1. ## tan[x - (pi/4)]

If cos(x) = 1/4, x in quadrant 4, find the exact value of
tan[x - (pi/4)]

2. Hello,
Originally Posted by magentarita
If cos(x) = 1/4, x in quadrant 4, find the exact value of
tan[x - (pi/4)]
If x is in quadrant 4, then sin(x)<0.
We know that $\displaystyle \sin^2(x)+\cos^2(x)=1$.
Hence $\displaystyle \sin^2(x)=1-\cos^2(x)=1-\tfrac 1{16}=\tfrac{15}{16}$.

$\displaystyle \implies \sin(x)=\pm \sqrt{\tfrac{15}{16}}$
Since the sine is negative, $\displaystyle \sin(x)=-\sqrt{\tfrac{15}{16}}=-\tfrac{\sqrt{15}}4$

---------------------------------
Method #1
$\displaystyle \tan(x-\tfrac \pi 4)=\frac{\sin(x-\tfrac \pi 4)}{\cos(x-\tfrac \pi 4)}$

Difference formula for cosine and sine (using $\displaystyle \sin(\tfrac \pi 4)=\cos(\tfrac \pi 4)=\tfrac{\sqrt{2}}{2}$) :

$\displaystyle \cos(x-\tfrac \pi 4)=\cos(x)\cos(\tfrac \pi 4)+\sin(x)\sin(\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (\cos(x)+\sin(x))=\tfrac{\sqrt{2}}{2} (\tfrac 14+\sin(x))$

$\displaystyle \sin(x-\tfrac \pi 4)=\sin(x)\cos(\tfrac \pi 4)-\cos(x)\sin(\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (\sin(x)-\cos(x))=\tfrac{\sqrt{2}}{2} (\sin(x)-\tfrac 14)$

Thus :
$\displaystyle \cos(x-\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (-\tfrac{\sqrt{15}}4+\tfrac 14)=\tfrac{\sqrt{2}}{8} (-\sqrt{15}+1)$

and $\displaystyle \sin(x-\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (-\tfrac{\sqrt{15}}4-\tfrac 14)=\tfrac{\sqrt{2}}{8} (-\sqrt{15}-1)$

$\displaystyle \implies \tan(x-\tfrac \pi 4)=\frac{{\color{red}\tfrac{\sqrt{2}}{8}} (-\sqrt{15}-1)}{{\color{red}\tfrac{\sqrt{2}}{8}} (-\sqrt{15}+1)}=\frac{\sqrt{15}+1}{\sqrt{15}-1}$

-------------------------------------

Method #2

Using difference formula for tangent :

$\displaystyle \tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}$

...

3. ## Fantastic!

Originally Posted by Moo
Hello,

If x is in quadrant 4, then sin(x)<0.
We know that $\displaystyle \sin^2(x)+\cos^2(x)=1$.
Hence $\displaystyle \sin^2(x)=1-\cos^2(x)=1-\tfrac 1{16}=\tfrac{15}{16}$.

$\displaystyle \implies \sin(x)=\pm \sqrt{\tfrac{15}{16}}$
Since the sine is negative, $\displaystyle \sin(x)=-\sqrt{\tfrac{15}{16}}=-\tfrac{\sqrt{15}}4$

---------------------------------
Method #1
$\displaystyle \tan(x-\tfrac \pi 4)=\frac{\sin(x-\tfrac \pi 4)}{\cos(x-\tfrac \pi 4)}$

Difference formula for cosine and sine (using $\displaystyle \sin(\tfrac \pi 4)=\cos(\tfrac \pi 4)=\tfrac{\sqrt{2}}{2}$) :

$\displaystyle \cos(x-\tfrac \pi 4)=\cos(x)\cos(\tfrac \pi 4)+\sin(x)\sin(\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (\cos(x)+\sin(x))=\tfrac{\sqrt{2}}{2} (\tfrac 14+\sin(x))$

$\displaystyle \sin(x-\tfrac \pi 4)=\sin(x)\cos(\tfrac \pi 4)-\cos(x)\sin(\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (\sin(x)-\cos(x))=\tfrac{\sqrt{2}}{2} (\sin(x)-\tfrac 14)$

Thus :
$\displaystyle \cos(x-\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (-\tfrac{\sqrt{15}}4+\tfrac 14)=\tfrac{\sqrt{2}}{8} (-\sqrt{15}+1)$

and $\displaystyle \sin(x-\tfrac \pi 4)=\tfrac{\sqrt{2}}{2} (-\tfrac{\sqrt{15}}4-\tfrac 14)=\tfrac{\sqrt{2}}{8} (-\sqrt{15}-1)$

$\displaystyle \implies \tan(x-\tfrac \pi 4)=\frac{{\color{red}\tfrac{\sqrt{2}}{8}} (-\sqrt{15}-1)}{{\color{red}\tfrac{\sqrt{2}}{8}} (-\sqrt{15}+1)}=\frac{\sqrt{15}+1}{\sqrt{15}-1}$

-------------------------------------

Method #2

Using difference formula for tangent :

$\displaystyle \tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}$

...
Beautiful math work!