Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

Find the exact values of sin(u/2), cos (u/2), tan (u/2) using the half angle formulas?

cos u= 3/5, 0<u<pi/2

3 Answers

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  • Best Answer

    sin(u / 2) = sqrt( (1 - cos(u)) / 2)

    cos(u / 2) = sqrt( (1 + cos(u)) / 2)

    tan(u / 2) = sin(u/2) / cos(u/2) = sqrt( (1 - cos(u)) / (1 + cos(u)) )

    cos(u) = 3/5, and we're in Q1, which means that sin(u/2), cos(u/2) and tan(u/2) will be positive:

    sqrt( (1 + 3/5) / 2) =

    sqrt( (8/5) / 2) =

    sqrt( 4/5 ) =

    2 / sqrt(5) =

    2 * sqrt(5) / 5

    sin(u/2) = 2 * sqrt(5) / 5

    sqrt( (1 - 3/5) / 2) =

    sqrt( (2/5) / 2 ) =

    sqrt(1/5) =

    sqrt(5) / 5

    cos(u/2) = sqrt(5) / 5

    2 * sqrt(5) / 5 / (sqrt(5) / 5) =

    2

    tan(u/2) = 2

  • 9 years ago

    It's important to note that 0<u/2<pi/4, so u/2 is in the first quadrant, so all of its trig functions are positive.

    Now, draw u in standard position, with 3=x and 5=hypotenuse of a triangle, and find y=4 by the Pythagorean Theorem (noting y>0 from the picture). Actually this is not necessary here because the half-angle formulas can all use cosine of u, which you already have.

    sin(u/2)=sqrt((1-cos u)/2)

    =sqrt((2/5)/2)

    =sqrt(1/5)

    =1/sqrt(5).

    cos(u/2)=sqrt((1+cos u)/2)

    =sqrt((8/5)/2)

    =sqrt(4/5)

    =2/sqrt(5).

    The tangent is the ratio of these, or tan(u/2)=1/2, but you can also get it from

    tan(u/2)=(1-cos u)/(sin u)

    =(1-3/5)/(4/5)

    =(2/5)/(4/5)

    =2/4

    =1/2.

    I used my triangle drawing there, so if the question asked only about the tangent, I would have drawn the triangle, but since it asked about sine and cosine of u/2 first, tan(u/2) would follow.

    --MD

  • 3 years ago

    cos(u)=3/5 as a results of fact: cos^2(x)=(a million/2)+(a million/2)cos(2x) enable x=u/2, leaving: cos^2(u/2)=(a million/2)+(a million/2)cos(u) and because you comprehend that cos(u)=3/5, you have: cos^2(u/2)=(a million/2)+(a million/2)(3/5) think of you have the rest from right here?

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