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Math Help - Reference Angles?

  1. #1
    Junior Member
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    Question Reference Angles?

    Find the measurement of the reference angle of an angle measuring 5 pi divided by 4.
    Answears
    a. pi/2
    b.-pi/2
    c.pi/4
    d. - pi/4

    step by step explanation is 100% wanted and I will give thanks!
    & just so I don't make another board. ..
    Write cos 6x cos x - sin 6x sin x in terms of a single.. trigonometric function.. i only got to cos 7x - sin 7x
    and write 2 cos(squared) 2 theta - 1 in terms of a single trigonometric function.
    Use the half angle identity to evaluate tan 67.5degrees.
    if i do tan(67.5dividedby 2) = sin67.5divided by 1+ cos67.5 how do i get
    a.2squareroot(2)-2
    b.squareroot(2) +1
    c.squareroot(2) - 1
    d. 2 - squareroot(2)
    Find the exact value of cos 2 theta given that sin theta = - 3 divided by 5 and theta is in quadrant III,.
    and last one i promiss after i get help with these i will try to help others to warm me up for my finals!
    a. - 7/25
    b. 7/25
    c. 17/25
    d. -17/25
    please don't just give the answear but explain things! THAnKS!!
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Trigonometry

    Hello j0nath0n3
    Quote Originally Posted by j0nath0n3 View Post
    Find the measurement of the reference angle of an angle measuring 5 pi divided by 4.
    Reference angle: look here Reference Angle: How to find the reference angle as a positive acute angle
    Quote Originally Posted by j0nath0n3 View Post
    Write cos 6x cos x - sin 6x sin x in terms of a single.. trigonometric function
    Use \cos A \cos B - \sin A \sin B = \cos (A+B) and the answer is simply \cos 7x
    Quote Originally Posted by j0nath0n3 View Post
    write 2 cos(squared) 2 theta - 1 in terms of a single trigonometric function.
    \cos 2A = 2\cos^2 A -1

    So \cos 4A = 2\cos^2 2A - 1

    Replace A by \theta in this second identity.

    Quote Originally Posted by j0nath0n3 View Post
    Use the half angle identity to evaluate tan 67.5degrees.
    67.5 \times 2 = 135 and 135 = 180 - 45

    So work from \tan 135 = -\tan 45 =-1, using \tan \theta = \frac{2t}{1-t^2} and solve for t.

    Quote Originally Posted by j0nath0n3 View Post
    Find the exact value of cos 2 theta given that sin theta = - 3 divided by 5 and theta is in quadrant III
    Use \cos 2\theta = 1-2\sin^2\theta and just plug the number -\frac{3}{5} in.

    Grandad
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  3. #3
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    Exclamation

    where do i plug -3/5 in? and i dont get how you do multiply 67.5 ?? Thanks for all the others though!
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  4. #4
    MHF Contributor
    Grandad's Avatar
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    Trigonometry

    Hello j0nath0n3
    Originally Posted by j0nath0n3
    Use the half angle identity to evaluate tan 67.5degrees.
    and

    So work from , using and solve for .
    Let \tan 67.5^o = t.

    Then \tan 135^o = \tan (2 \times 67.5^o) = \frac{2t}{1-t^2} = -1.

    \Rightarrow 2t = -1(1-t^2) = t^2 - 1

    \Rightarrow t^2 -2t - 1 = 0

    Factorise and solve for t, bearing in mind that \tan 67.5^o > 0

    Originally Posted by j0nath0n3
    Find the exact value of cos 2 theta given that sin theta = - 3 divided by 5 and theta is in quadrant III
    Use and just plug the number in.
    \sin\theta = -\frac{3}{5}

    \Rightarrow \cos 2\theta = 1 - 2\sin^2\theta

    = 1 - 2 \times \left(-\frac{3}{5}\right)^2

    = 1 - 2\times \frac{9}{25}

    = \frac{7}{25}

    Grandad
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