to multiply top and bottom of the fraction by cos⁡x\cos x.

Thank you. I never would have known to multiply the top and bottom by cos(x)
(edited 7 years ago)

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