Question

Sin7π/12*cos π/12-sinπ/12*cis7π/12

Answer 1

Sin7π/12*cos π/12-sinπ/12*cis7π/12=sin(7π/12-π/12)=sin6π/12=sinπ/2=1

Answer 2

To simplify the expression Sin(7π/12)*Cos(π/12) - Sin(π/12)*Cis(7π/12), we need to use trigonometric identities and simplify the values.

Let's start with the trigonometric identities:

1. Sin(a) * Cos(b) = (1/2) * [Sin(a + b) + Sin(a - b)]
2. Cis(c) = Cos(c) + i * Sin(c), where "i" represents the imaginary unit.

Now, we can proceed with the simplification:

1. Sin(7π/12) * Cos(π/12): Using identity 1 with a = 7π/12 and b = π/12: (1/2) * [Sin(7π/12 + π/12) + Sin(7π/12 - π/12)] (1/2) * [Sin(8π/12) + Sin(6π/12)] (1/2) * [Sin(2π/3) + Sin(π)] (1/2) * [√3/2 + 0] (1/2) * √3/2 √3/4

2. Sin(π/12) * Cis(7π/12): Using identity 2: Sin(π/12) * [Cos(7π/12) + i * Sin(7π/12)] Sin(π/12) * Cos(7π/12) + i * Sin(π/12) * Sin(7π/12)

Now, we'll calculate the values of Sin(π/12) and Cos(7π/12):

• Sin(π/12): We can use the half-angle formula for sine: Sin(π/12) = √[(1 - Cos(π/6)) / 2] = √[(1 - √3/2) / 2] = √[(2 - √3) / 4]

• Cos(7π/12): We can use the difference formula for cosine: Cos(7π/12) = Cos(π - 5π/12) = -Cos(5π/12)

  Now, we need to calculate Cos(5π/12): Cos(5π/12) = √[(1 + Cos(π/6)) / 2] = √[(1 + √3/2) / 2] = √[(2 + √3) / 4]

So, Sin(π/12) * Cis(7π/12) becomes:

√[(2 - √3) / 4] * [√[(2 + √3) / 4] + i * Sin(7π/12)]

Now, we can plug in the value of Sin(7π/12) from before:

√[(2 - √3) / 4] * [√[(2 + √3) / 4] + i * (1/2) * √3]

Finally, we can combine the real and imaginary parts:

Real part: √[(2 - √3) / 4] * √[(2 + √3) / 4] = √[(2^2 - 3) / 4] = √[(1/4)] = 1/2

Imaginary part: √[(2 - √3) / 4] * (1/2) * √3 = (1/2) * √3 * √[(2 - √3) / 4]

So the simplified expression is:

(√3/4) - (1/2 + (1/2) * √3 * √[(2 - √3) / 4]) * i

This is the simplified form of the expression Sin(7π/12)*Cos(π/12) - Sin(π/12)*Cis(7π/12).