Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Trigonometric Identity?

(cos^2x) / (1+2sin(x)-3sin^2(x)) = (1 + sin(x)) / (1 + 3sin(x))

Please show ALL the steps involved so that I can understand it better.

Thank you in advance.

2 Answers

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  • 1 decade ago
    Favorite Answer

    LHS = cos²(x) / [1 + 2sin(x) - 3sin²(x)]

    = [1 - sin²(x)] / [(1 + 3sin(x))(1 - sin(x))]

    = [(1 - sin(x))(1 + sin(x))] / [(1 + 3sin(x))(1 - sin(x))]

    = [1 + sin(x)] / [1 + 3sin(x)]

    = RHS qed

  • 1 decade ago

    i am working on the left hand side

    numerator

    cos^2(x)=1-sin^2(x)

    1-sin^2(x)=(1-sin(x))(1+sin(x)) (difference of squares)

    the denominator is a trigonometric quadratic equation

    so factor it

    (1+2sin(x)-3sin^2(x))= -(3sin^2(x)-2sin(x)-1)

    -(3sin^2(x)-2sin(x)-1)=-(3sin^2(x)-3sin(x)+sin(x)-1)

    =-( 3sin(x)(sin(x)-1)+1(sin(x)-1))

    =-(3sin(x)+1)(sin(x)-1)

    multiply the second bracket by the negative sign

    =(3sin(x)+1)(1-sin(x))

    now the left hand side becomes

    (1-sin(x))(1+sin(x))/(3sin(x)+1)(1-sin(x))

    =(1+sin(x))/(3sin(x)+1)

    there fore left hand side = right hand side

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