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Maths C3 - Trigonometry... Help??

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Original post by Philip-flop
C3 EXE7A Q13.png

for part (e) where do I go from?...

tanθ−11+tanθ=6tanθ \frac{tan \theta -1}{1+tan \theta} =6tan \theta

Have you tried anything from here? Do some manipulation of the equation and see what happens.

Post our working if you get stuck.
Original post by notnek
Have you tried anything from here? Do some manipulation of the equation and see what happens.

Post our working if you get stuck.


I still couldn't manage to work it out so I moved on to the other questions :frown:
Original post by Philip-flop
I still couldn't manage to work it out so I moved on to the other questions :frown:


Multiply both sides by

1+tanθ 1 + tan\theta

Then you reaarange and eventually you have something in the form

atan2θ+btanθ+c=0 a tan^2\theta + b tan \theta + c = 0

Where a, b and c are constants,

Now you factorise or use quadratic fotmula
Original post by asinghj
Multiply both sides by

1+tanθ 1 + tan\theta

Then you reaarange and eventually you have something in the form

atan2θ+btanθ+c=0 a tan^2\theta + b tan \theta + c = 0

Where a, b and c are constants,

Now you factorise or use quadratic fotmula


OMG thank you so much!! Managed to work it out now :smile: :smile: :smile:
Where have I gone wrong for part (b)?...
C3 EXE7A Q21.png

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Photo 08-10-2016, 16 49 54.jpg315.3KB
Original post by Philip-flop
Where have I gone wrong for part (b)?...
C3 EXE7A Q21.png

Attachment not found


Recheck how you got from 3rd line to 4th line
Original post by asinghj
Recheck how you got from 3rd line to 4th line

Thank you so much!! I was stressing about this question so much to the point where I couldn't even think straight! I got there in the end :tongue:
Can someone help me? I'm stuck on part (h) and (i)...

C3 EXE7B Q3 .png

For part (h) I've done....

tanθsec2θ−2 \frac{tan \theta}{sec^2 \theta - 2}

=tanθ(1cos2θ−2) = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}



For part (i) I've done...

sin4θ−2sin2θcos2θ+cos4θ sin^4 \theta -2sin^2 \theta cos^2 \theta + cos^4 \theta

=(sin2θ−cos2θ)2 = (sin^2 \theta - cos^2 \theta)^2 <<<or should I re-arrange this line to give...(cos2θ−sin2θ)2(cos^2 \theta - sin^2 \theta)^2 ???

=(1−cos2θ−cos2θ)2 = (1 - cos^2 \theta - cos^2 \theta)^2

=(1−2cos2θ)2 = (1 - 2cos^2 \theta)^2 <<< I know the double angle formula... cos2θ=2cos2θ−1 cos2 \theta = 2cos^2 \theta -1 ... but this can't be used here, right?


Don't know where to go from there for either of them though :frown:


Edit: I've managed to work out part (i) now by rearranging the second line from my workings above :smile:
(edited 7 years ago)
Original post by Philip-flop
Can someone help me? I'm stuck on part (h) and (i)...


For h, multiply top and bottom by cos⁡2(θ)\cos^2(\theta) and use identities for double angles. Also it will be useful to rewrite sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta) as 12[2sin⁡(θ)cos⁡(θ)]\frac{1}{2}[2\sin(\theta)\cos(\theta)]
Original post by Philip-flop
Can someone help me? I'm stuck on part (h) and (i)...

C3 EXE7B Q3 .png

For part (h) I've done....

tanθsec2θ−2 \frac{tan \theta}{sec^2 \theta - 2}

=tanθ(1cos2θ−2) = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}



For part (i) I've done...

sin4θ−2sin2θcos2θ+cos4θ sin^4 \theta -2sin^2 \theta cos^2 \theta + cos^4 \theta

=(sin2θ−cos2θ)2 = (sin^2 \theta - cos^2 \theta)^2 <<<or should I re-arrange this line to give...(cos2θ−sin2θ)2(cos^2 \theta - sin^2 \theta)^2 ???

