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15 April, 17:36

How do i solve 2cos^2 (5pi/12) - 1?

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  1. 15 April, 17:48
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    1. exact value of arcsin (sin (5pi/4))

    answer : 5pi/4

    because arcsin is the inverse operation of sin

    so doing sin then arcsin is the same as doing nothing!

    2. cos (5pi/12)

    we can use the addition formulae.

    eg 5/12 = 2/12 + 3/12

    cos (5pi/12) = cos (2pi/12 + 3pi/12)

    = cos (pi/6 + pi/4)

    = cos (pi/6) cos (pi/4) - sin (pi/6) sin (pi/4)

    = [1/2][1/sqrt (2) ] - [1/2][ (1/2) sqrt (3) ]

    = [1/4][sqrt (2) - sqrt (3) ]

    = - [1/4][sqrt (3) - sqrt (2) ].

    3. sin (5pi/8) = cos (pi/8)

    because

    sin (x) = cos (x-pi/2) = cos (x-4pi/8).

    Using cos (2x) = 2cos^2 (x) - 1 we get

    cos^2 (x) = (1/2) [cos (2x) + 1]

    cos^2 (pi/8) = (1/2) [cos (pi/4) + 1]

    = (1/2) [1/sqrt (2) + 1]

    = (1/4) [sqrt (2) + 2].

    answer : cos (pi/8) = (1/2) sqrt ((2+sqrt (2)).

    4. sin (arcsin (3/5) - arccos (3/5)).

    Imagine a 3-4-5 right triangle with base 4 units, opposite 3 units and hypotenuse 5 units.

    If base angle is A and the vertex angle is B then

    sinA = cosB = 3/5

    A = arcsin (3/5) and B = arccos (3/5)

    sinB = cosA = 4/5.

    use the addition formula again:

    sin (arcsin (3/5) - arccos (3/5)).

    = sin (A-B)

    = sinAcosB - cosAsinB

    = (3/5) (3/5) - (4/5) (4/5)

    = (1/25) (9 - 16)

    = - 7/25.
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