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Math Help - angles and trigonometic functions

  1. #1
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    angles and trigonometic functions

    Find the exact value of sin 30 - sin 45 cos 45 i dnt unnderstand how u find the value
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    Quote Originally Posted by missjky View Post
    Find the exact value of sin 30 - sin 45 cos 45 i dnt unnderstand how u find the value
    You have to use your knowledge of trig exact values and your use of trig identites

    \sin(\psi)\cos(\psi)=\frac{1}{2}\sin(2\psi)

    \therefore\sin(45)\cos(45)=\frac{1}{2}\sin(90)=\fr  ac{1}{2}

    and it is commonly known that \sin(30)=\frac{1}{2}

    So we have

    \frac{1}{2}-\frac{1}{2}=0
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  3. #3
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    Quote Originally Posted by missjky View Post
    Find the exact value of sin 30 - sin 45 cos 45 i dnt unnderstand how u find the value
    In a 30-60-90 deg. right triangle, sin 30 = 1/2

    In a 45-45-90 deg. right triangle, sin 45 = cos 45 = \frac{\sqrt2}{2}

    \sin 30-\sin 45 \cos45 = \frac{1}{2}-(\frac{\sqrt2}{2})(\frac{\sqrt2}{2})=\frac{1}{2}-\frac{2}{4}=0
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    [quote=masters;155321]

    thanks so much and also the exact value of cos
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    MHF Contributor Mathstud28's Avatar
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    [quote=missjky;155322]
    Quote Originally Posted by Mathstud28 View Post
    thanks oso much i understand and also
    the exact value of cos
    \cos\bigg(\frac{\pi}{6}\bigg)=\frac{\sqrt{3}}{2}
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    [quote=Mathstud28;155325]
    Quote Originally Posted by missjky View Post
    \cos\bigg(\frac{\pi}{6}\bigg)=\frac{\sqrt{3}}{2}

    If θ is an acute angle in standard position and cot θ = 12/5 how do u find sin θ
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  7. #7
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    \cot \theta = \frac{12}{5}

    I don't know what is meant by 'standard position' but I'm assuming that the angle is restricted to Quadrant I.

    If you recall the definition that \cot \theta = \frac{\text{adjacent}}{\text{opposite}}, then you can draw a right angled triangle with a base of 12 and a height of 5 with \theta as your angle.

    Recall that \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}. We already have the 'opposite' side, i.e. 5. We just need the hypotenuse which you can easily calculate with Pythagoras' Theorem.
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    Quote Originally Posted by o_O View Post
    \cot \theta = \frac{12}{5}

    I don't know what is meant by 'standard position' but I'm assuming that the angle is restricted to Quadrant I.

    If you recall the definition that \cot \theta = \frac{\text{adjacent}}{\text{opposite}}, then you can draw a right angled triangle with a base of 12 and a height of 5 with \theta as your angle.

    Recall that \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}. We already have the 'opposite' side, i.e. 5. We just need the hypotenuse which you can easily calculate with Pythagoras' Theorem.
    i dnt no how to do that this is my first time taking trig plus its online
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    Quote Originally Posted by missjky View Post
    i dnt no how to do that this is my first time taking trig plus its online
    the asnswr is either
    12/13
    5/13
    13/12
    13/5
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  10. #10
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    Do you know what \cot \theta and \sin \theta stand for in terms of a triangle? You should be able to figure it out.

    Here's a diagram to help illustrate what I'm trying to say. Solve for h and you should be able to find \sin \theta
    Attached Thumbnails Attached Thumbnails angles and trigonometic functions-cot.jpg  
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  11. #11
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    Quote Originally Posted by o_O View Post
    Do you know what \cot \theta and \sin \theta stand for in terms of a triangle? You should be able to figure it out.

    Here's a diagram to help illustrate what I'm trying to say. Solve for h and you should be able to find \sin \theta
    would it b 5/13
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  12. #12
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