Double angle question

**RAz** · February 14th 2010, 11:40 PM

The question is

If tanA=4 and tabB=3/5, and A and B are acute angles, prove that A-B=pi/4 (pi in radians, so really it's 45degrees).

My answer is:

$tan(x-y)=\frac {tan4-tan \frac {3}{5}}{1 + tan4 * tan \frac {3}{5}}<br /> =1<br /> \therefore tan(x-y)=1=\frac {\pi}{4}$

I have a feeling there is more to add. My lecturer said something about 135 being a possible answer, and I was wondering why the question included "acute"; as in first quadrant?

**Grandad** · February 15th 2010, 01:08 AM

Hello RAz

Originally Posted by RAz

The question is

If tanA=4 and tabB=3/5, and A and B are acute angles, prove that A-B=pi/4 (pi in radians, so really it's 45degrees).

My answer is:

$\color{red}tan(x-y)=\frac {tan4-tan \frac {3}{5}}{1 + tan4 * tan \frac {3}{5}}\color{black}<br /> =1<br /> \therefore tan(x-y)=1\color{red}=\frac {\pi}{4}$

I have a feeling there is more to add. My lecturer said something about 135 being a possible answer, and I was wondering why the question included "acute"; as in first quadrant?

Your general method is OK, but your use of notation is very sloppy. Say what you mean! What you mean, of course, is:

$\tan(A-B)=\frac {\tan A-\tan B}{1 + \tan A \tan B} = \frac{4-\tfrac35}{1+4\cdot\tfrac35}= 1$

$\Rightarrow A-B = \frac{\pi}{4}$

I can't see where the possible angle of $135^o$ might come from. $\tan135^o = -1$ , not $+1$ . And if $A$ and $B$ are acute with $\tan A > \tan B$ , then $A>B$ . So $A-B$ is also acute. Hence, $\tan(A-B) = 1\Rightarrow A-B = 45^o$ .

Grandad

**Punch** · February 15th 2010, 05:24 AM

If your teacher meant $225^\circ$ , I can see the possibility but definitely not $135^\circ$

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