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Thread: tan[x - (pi/4)]

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

    If cos(x) = 1/4, x in quadrant 4, find the exact value of
    tan[x - (pi/4)]
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  2. #2
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    Hello,
    Quote Originally Posted by magentarita View Post
    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 :

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

    ...
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  3. #3
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    Smile Fantastic!

    Quote Originally Posted by Moo View Post
    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!
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