Multiple Choice Question (MCQ)

If \cos A + 2\cos B + \cos C = 2 then a, b, c are in

  1. 2b=a+c

  2. ×

    b^2=ac

  3. ×

    a=b=c

  4. ×

    None of these

Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 1 week ago


Answer 1

  1. This method uses trigonometric identities and Helen’s Formula

    Given \cos A + 2 \cos B + \cos C = 2

    ⇒ \cos A + \cos C = 2(1 – \cos B)

    Using Sum to Product Identity

    \cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

    and Double angle identity

    \cos2 \alpha = 1-2 \sin^2 \alpha

    ⇒ 2 \cos\dfrac{A+C}{2} \cdotp \cos\dfrac{A-C}{2} = 4 \sin^2(\dfrac{B}{2} )

    Since the sum of the three internal angles is equal to \pi

    \therefore \dfrac{A+C}{2} = \dfrac{\pi}{2}-\dfrac{B}{2}


    ⇒ 2 \sin\dfrac{B}{2}\cos\dfrac{A-C}{2} = 4\sin^2 \dfrac{B}{2}

    ⇒ \cos\dfrac{A-C}{2} = 2\sin \dfrac{B}{2}
    (1)
    ⇒ \cos\dfrac{A-C}{2} = 2\cos\dfrac{A+C}{2}

    ⇒ \cos\dfrac{A-C}{2} – \cos\dfrac{A+C}{2} = \cos\dfrac{A+C}{2}

    ⇒ 2\sin \dfrac{A}{2}\sin \dfrac{C}{2} = \sin\dfrac{B}{2}
    (2)

    Using Helen formula


    helen’s formula

    \because A =\dfrac{1}{2} (a+b+c)r

    r = \dfrac{A}{s}

    = \dfrac{ \sqrt{s(s-a)(s-b)(s-c)}}{s}


    =\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s} }

    in which A is the area of the triangle, s = \dfrac{a+b+c}{2}

    \sin \dfrac{A}{2} =\dfrac{r}{\sqrt{r^2+(s-a)^2} }

    =\dfrac{1}{\sqrt{1+\dfrac{(s-a)^2}{r^2} } }

    =\dfrac{1}{\sqrt{1+\dfrac{s(s-a)}{(s-b)(s-c)} } }

    =\dfrac{\sqrt{(s-b)(s-c)} }{\sqrt{s^2-(b+c)s+bc+s^2-sa}}

    =\dfrac{\sqrt{(s-b)(s-c)} }{\sqrt{2s^2-(a+b+c)s+bc}}

    =\dfrac{\sqrt{(s-b)(s-c)} }{\sqrt{bc}}

    Similarly,

    \sin \dfrac{B}{2} = \dfrac{\sqrt{(s-a)(s-c)} }{\sqrt{ac}}

    \sin \dfrac{C}{2} = \dfrac{\sqrt{(s-a)(s-b)} }{\sqrt{ab}}

    Substituting to (2) yields

    ⇒ 2(s – b) = b

    ⇒ a + b + c – 2b = b

    ⇒ a + c = 2b

Steven Zheng Steven Zheng posted 1 week ago


Answer 2

  1. This method uses trigonometric identities and the Law of Sines

    Following (1) of the method 1

    \cos\dfrac{A-C}{2} = 2\sin\dfrac{B}{2}
    (1)

    Multiplying 2\cos \dfrac{B}{2} with two sides of the equation

    2\cos\dfrac{A-C}{2} \cos \dfrac{B}{2} = 2\cdotp 2\sin\dfrac{B}{2}\cos \dfrac{B}{2}

    2\cos\dfrac{A-C}{2} \sin \dfrac{A+C}{2} = 2\cdotp 2\sin B

    Using Product to sum identity,

    \cos \alpha\sin \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) - \sin(\alpha - \beta) ]

    \sin A+\sin C=2\sin B
    (2)

    Using the Law of Sines,

    \sin A=\dfrac{a}{2R}

    \sin B=\dfrac{b}{2R}

    \sin C=\dfrac{c}{2R}

    Substituting to (2) yields


    2b=a+c

Steven Zheng Steven Zheng posted 1 week ago

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