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

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.