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  1. #1
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    trig

    given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90-Θ)

    thank you
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by tangerine_tomato View Post
    given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90-Θ)

    thank you
    You can observe that \csc(90^o-\vartheta)=\sec(\vartheta). But if you didn't know that, how would you go about solving this problem?

    Note 3 things:

    \sec(\varphi)=\frac{1}{\cos(\varphi)}
    \csc(\varphi)=\frac{1}{\sin(\varphi)} and \sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)

    Thus,

    \csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}

    Simplifying, we get:

    \csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet  a)

    Since \tan(\vartheta)=2, we can conclude that \cos(\vartheta)=\frac{1}{\sqrt{5}}

    Thus, \csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt  {5}}

    Does this make sense?

    --Chris
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You can observe that \csc(90^o-\vartheta)=\sec(\vartheta). But if you didn't know that, how would you go about solving this problem?

    Note 3 things:

    \sec(\varphi)=\frac{1}{\cos(\varphi)}
    \csc(\varphi)=\frac{1}{\sin(\varphi)} and \sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)

    Thus,

    \csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}

    Simplifying, we get:

    \csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet  a)

    Since \tan(\vartheta)=2, we can conclude that \cos(\vartheta)=\frac{1}{\sqrt{5}}

    Thus, \csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt  {5}}

    Does this make sense?

    --Chris
    Two "var"s in one post...you..you've made me proud
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    You can observe that \csc(90^o-\vartheta)=\sec(\vartheta). But if you didn't know that, how would you go about solving this problem?

    Note 3 things:

    \sec(\varphi)=\frac{1}{\cos(\varphi)}
    \csc(\varphi)=\frac{1}{\sin(\varphi)} and \sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)

    Thus,

    \csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}

    Simplifying, we get:

    \csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet  a)

    Since \tan(\vartheta)=2, we can conclude that \cos(\vartheta)=\frac{1}{\sqrt{5}}

    Thus, \csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt  {5}}

    Does this make sense?

    --Chris
    i dont understand that stuff in the 3rd line after the word thus

    is that cause of the complementary angle theorem

    and i dont understand anything after that

    but im too tired ATM to make much effort to understand it
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  5. #5
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    Quote Originally Posted by tangerine_tomato View Post
    i dont understand that stuff in the 3rd line after the word thus

    is that cause of the complementary angle theorem

    and i dont understand anything after that

    but im too tired ATM to make much effort to understand it
    It all boils down to finding \cos \theta from \tan \theta = 2.

    Half the solution is given above. The other half is {\color{red}-}\sqrt{5}.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mr fantastic View Post
    It all boils down to finding \cos \theta from \tan \theta = 2.

    Half the solution is given above. The other half is {\color{red}-}\sqrt{5}.
    Shoot...darn negative sign...

    ...wait...how would we know for sure it was in the first quadrant instead of the third quadrant or vice versa? I didn't see any restriction on the angle...

    EDIT: Just realized what he meant...never mind...
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