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Math Help - Exact Values with the different trig functions

  1. #1
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    Exclamation Exact Values with the different trig functions

    okay so im having problems with this one specific math problem:

    sin 30 cos 60 + sin 60 cos 30

    I know that sin = y/r and that cos= x/r but what exactly does radius (r) equal??? doesnt it equal 1?

    its a simple question, but its bugging me because i can't seem to get the right answer, which is 1.


    ANY HELP??? thanks!!!!
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  2. #2
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    Quote Originally Posted by hazel31304trighelp View Post
    okay so im having problems with this one specific math problem:

    sin 30 cos 60 + sin 60 cos 30

    I know that sin = y/r and that cos= x/r but what exactly does radius (r) equal??? doesnt it equal 1?

    its a simple question, but its bugging me because i can't seem to get the right answer, which is 1.


    ANY HELP??? thanks!!!!
    here's a big hint for you ...

    \textcolor{red}{\sin(a)\cos(b) + \sin(b)\cos(a) = \sin(a+b)}<br />
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  3. #3
    MHF Contributor matheagle's Avatar
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    You can just plug in these values into any basic calculator.
    I use a $10 TI-30, which suits all my needs.
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  4. #4
    Ife
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    Quote Originally Posted by hazel31304trighelp View Post
    okay so im having problems with this one specific math problem:

    sin 30 cos 60 + sin 60 cos 30

    I know that sin = y/r and that cos= x/r but what exactly does radius (r) equal??? doesnt it equal 1?

    its a simple question, but its bugging me because i can't seem to get the right answer, which is 1.


    ANY HELP??? thanks!!!!
    I will try to answer this as best as possible. Think of an equilateral triangle, with length of side equal 1. If this was bisected, we would get 2 right triangles, with hypothenuse of 1, and with interior angles 60, and 30. Let this be the basis of the calculation of the sine and cos exact values in a unit circle. We know that a unit circle has a radius of 1. So if one of those right triangles was at the centre of a unit circle, the hypothenuse would be equal to the radius, and would have a length of 1. Sketch it out on paper. At 60 degrees, the length of the adjacent leg of the triangle would be 1/2. (remember we bisected the the triangle of side = 1, so the length of side is now 1/2. The third side (opposite the angle) would be root 3/2. Use pythagoras' theorem to verify. 1 sq = 1/2 sq + side sq. Therefore: 1-1/4 = side sq, = 3/4, therefore root of 3 / 2 is the length of the 3 rd arm. Since Sin x = y/r, when x is 60, sin 60 = (root 3 / 2 ) /1. cos 60 = (1/2)/1. And for the 30 degrees angle, just rearrange ur triangle, so the y value would be 1/2, and the x value would be root 3/2. If you need more help, feel free to ask. I hope you follow
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  5. #5
    MHF Contributor matheagle's Avatar
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    Using...

    Quote Originally Posted by skeeter View Post
    here's a big hint for you ...

    \textcolor{red}{\sin(a)\cos(b) + \sin(b)\cos(a) = \sin(a+b)}<br />
    You should get \sin(90)=1 assuming you are talking degrees here.
    Or just use your calculator, but make sure it's set to degrees and not radians.
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