=(1−cos2θ−cos2θ)2 = (1 - cos^2 \theta - cos^2 \theta)^2

=(1−2cos2θ)2 = (1 - 2cos^2 \theta)^2 <<< I know the double angle formula... cos2θ=2cos2θ−1 cos2 \theta = 2cos^2 \theta -1 ... but this can't be used here, right?


Don't know where to go from there for either of them though :frown:


Edit: I've managed to work out part (i) now by rearranging the second line from my workings above :smile:

For h), the quickest way is to use the tan/sec identity on the denominator and then compare what you've got with

tan⁡2θ≡2tan⁡θ1−tan2θ\displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}
Original post by RDKGames
For h, multiply top and bottom by cos⁡2(θ)\cos^2(\theta) and use identities for double angles. Also it will be useful to rewrite sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta) as 12[2sin⁡(θ)cos⁡(θ)]\frac{1}{2}[2\sin(\theta)\cos(\theta)]

Ok so this is what I've done for part (h) now...

tanθsec2θ−2 \frac{tan \theta}{sec^2 \theta - 2}

=tanθ(1cos2θ−2) = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}

=tanθcos2θ1−2cos2θ = \frac{tan \theta cos^2 \theta}{1 - 2cos^2 \theta}

=tanθcos2θcos2θ = \frac{tan \theta cos^2 \theta}{cos2 \theta}

But then where from here? :frown:

Original post by notnek
For h), the quickest way is to use the tan/sec identity on the denominator and then compare what you've got with

tan⁡2θ≡2tan⁡θ1−tan2θ\displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}

Ok, following your way for part (h) I've done...

tanθsec2θ−2 \frac{tan \theta}{sec^2 \theta - 2}

=tanθ1+tan2θ−2 = \frac{tan \theta}{1+tan^2 \theta -2}

=tanθtan2θ−1 = \frac{tan \theta}{tan^2 \theta -1}

But then I'm having trouble comparing it to the double angle formula... tan⁡2θ≡2tan⁡θ1−tan2θ\displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}

I can only vaguely see why it becomes... =−12tan2θ = -\frac{1}{2}tan2 \theta :frown:

Is it because from the identity tan2θ≡2tan⁡θ1−tan2θtan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta} you would have to times by -1 and divide by 2 to get the identity into the same form as =tanθtan2θ−1 = \frac{tan \theta}{tan^2 \theta -1}???
(edited 7 years ago)
Original post by Philip-flop
Ok so this is what I've done for part (h) now...

tanθsec2θ−2 \frac{tan \theta}{sec^2 \theta - 2}

=tanθ(1cos2θ−2) = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}

=tanθcos2θ1−2cos2θ = \frac{tan \theta cos^2 \theta}{1 - 2cos^2 \theta}

=tanθcos2θcos2θ = \frac{tan \theta cos^2 \theta}{cos2 \theta}

But then where from here? :frown:


Ok, following your way for part (h) I've done...

tanθsec2θ−2 \frac{tan \theta}{sec^2 \theta - 2}

=tanθ1+tan2θ−2 = \frac{tan \theta}{1+tan^2 \theta -2}

=tanθtan2θ−1 = \frac{tan \theta}{tan^2 \theta -1}

But then I'm having trouble comparing it to the double angle formula... tan⁡2θ≡2tan⁡θ1−tan2θ\displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}

tanθtan2θ−1 \displaystyle \frac{tan \theta}{tan^2 \theta -1}

Multiply this by 2:

2tanθtan2θ−1\displaystyle \frac{2tan \theta}{tan^2 \theta -1}

Then multiply it by -1, which is the same as multiplying the denominator by -1:

2tanθ1−tan2θ\displaystyle \frac{2tan \theta}{1-tan^2 \theta}

Does that help?

A lot of this exercise is about comparing expressions with the double angle identities. This question is one of the trickier ones.
Original post by Philip-flop

Is it because from the identity tan2θ≡2tan⁡θ1−tan2θtan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta} you would have to times by -1 and divide by 2 to get the identity into the same form as =tanθtan2θ−1 = \frac{tan \theta}{tan^2 \theta -1}???

