Question

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

Answer 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.