How do i find the exact value of (sin -π/8)?

Question

i need to know how to find fid the exact answer of SIN (-π/8)

if anyone can give me a step by step action i would be thnakful

Answer 1

use a graphing calculater :-) thats how i used to do it. i forgot the actual steps. sorry.

Answer 2

As you know sin2x = 2sinxcosx = 2sinx√[1-sin²x]
So just put in x = -π/8
sin -π/4 = 2sin(-π/8)√[1-sin²(-π/8)]
As sin(-π/8) is going to be a pain to write let sin(-π/8) = a
-1/√2 = 2a√[1-a²]
Squaring both sides
1/2 = 4a²(1-a²)
1 = 8a²-8a^4
0 = -1+8a²-8a^4
This is a quadratic in a². Using the quadratic formula:
a² = ( -8 ± √[8²-4×(-1)×(-8)] ) / (2×8)
a² = ( -8 ± √[32] ) / (16)
a² = ( -8 ± √[32] ) / (16)
a² = ( -8 ± 4√[2] ) / (16)
a² = -1/2 ± √[2]/4
Ignoring the negative square root as -1/2 - √[2]/4 is less than zero.
a² = -1/2 + √[2]/4
So
a = sin(-π/8) = - √[-1/2 + √[2]/4]
Taking the negative root as the angle is in the 4th quadrant.

Answer 3

use the half angle formula

sin^2(a) = 1/2(1-cos(2a))
sin^2(-pi/8)= 1/2(1-sqrt(2)/2)
sin(-pi/8)=-sqrt(1/2(1-sqrt(2)/2)) ;since (-pi/8) is in the 4th quad