That's correct - you got it before my reply :smile:
Original post by notnek
tanθtan2θ−1 \displaystyle \frac{tan \theta}{tan^2 \theta -1}

Multiply this by 2:

2tanθtan2θ−1\displaystyle \frac{2tan \theta}{tan^2 \theta -1}

Then multiply it by -1, which is the same as multiplying the denominator by -1:

2tanθ1−tan2θ\displaystyle \frac{2tan \theta}{1-tan^2 \theta}

Does that help?

A lot of this exercise is about comparing expressions with the double angle identities. This question is one of the trickier ones.

Thank you so much! I had a feeling I might have to compare it to that double-angle formula, but it took me a while to realise it at first! :colondollar:
Original post by Philip-flop
Ok so this is what I've done for part (h) now...

tanθsec2θ−2 \frac{tan \theta}{sec^2 \theta - 2}

=tanθ(1cos2θ−2) = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}

=tanθcos2θ1−2cos2θ = \frac{tan \theta cos^2 \theta}{1 - 2cos^2 \theta}

=tanθcos2θcos2θ = \frac{tan \theta cos^2 \theta}{cos2 \theta}

But then where from here? :frown:

If you turn tan into sine over cosine then you are left with sin(x)cos(x). As I mentioned above, manipulating this slightly will enable you to use double angle formulae.
Original post by RDKGames
If you turn tan into sine over cosine then you are left with sin(x)cos(x). As I mentioned above, manipulating this slightly will enable you to use double angle formulae.

Omg yes!! Thank you!! I've managed to get there now :smile: I seized up at first, but then re-arranged tan like you said and it was pretty simple from there. Really appreciate the help :smile:
Ok, so I spent way too much time trying to solve this question last night without any luck....
C3 EXE7B Q9.png
I've tried comparing tanθ2 tan \frac{\theta}{2} to the double angle formula for tan2θ=2tanθ1−tan2θ tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

I can see that tan2θ tan 2\theta needs to be divided by 4 in order to get tanθ2 tan \frac{\theta}{2}

so this means...
tan2θ=2tanθ1−tan2θ tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

Divide by 4 gives...

tanθ2=2tanθ4−4tan2θ tan \frac{\theta}{2} = \frac{2tan \theta}{4-4tan^2 \theta} .... right??

And then I have to sub tanθ=34tan \theta = \frac{3}{4} into the equation I've just found above??
Original post by Philip-flop
Ok, so I spent way too much time trying to solve this question last night without any luck....
C3 EXE7B Q9.png
I've tried comparing tanθ2 tan \frac{\theta}{2} to the double angle formula for tan2θ=2tanθ1−tan2θ tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

I can see that tan2θ tan 2\theta needs to be divided by 4 in order to get tanθ2 tan \frac{\theta}{2}

so this means...
tan2θ=2tanθ1−tan2θ tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

Divide by 4 gives...

tanθ2=2tanθ4−4tan2θ tan \frac{\theta}{2} = \frac{2tan \theta}{4-4tan^2 \theta} .... right??

And then I have to sub tanθ=34tan \theta = \frac{3}{4} into the equation I've just found above??


You are not dividing tan by 4, you are dividing the angle. So 2θ↦θ2⇒tan⁡(θ2)=2tan⁡(θ4)1−tan⁡2(θ4)\displaystyle 2\theta \mapsto \frac{\theta}{2} \Rightarrow \tan(\frac{\theta}{2})=\frac{2 \tan(\frac{\theta}{4})}{1-\tan^2(\frac{\theta}{4})}
(edited 7 years ago)
Original post by RDKGames
You are not dividing tan by 4, you are dividing the angle. So 2θ↦θ2⇒tan⁡(θ2)=2tan⁡(θ4)1−tan⁡2(θ4)\displaystyle 2\theta \mapsto \frac{\theta}{2} \Rightarrow \tan(\frac{\theta}{2})=\frac{2 \tan(\frac{\theta}{4})}{1-\tan^2(\frac{\theta}{4})}


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:
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:


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.

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