# Use trigonometric identities to calcuate sin(345)

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We have to find sin 345.

sin 345 = sin ( 360 - 15) = - sin 15

Now cos 30 = (cos 15)^2 - ( sin 15)^2

=> cos 30 = 1 - 2*(sin 15)^2

=> 1 - 2*(sin 15)^2 = cos 30

=> - 2*(sin 15)^2 = cos 30 - 1

=> 2 (sin 15)^2 = 1 - cos 30

=> 2 (sin 15)^2 = 1 - sqrt 3/2

=> (sin 15)^2 = 1/2 - sqrt 3/4

=> (sin 15) = sqrt (1/2 - sqrt 3/4)

=> sin 15 = sqrt (1/2 - sqrt 3/4)

**Therefore sin 345 = - (sqrt (1/2 - sqrt 3/4))**

We need to use the trigonometric identities to find the values of sin(345).

First we need to rewrite 345 as a product of two common angles.

==> 345 = 120 + 225

We know that 120 = (60 degrees in the second quadrant)

Also we know that 225 = (45 degrees into the 3rd quadrant).

==> sin345= sin(120+225)

We know that:

sin(a+b) = sina*cosb + cosa*sinb.

==> sin(120+225) = sin120*cos225+sin225*cos120.

= sqrt3/2 * -1/sqrt2 + -1/sqrt2* -1/2

= -sqrt3/2sqrt2 + 1/2sqrt2

= (1-sqrt3)/2sqrt2

**==> sin(345) = (1-sqrt3)/2sqrt2**

sin345deg = sin(360-15) = -sin15degree..

We know that sin30 = sin(15+15) = 2sin15cos15, as sin2A = 2sinAcosA.

sin30 = 1/2 = 2x*sqrt(1-x^2), where x=sin15.

(1/2)^2 = 4x^2(1-x^2)

1/4 = 4x^2-4x^4.

4x^4-4x^2+1/4 = 0. We multiply by 4.

16x^4-16x^2+1 = 0, which is a quadratic equation in x^2.

x^2 = (16+sqrt (16^2- 4*16)}/(2*16) = (2- or sqrt3)/4.

Therefore x = sqrt{(2-sqrt3)/4} which is positive as sin15 > 0.

x = 0.2588 nearly.

Therefore sin345 = -sin 15 = -0.2588 nearly.