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Kason Snow
Mathematics
15 April, 17:36
How do i solve 2cos^2 (5pi/12) - 1?
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Miya Li
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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