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Math Help - more identities.(in a hurry)

  1. #1
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    more identities.(in a hurry)

    need the work already have answers.

    1)what is this equivalent to cos(180degrees + B) answer:-cosB


    2)use an identity to find an exact value for sin 105degrees.

    answer:root6+root2/4

    3)use an identity to find an exact value for 22.5degrees.

    answer:root2-1
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  2. #2
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    Quote Originally Posted by Dave19 View Post
    need the work already have answers.

    1)what is this equivalent to cos(180degrees + B) answer:-cosB
    \cos (180^{\circ} + B) = \cos 180^{\circ} \cos B - \sin 180^{\circ} \sin B

    But \cos 180^{\circ} = -1 and \sin 180^{\circ} = 0

    \cos (180^{\circ} + B) = \cos 180^{\circ} \cos B - \sin 180^{\circ} \sin B = (-1) \cos B = -\cos B<br />

    2)use an identity to find an exact value for sin 105degrees.

    answer:root6+root2/4

    \sin 105^{\circ} = \sin (90^{\circ} + 15^{\circ}) = \cos 15^{\circ}

    Use the double angle formula: \cos 2x = 2cos^2x - 1

    put x = 15^{\circ}, since you already know \cos 30^{\circ} = \frac{\sqrt{3}}2.

    3)use an identity to find an exact value for 22.5degrees.

    answer:root2-1
    DO you mean sin of 22.5 degrees or cosine of 22.5 degrees? Whoat does the phrase "exact value of 22.5 degrees mean"?
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  3. #3
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    sorry its tan and can you try my other questions if you can find them.
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  4. #4
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    its under urgent homework help i think
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  5. #5
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    Quote Originally Posted by Dave19 View Post
    sorry its tan and can you try my other questions if you can find them.
    \tan 2x = \frac{2\tan x}{1 - \tan^2 x} and \tan 45 = 1

    Now if x = 22.5, 2x = 45. So

    \tan 2x = \tan 45 = 1 = \frac{2\tan x}{1 - \tan^2 x}

    1 = \frac{2\tan x}{1 - \tan^2 x}

    Now solve the above equation for tan x. It is a quadratic equation which you can easily solve.
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  6. #6
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    i dont no how that will equal root2-1
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  7. #7
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    Quote Originally Posted by Isomorphism View Post
    \tan 2x = \frac{2\tan x}{1 - \tan^2 x} and \tan 45 = 1

    Now if x = 22.5, 2x = 45. So

    \tan 2x = \tan 45 = 1 = \frac{2\tan x}{1 - \tan^2 x}

    1 = \frac{2\tan x}{1 - \tan^2 x}

    Now solve the above equation for tan x. It is a quadratic equation which you can easily solve.
    1 = \frac{2\tan x}{1 - \tan^2 x} \Rightarrow 1 - \tan^2 x = 2\tan x

    If it is hard for you to see, put u =\tan x,

    1 - \tan^2 x = 2\tan x \Rightarrow 1 - u^2 = 2u \Rightarrow u^2 +2u -1 = 0

    Now try solving it. Come on show us some effort... You want to learn, dont you?

    Ask me if you dont know...
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