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UnbelievableDreams

  • one year ago

Confusing math question

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  1. anonymous
    • one year ago
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    ill help you

  2. UnbelievableDreams
    • one year ago
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    Choose the correct expression that completes the identity. sec^2 (x) - tan^2(x)

  3. anonymous
    • one year ago
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    im sorry this is above my math intelligence

  4. Nnesha
    • one year ago
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    sec^2 = what identity ?

  5. Nnesha
    • one year ago
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    do you have sin ,cos identities notes ?

  6. dan815
    • one year ago
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    what are the choices

  7. dan815
    • one year ago
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    |dw:1438637947361:dw|

  8. dan815
    • one year ago
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    |dw:1438637971118:dw|

  9. UnbelievableDreams
    • one year ago
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    No, I don't have notes. I am so confused.

  10. Nnesha
    • one year ago
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    now u got the answer if you apply identities that would be easy \[\huge\rm tan^2 +1=\sec^2\]

  11. Nnesha
    • one year ago
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    replace sec^2 by tan^2 +1 \[\huge\rm \color{reD}{sec^2(x)}-\tan^2(x)\] \[\huge\rm \color{reD}{tan^2+1 }-\tan^2\]

  12. dan815
    • one year ago
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    u need to know only the basic ones sin^2(x)+cos^2(x)=1 and sec(x)=1/cos(x) csc(x)=1/sin(x) cot(x)=1/tan(x) there are the basic ones to know

  13. UnbelievableDreams
    • one year ago
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    I am still confused. :(

  14. freckles
    • one year ago
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    do you know the Pythagorean identity(ies)?

  15. freckles
    • one year ago
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    \[\sin^2(x)+\cos^2(x)=1 \text{ do you know the one mentioned by dan ?}\]

  16. UnbelievableDreams
    • one year ago
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    No.

  17. freckles
    • one year ago
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    Do you know how to graph the equation x^2+y^2=1? This is the circle with center (0,0) and radius 1. Looks like this: |dw:1438638836144:dw| say we pick a random point (a,b) on the circle: |dw:1438638871637:dw| now we draw a right triangle |dw:1438638917068:dw| we know the relationship between the side of those right triangle to be \[a^2+b^2=1^2 \\ \text{ or } \\a^2+b^2=1 \\ \text{ the 1 comes from the radius of the circle being 1 }\] now we could write this same equation in terms of theta by using the drawing

  18. freckles
    • one year ago
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    \[\cos(\theta)=\frac{a}{1} \text{ or } \cos(\theta)=a \\ \sin(\theta)=\frac{b}{1} \text{ or } \sin(\theta)=b \\ \text{ squaring both sides gives } \\ (\cos(\theta))^2=a^2 \text{ also written as } \cos^2(\theta)=a^2 \\ (\sin(\theta))^2=b^2 \text{ also written as } \sin^2(\theta)=b^2 \\ \text{ so replacing the items \in } a^2+b^2=1 \\ \text{ we have } \cos^2(\theta)+\sin^2(\theta)=1 \]

  19. freckles
    • one year ago
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    you can get the identity that nnesha mentioned from this if you divide both sides by cos^2(theta)

  20. freckles
    • one year ago
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    \[(\cos(\theta))^2+(\sin(\theta))^2=1 \\ \text{ dividing both sides by } (\cos(\theta))^2 \\ \frac{(\cos(\theta))^2+(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1}{(\cos(\theta))^2} \\ \text{ separating fraction on \left hand side } \\ \frac{(\cos(\theta))^2}{(\cos(\theta))^2}+\frac{(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1^2}{(\cos(\theta))^2} \text{ note: notice I replace } 1 \text{ with } 1^2 \\ \text{ now using law of exponents } \\ (\frac{\cos(\theta)}{\cos(\theta)})^2+(\frac{\sin(\theta)}{\cos(\theta)})^2=(\frac{1}{\cos(\theta)})^2 \\ \text{ now there were ratios mentioned by dan that were equivalent \to some you see here }\] for example you know \[\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta) \\ \text{ and } \frac{1}{\cos(\theta)}=\sec(\theta)\]

  21. freckles
    • one year ago
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    \[\text{ also you know } \frac{\cos(\theta)}{\cos(\theta)}=1 \text{ on it's domain } \\ \text{ so we have } 1^2+(\tan(\theta))^2=(\sec(\theta))^2 \\\ \text{ which can be written as } \\ 1+\tan^2(\theta)=\sec^2(\theta)\]

  22. freckles
    • one year ago
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    this is the Pythagorean identity you want to apply

  23. UnbelievableDreams
    • one year ago
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    Ah, I see.

  24. UnbelievableDreams
    • one year ago
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    Can you help me another one?

  25. freckles
    • one year ago
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    I can try

  26. UnbelievableDreams
    • one year ago
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    It is similar to this question. But I like to see step by step. 1-tan^2x= ___________

  27. freckles
    • one year ago
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    do you want to write it in terms of sec or use difference of squares

  28. freckles
    • one year ago
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    \[a^2-b^2=(a-b)(a+b) \text{ is difference of squares in factored form }\]

  29. freckles
    • one year ago
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    if you want to write it in terms of sec you can use the identity we just mentioned \[1+\tan^2(\theta)=\sec^2(\theta) \\ \text{ subtract } 1 \text{ on both sides } \\ \tan^2(\theta)=\sec^2(\theta)-1 \\ \text {replace } \tan^2(\theta) \text{ with } \sec^2(\theta)-1 \]

  30. UnbelievableDreams
    • one year ago
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    The answer is 1/ sec^2x

  31. UnbelievableDreams
    • one year ago
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    How do you get sec?

  32. freckles
    • one year ago
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    did you replace tan^2(theta) with sec^2(theta)-1 ?

  33. freckles
    • one year ago
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    \[1-\tan^2(\theta) \\ \text{ replace } \tan^2(\theta) \text{ with } \sec^2(\theta)-1 \\ 1-(\sec^2(\theta)-1))\]

  34. freckles
    • one year ago
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    distribute and combine like terms but this shouldn't give you 1/sec^2(x)

  35. UnbelievableDreams
    • one year ago
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    Sec^2x - 1 ?

  36. freckles
    • one year ago
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    nope you distribute \[1-\tan^2(x) \\ \text{ replace } \tan^2(x) \text{ with } \sec^2(x)-1 \\ \text{ which gives } \\ 1-(\sec^2(x)-1) \\ \text{ now distribute } \\1-\sec^2(x)+1 \\ \text{ combine like terms } (1+1)-\sec^2(x)\]

  37. UnbelievableDreams
    • one year ago
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    Ah, I see.

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