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Thread: Trigonometric Expression

  1. #1
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    Trigonometric Expression

    Simplify the trigonometric expression.

    csc 40(sin 40 + sin 80 + sin 120).

    The answer is 4 cos^2 (10).

    My Work:

    I have no idea how to reach the answer without using my calculator.

    csc 40 = 1/sin 40

    1/sin 40 = 1.5557238268

    sin 40 + sin 80 + sin 120 = 2.4936207664

    Multiply:

    (1.5557238268)(2.4936207664) =

    3.8793852412 which equals

    4 cos^2 (10).

    How can I reach the same answer without use of my calculator?

    Thank you.
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  2. #2
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    Re: Trigonometric Expression

    You have sin(40)= sin(4(10)), sin(80)= sin(8(10)), and sin(120)= sin(12(10)).

    So I would recommend some trig identities:
    sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)
    cos(x+ y)= cos(x)cos(y)- sin(x)sin(y)

    Setting x= y= t, sin(2t)= 2 sin(t)cos(t) and cos(2t)= cos^2(t)- sin^2(t).


    sin(3t)= sin(2t+ t)= sin(t)cos(2t)+ cos(t)sin(2t)= sin(t)(cos^2(t)- sin^2(t))+ cos(t)(2sin(t)cos(t))= sin(t)cos^2(t)- sin^3(t)+ 2sin(t)cos^2(t)= 3sin(t)cos^2(t)- sin^3(t).
    With t= 40, sin(80)= 2 sin(40)cos(40) and sin(120)= 3 sin(40)cos^2(40)- sin^3(40).

    With t= 10, sin(20)= 2sin(10)cos(10) and cos(20)= cos^2(10)- sin^2(10).
    sin(40)= 2sin(20)cos(20)= 2sin(10)cos(10)(cos^2(10)- sin^2(10))= 2sin(10)cos^3(10)- 2sin^3(10)cos(10) and
    cos(40)= cos^2(20)- sin^2(20)= (cos^2(10)- sin^2(10))^2- (2sin(10)cos(10)^2= cos^4(10)- 2cos^2(10)sin^2(10)+ sin^4(10)- 4sin^2(10)cos^2(10)=
    cos^4(10)- 6cos^2(10)sin^2(10)+ sin^4(10)

    Keep going!
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  3. #3
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    Re: Trigonometric Expression

    Quote Originally Posted by HallsofIvy View Post
    You have sin(40)= sin(4(10)), sin(80)= sin(8(10)), and sin(120)= sin(12(10)).

    So I would recommend some trig identities:
    sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)
    cos(x+ y)= cos(x)cos(y)- sin(x)sin(y)

    Setting x= y= t, sin(2t)= 2 sin(t)cos(t) and cos(2t)= cos^2(t)- sin^2(t).


    sin(3t)= sin(2t+ t)= sin(t)cos(2t)+ cos(t)sin(2t)= sin(t)(cos^2(t)- sin^2(t))+ cos(t)(2sin(t)cos(t))= sin(t)cos^2(t)- sin^3(t)+ 2sin(t)cos^2(t)= 3sin(t)cos^2(t)- sin^3(t).
    With t= 40, sin(80)= 2 sin(40)cos(40) and sin(120)= 3 sin(40)cos^2(40)- sin^3(40).

    With t= 10, sin(20)= 2sin(10)cos(10) and cos(20)= cos^2(10)- sin^2(10).
    sin(40)= 2sin(20)cos(20)= 2sin(10)cos(10)(cos^2(10)- sin^2(10))= 2sin(10)cos^3(10)- 2sin^3(10)cos(10) and
    cos(40)= cos^2(20)- sin^2(20)= (cos^2(10)- sin^2(10))^2- (2sin(10)cos(10)^2= cos^4(10)- 2cos^2(10)sin^2(10)+ sin^4(10)- 4sin^2(10)cos^2(10)=
    cos^4(10)- 6cos^2(10)sin^2(10)+ sin^4(10)

    Keep going!
    This problem takes too long to type. It is super tedious.
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  4. #4
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    Re: Trigonometric Expression

    \sin  80{}^{\circ}+\sin  40{}^{\circ} + \sin  120{}^{\circ} =\sin  80{}^{\circ}+2 \sin  80{}^{\circ} \cos  40{}^{\circ}=

    \sin  80{}^{\circ}(1+2 \cos  40{}^{\circ})
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  5. #5
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    Re: Trigonometric Expression

    Quote Originally Posted by Idea View Post
    \sin  80{}^{\circ}+\sin  40{}^{\circ} + \sin  120{}^{\circ} =\sin  80{}^{\circ}+2 \sin  80{}^{\circ} \cos  40{}^{\circ}=

    \sin  80{}^{\circ}(1+2 \cos  40{}^{\circ})
    Can you please complete this problem for me? It will help me solve similar questions in my textbook.
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  6. #6
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    Re: Trigonometric Expression

    \frac{\sin  80 +(\sin  40 + \sin  120)}{\sin  40}=

    \frac{\sin  80 (1+ 2 \cos  40)}{\sin  40}=

    4 \cos  40 \left(\frac{1}{2} +\text{  }\cos  40\right)=

    4 \cos  40 ( \cos  60 + \cos  40)=

    4 \cos  40\text{  }2 \cos  50 \cos  10=

    8 \sin  50 \cos  50 \cos  10=

    4 \sin  100 \cos  10=

    4 \cos ^2 10
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  7. #7
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    Re: Trigonometric Expression

    Quote Originally Posted by Idea View Post
    \frac{\sin  80 +(\sin  40 + \sin  120)}{\sin  40}=

    \frac{\sin  80 (1+ 2 \cos  40)}{\sin  40}=

    4 \cos  40 \left(\frac{1}{2} +\text{  }\cos  40\right)=

    4 \cos  40 ( \cos  60 + \cos  40)=

    4 \cos  40\text{  }2 \cos  50 \cos  10=

    8 \sin  50 \cos  50 \cos  10=

    4 \sin  100 \cos  10=

    4 \cos ^2 10
    Thank you so much.
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