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Math Help - Fiding the exact value on sum and difference identities?

  1. #1
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    Fiding the exact value on sum and difference identities?

    Can anyone help me solve these two problem as I am confused on how to them?

    1. Find the exact value
    cos(x+y) if tan x=5/3 and sin y= 1/3

    2.sec(x-y) if csc x=5/3 and tan y=5/12

    Can anyone show me how to solve these problems what I am having trouble understanding is what to do with the tan.

    For example on number 1 I know you can use sin y to find cos y using square root 1-sin^2 then use cos a cos b+ sin a sin b

    But I am confused can anyone show me how to work one of these problems out please?
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    Quote Originally Posted by homeylova223 View Post
    Can anyone help me solve these two problem as I am confused on how to them?

    1. Find the exact value
    cos(x+y) if tan x=5/3 and sin y= 1/3

    need more info ... quad I or III for angle x ? quad I or II for angle y ?

    2.sec(x-y) if csc x=5/3 and tan y=5/12

    same quadrant info for angles x and y needed here, also.
    ...
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    Both x and y use quadrant one
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    Keep in mind that this solution is for when it's in Quadrant I, else you will have different signs for the trigonometric ratios.

    For #1, we have the identity: cos(x+y) = cos x * cosy - sin x * sin y
    so all we have to find are these individual components.

    For angle x, we are given tan(x) = 5/3. Draw out the triangle. tan x = O/A (opposite over adjacent)


    Find ? using Pythagoras. we then know that cos(x) = A/H (adjacent over hypotenuse). ie. 3/?
    and that sin(x) = O/H (opposite over hypotenuse) ie. 5/?

    Repeat this process for the other piece of information you have for sin y = 1/3, a diagram with y as your angle and given side length embedded in the ratio. hope that helps.
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    \cos(x+y) = \cos{x}\cos{y} - \sin{x}\sin{y}

    sketch a couple of reference triangles in quad I for angles x and y

    \tan{x} = \dfrac{5}{3} = \dfrac{opp}{adj}

    \cos{x} = \dfrac{3}{\sqrt{34}} , \sin{x} =  \dfrac{5}{\sqrt{34}}<br />

    \sin{y} = \dfrac{1}{3} = \dfrac{opp}{hyp}<br />

    \cos{y} = \dfrac{\sqrt{8}}{3}

    sub the necessary values into the sum identity and evaluate.
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    I have a quick question I did cos(x+y)= cos (3/square root 34) cos (square root 8/3)- sin (5/square root 34) sin (1/3)
    I end up getting .999 but the answer is 12 square root 17- 5 square root 34 / 102. I am not sure what I did to get wrong answer.
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    Quote Originally Posted by homeylova223 View Post
    I have a quick question I did cos(x+y)= cos (3/square root 34) cos (square root 8/3) - sin (5/square root 34) sin (1/3) no, no, - no, no
    it's \cos{x} = \dfrac{3}{\sqrt{34}} , not \cos\left( \dfrac{3}{\sqrt{34}}\right)


    \cos(x+y) = \dfrac{3}{\sqrt{34}} \cdot \dfrac{\sqrt{8}}{3} - \dfrac{5}{\sqrt{34}} \cdot \dfrac{1}{3}
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    I have one more question? In my second problem I have sec(x-y) if csc x=5/3 and tan y=5/12

    So figuring out csc=5/3 tany=12/5 secy=13/5 cscy= 13/12 secx=5/4

    I did (5/4)(13/5)+(5/3)(13/12) and I got 5.23 but the correct answer is 65/56 can anyone help me figure out what I did wrong?
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    Quote Originally Posted by homeylova223 View Post
    I have one more question? In my second problem I have sec(x-y) if csc x=5/3 and tan y=5/12

    So figuring out csc=5/3 tany=12/5 secy=13/5 cscy= 13/12 secx=5/4

    I did (5/4)(13/5)+(5/3)(13/12) and I got 5.23 but the correct answer is 65/56 can anyone help me figure out what I did wrong?
    \sec(x-y) = \dfrac{1}{\cos(x-y)} = \dfrac{1}{\cos{x}\cos{y} + \sin{x}\sin{y}}

    \csc{x} = \dfrac{5}{3} , \sin{x} = \dfrac{3}{5} , \cos{x} = \dfrac{4}{5}

    \csc{y} = \dfrac{13}{12} , \sin{x} = \dfrac{12}{13} , \cos{x} = \dfrac{5}{13}
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