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Math Help - 2 Homework Problems

  1. #1
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    2 Homework Problems

    I'd like to thank Chris for the help on my last post. I just have two more problems I could use assistance on and I'll be all set.

    Find the values of all trigonometric functions of theta if tan (theta) = the square root of 7/3 and sec(theta) = 4/3 (Just the 7 is the square root)

    Let x=3sin(theta), -pi/2<x<pi/2. Simplify the expression: x/square foot of 9 + x squared. (whole bottom is under square root.) I have put 3sin(theta) in for the x and the x squared, is this the right step? I just cannot get the hang of this! Thanks in advance for the help!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ChargersFan84 View Post
    I'd like to thank Chris for the help on my last post. I just have two more problems I could use assistance on and I'll be all set.

    Find the values of all trigonometric functions of theta if tan (theta) = the square root of 7/3 and sec(theta) = 4/3 (Just the 7 is the square root)
    Constructing a triangle, we see that \text{opp}=\sqrt{7}, \text{adj}=3 and \text{hyp}=4

    You can now easily find the values of the 4 remaining trigonometric functions.

    Let x=3sin(theta), -pi/2<x<pi/2. Simplify the expression: x/square foot of 9 + x squared. (whole bottom is under square root.) I have put 3sin(theta) in for the x and the x squared, is this the right step? I just cannot get the hang of this! Thanks in advance for the help!
    Substituting this into the expression, we get \frac{3\sin\vartheta}{\sqrt{9-9\sin^2\vartheta}}=\frac{3\sin\vartheta}{3\sqrt{1-\sin^2\vartheta}}=\frac{\sin\vartheta}{\sqrt{\cos^  2\vartheta}}=\dots

    Can you take it from here?

    --Chris
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  3. #3
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    Thank you so much. I really appreciate it.
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  4. #4
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    Quote Originally Posted by ChargersFan84 View Post
    I'd like to thank Chris for the help on my last post. I just have two more problems I could use assistance on and I'll be all set.

    Find the values of all trigonometric functions of theta if tan (theta) = the square root of 7/3 and sec(theta) = 4/3 (Just the 7 is the square root)

    Let x=3sin(theta), -pi/2<x<pi/2. Simplify the expression: x/square foot of 9 + x squared. (whole bottom is under square root.) I have put 3sin(theta) in for the x and the x squared, is this the right step? I just cannot get the hang of this! Thanks in advance for the help!
    hey mate, we are given that,

    tan(A) = sqrt(7)/3 and sec(A) = 4/3

    okay, start with tan(A),
    tan(A) = sin(A)/cos(A) = sin(A)*(1/cos(A)) = sin(A) * sec(A)

    Therefore

    sqrt(7)/3 = sin(A) * (4/3)
    thus sin(A) = sqrt(7)/4
    which you then can solve using your calculator (there may be a Identity Triangle to gain a direct value of A - I'll have a play around and if I find anything I'll let you know,

    Hope this helps,

    David
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