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Hi,

I was doing a question where you had to prove 2tanx/(1+tan^2x)=sin2x which I did, and then it said:

Hence or otherwise find the exact value of tan 15 in the form of a+b√3, where a and b are integers.

I decided to use the equation I just proved and then using that I ended up with a quadratic equation to solve, and got tan15=2+-√3. However, the answer is just 2-√3, which is correct, but how am I meant to show this when I have got two possible solutions using the quadratic formula?

Thanks!

I was doing a question where you had to prove 2tanx/(1+tan^2x)=sin2x which I did, and then it said:

Hence or otherwise find the exact value of tan 15 in the form of a+b√3, where a and b are integers.

I decided to use the equation I just proved and then using that I ended up with a quadratic equation to solve, and got tan15=2+-√3. However, the answer is just 2-√3, which is correct, but how am I meant to show this when I have got two possible solutions using the quadratic formula?

Thanks!

sweet_gurl

Hi,

I was doing a question where you had to prove 2tanx/1+tan^2x=sin2x which I did, and then it said:

Hence or otherwise find the exact value of tan 15 in the form of a+b√3, where a and b are integers.

I decided to use the equation I just proved and then using that I ended up with a quadratic equation to solve, and got tan15=2+-√3. However, the answer is just 2-√3, which is correct, but how am I meant to show this when I have got two possible solutions using the quadratic formula? Or is there a better way of doing this question while still making use of the equation I proved?

Thanks!

I was doing a question where you had to prove 2tanx/1+tan^2x=sin2x which I did, and then it said:

Hence or otherwise find the exact value of tan 15 in the form of a+b√3, where a and b are integers.

I decided to use the equation I just proved and then using that I ended up with a quadratic equation to solve, and got tan15=2+-√3. However, the answer is just 2-√3, which is correct, but how am I meant to show this when I have got two possible solutions using the quadratic formula? Or is there a better way of doing this question while still making use of the equation I proved?

Thanks!

This is an entirely different and independent method:

To see tan 15° [Pencil and ruler method!]

draw an equilateral triangle ABC, and drop a perpendicular from B to the middle of AC. (M)

Produce MB to D, so that AB=BD. Now ADM = 15° and MAD = 75°

Let AB=2, so AM = 1, so BM=√3, and BD = 2.

Tan 15°= Tan ADM = 1/(2+√3)

Rationalise the denominator to get 2-√3

Just use use tan15= tan(45-30) and simplify. You get the right answer.

i'm not sure where you've gone wrong coz i dont understand the LHS of your identity. It looks like 2tanx/1 but thats just 2tanx??

i'm not sure where you've gone wrong coz i dont understand the LHS of your identity. It looks like 2tanx/1 but thats just 2tanx??

I just edited it, sorry it wasn't very clear.

I know you can do it in other ways such as by tan15= tan(45-30), but I just wanted to know how I could do it using the identity?

sweet_gurl

Hi,

I was doing a question where you had to prove 2tanx/1+tan^2x=sin2x which I did, and then it said:

Hence or otherwise find the exact value of tan 15 in the form of a+b√3, where a and b are integers.

I decided to use the equation I just proved and then using that I ended up with a quadratic equation to solve, and got tan15=2+-√3. However, the answer is just 2-√3, which is correct, but how am I meant to show this when I have got two possible solutions using the quadratic formula?

Thanks!

I was doing a question where you had to prove 2tanx/1+tan^2x=sin2x which I did, and then it said:

Hence or otherwise find the exact value of tan 15 in the form of a+b√3, where a and b are integers.

I decided to use the equation I just proved and then using that I ended up with a quadratic equation to solve, and got tan15=2+-√3. However, the answer is just 2-√3, which is correct, but how am I meant to show this when I have got two possible solutions using the quadratic formula?

Thanks!

on (0,90) tan increases. Note also that tan45=1.

Now 2+rt(3) is >1 so it is tan x for an x greater than 45.

Yeah what he said. Because on the LHS tan15 has been squared, then you have to test/check your solutions to make sure they fit. tan15 has to be less than tan 45.

Its like some of those quadratics you get when sinx= 2+-√3. You know it cant be 2+√3 as its greater than 1. With tan you need to be more careful.

Its like some of those quadratics you get when sinx= 2+-√3. You know it cant be 2+√3 as its greater than 1. With tan you need to be more careful.

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can someone please explain what principle domain is and why the answer is a not c?Maths

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