1. Evaluate : sin 18° ci (1) cos 72°​

Question

1. Evaluate :
sin 18°
ci
(1)
cos 72°​

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Mia 1 month 2021-08-19T08:03:16+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-08-19T08:04:16+00:00

    How can I find the value of sin18°?

    Work the way you want from anywhere.

    The explanation in video :

    To find the value of sin 18°

    Let A = 18°

    Therefore, 5A = 90°

    ⇒ 2A + 3A = 90˚

    ⇒ 2A= 90˚ – 3A

    Taking sine on both sides, we get

    sin 2A = sin (90˚ – 3A) = cos 3A we know , sin(90 – A) = cos A

    ⇒ 2 sin A cos A = 4cos^3 A – 3 cos A using formulas for sin 2A and cos 3A

    ⇒ 2 sin A cos A – 4cos^3A + 3 cos A = 0

    ⇒ cos A (2 sin A – 4 cos^2 A + 3) = 0

    Dividing both sides by cos A = cos 18˚ ≠ 0, we get

    ⇒ 2 sin θ – 4 (1 – sin^2 A) + 3 = 0

    ⇒ 4 sin^2 A + 2 sin A – 1 = 0, which is a quadratic in sin A

    Using SHRI DHAR ACHRYA formula

    x = [-b ± √(b^2 – 4ac)] / 2a

    18° lies in

    Question is Sin18° = ?

    Let x = 18°

    so, 5x = 90°

    now we can write

    2x + 3x = 90°

    so 2x = 90° – 3x

    now taking sin both side we can write

    sin2x = sin(90°-3x)

    sin2x = cos3x [as we know, sin(90°-3x) = Cos3x ]

    so evaluating we can write

    2sinxcosx =4cos³x – 3cosx [as we know, Cos3x = 4cos³x – 3cosx, This I will explain later ]

    Now, 2sinxcosx – 4cos³x + 3cosx = 0

    cosx(2sinx – 4cos²x + 3) = 0

    Now divding both side by cosx we get,

    2sinx – 4cos²x + 3 = 0

    2sinx – 4(1-sin²x) + 3 = 0 [as we know, cos²x = (1-sin²x), by sin²x + cos²x = 1 ]

    2sinx – 4 + 4sin²x + 3 = 0

    2sinx + 4sin²x – 1 = 0

    we can write it as,

    4sin²x + 2sinx – 1 = 0

    Now apply Sridhar Acharya Formula Here,

    ax² + bx + c = 0

    so, x = (-b ± √(b² – 4ac))/2a

    now applying it in the equation

    sinx =

    (-2 ± √(2² – 44(-1)))/2.(4)

    sinx = (-2 ± √(4 +16))/8

    sinx = (-2 ± √20)/8

    sinx = (-2 ± 2.√5) / 8

    sin x = 2(-1 ± √5 ) / 8

    sin x = (-1 ± √5)/4

    sin18° = (-1 ± √5)/4 = cos72°

    Thanks

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