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Math Help - cos35sin55 + cos55sin35

  1. #1
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    cos35sin55 + cos55sin35

    ok the answer 1 apparently because i went on hotmath.com and it explained it like this

    cos35sin55 + cos55sin35

    sin^2 55 + cos ^2 35 =1

    ok i understand that

    but cos35 and sin 55 are equal and so are cos55 and sin35

    so i thought it would be like...1 + 1...i dont understand why that isn't correct.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by desperate_on_sunday_night View Post
    ok the answer 1 apparently because i went on hotmath.com and it explained it like this

    cos35sin55 + cos55sin35

    sin^2 55 + cos ^2 35 =1

    ok i understand that

    but cos35 and sin 55 are equal and so are cos55 and sin35

    so i thought it would be like...1 + 1...i dont understand why that isn't correct.
    huh?

    the original expression is the addition formula for sine:

    cos35sin55 + cos55sin35 = sin(55 + 35) = sin90 = 1


    neither sin^2 (55) nor cos^2 (35) are 1

    any questions?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by desperate_on_sunday_night View Post
    ok the answer 1 apparently because i went on hotmath.com and it explained it like this

    cos35sin55 + cos55sin35

    sin^2 55 + cos ^2 35 =1

    ok i understand that

    but cos35 and sin 55 are equal and so are cos55 and sin35

    so i thought it would be like...1 + 1...i dont understand why that isn't correct.
    I think what the poster meant was

    \cos(55)=\sin(90-55)=\sin(35)

    and \sin(55)=\cos(90-55)=\cos(35)

    Giving us

    \sin^2(55)+\cos^2(55)=1
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  4. #4
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    Hello, desperate_on_sunday_night!

    ok the answer 1 apparently because i went on hotmath.com and it explained it like this

    \cos35\sin55 + \cos55\sin35

    \sin^2\!55 + \cos^2\!35 \:=\:1

    ok i understand that . . . . you do?
    If that's all they wrote, very sloppy!!

    We're expected to visualize this right triangle . . .
    Code:
                            *
                         *  *
                  c   * 55 *
                   *        * a
                *           *
             * 35          *
          *  *  *  *  *  *  *
                    b

    \begin{array}{cccccc}\text{We see that:} &\cos35 & = & \frac{b}{c} &=& \sin 55^o \\<br />
\text{and that:} & \sin35 &=& \frac{a}{c} &=& \cos55^o\end{array}


    . We have: . \underbrace{\cos35}_{\downarrow}\sin55 + \cos55\underbrace{\sin35}_{\downarrow}
    Substitute: . \overbrace{\sin55}\sin55 + \cos55\overbrace{\cos55} \;\;=\;\;\sin^2\!55 + \cos^2\!55 \;\;=\;\;1

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