how to evaluate cos13pi/6 and sin37pi/4?

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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**

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 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**

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