how to evaluate cos13pi/6 and sin37pi/4?
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We have to evaluate cos 13pi/6 and sin 37pi/4
Now cos (x+ 2*pi) = cos x
cos 13pi/6 = cos ( pi/6 + 2*pi) = cos (pi/6) = sqrt 3 / 2
sin (x + pi) = -sin x
sin ( 37*pi/4) = sin (36*pi/4 + pi/4) = sin (9*pi + pi/4)
=> sin ( 4*pi + Pi + pi/4)
=> -sin (pi/4)
=> -1/sqrt 2
cos 13pi/6 = sqrt 3/2 and sin 37pi/4 = -1/sqrt 2
To evaluate cos13pi/6 and sin37pi/4.
We know that both cos and sin functions are 2pi periodic. So cos(2npi+x) = cosx and sin(2npi+x) = sinx, where n is apositive integerer.
Therefore cos(13pi/6) = cos (2pi + pi/6) = cos pi/6 = (1/2)*3^(1/2).
sin37pi/4 = sin(8pi+5pi/4) = sin(2*4pi +5pi) = sin(5pi/4) = sin(pi+pi/4) .
=> sin 37pi/4 = sin(pi + pi/4) .
=> sin37pi/4 = sin(pi + pi/4) = -sin pi/4, as sin (pi+x) = -sin x.
Therefore sin 37pi/4 = - sinpi/2 = - 1/2^(1/2) = -(1/2)*2^(1/2).
cos 13pi/6 = (1/2)3^(1/2) and sin 37pi/4 = -(1/2)2^(1/2).
We'll calculate cos 13pi/6.
We notice that we can write:
13pi/6 = pi/6 + 12pi/6
pi/6 + 12pi/6 = pi/6 + 2pi
We can substitute 2pi by 0, because 2pi and 0 are overlapping.
So,
cos 13pi/6 = cos (2pi + pi/6)
cos (2pi + pi/6) = cos pi/6
cos 13pi/6 = cos pi/6 = sqrt3/2
We'll calculate sin 37pi/4.
We notice that we can write:
37pi/4 = pi/4 + 36pi/4
pi/4 + 36pi/4 = pi/4 + 9pi
But 9pi = 4*2pi + pi
We'll substitute 2pi by 0.
4*2pi = 4*0 = 0
cos 37pi/4 = cos (pi/4 + pi)
cos (pi/4 + pi) = - cos pi/4
cos 37pi/4 = - cos pi/4 = - sqrt2/2
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