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Math Help - θ is Acute?

  1. #1
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    θ is Acute?

    Can anybody help me with this following question, I never heard of it before...until now:

    If tan θ = x where θ is acute, find in the terms of x:

    a) 2 sin θ.
    b) 3 cos θ.

    What does θ acute in this case represent?
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  2. #2
    MHF Contributor red_dog's Avatar
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    If \theta is acute then

    \sin\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta}  }

    \cos\theta=\frac{1}{\sqrt{1+\tan^2\theta}}
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  3. #3
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    Quote Originally Posted by red_dog View Post
    If \theta is acute then

    \sin\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta}  }

    \cos\theta=\frac{1}{\sqrt{1+\tan^2\theta}}
    Ah thank you for your help!

    Do you know where one could get best reference about this particular topic?
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  4. #4
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    Quote Originally Posted by Lites View Post
    Can anybody help me with this following question, I never heard of it before...until now:

    If tan θ = x where θ is acute, find in the terms of x:

    a) 2 sin θ.
    b) 3 cos θ.

    What does θ acute in this case represent?
    In terms of x...

    if tan\theta=x

    tan\theta=\frac{x}{1}=\frac{opposite}{adjacent} in a right-angled triangle.

    Using Pythagoras' theorem

    hypotenuse=\sqrt{adj^2+opp^2}=\sqrt{1+x^2}

    2sin\theta=2\frac{opp}{hyp}=\frac{2x}{\sqrt{1+x^2}  }

    3cos\theta=3\frac{adj}{hyp}=\frac{3}{\sqrt{1+x^2}}
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    In terms of x...

    if tan\theta=x

    tan\theta=\frac{x}{1}=\frac{opposite}{adjacent} in a right-angled triangle.

    Using Pythagoras' theorem

    hypotenuse=\sqrt{adj^2+opp^2}=\sqrt{1+x^2}

    2sin\theta=2\frac{opp}{hyp}=\frac{2x}{\sqrt{1+x^2}  }

    3cos\theta=3\frac{adj}{hyp}=\frac{3}{\sqrt{1+x^2}}
    Most helpful! Thank you for your kind explanation!
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