how do you find the standard form of
7(cos255degrees + isin255degrees) ?
we need cos 225° and sin225°
225 = 180+45 , it is in quadrant III and has the same numerical trig values as 45°, but in III both the sine and cosine are negative
sin 225 = - sin 45° = -1/√2
cos 225 = - cos 45° = -1/√2
so 7(cos255degrees + isin255degrees)
= 7(-1/√2 - i (1/√2))
= -7/√2 - (7/√2) i
now, just change all those 225's to 255's and redo the calculations.
To find the standard form of a complex number in the form a(cosθ + isinθ), we can use the trigonometric identities:
cosθ = (e^(iθ) + e^(-iθ)) / 2
sinθ = (e^(iθ) - e^(-iθ)) / (2i)
Let's apply these formulas to your complex number:
Given: 7(cos255° + isin255°)
Using the cosine formula, we get:
cos255° = (e^(i * 255°) + e^(-i * 255°)) / 2
Similarly, using the sine formula, we get:
sin255° = (e^(i * 255°) - e^(-i * 255°)) / (2i)
Now, let's substitute these values back into the original expression:
7(cos255° + isin255°)
= 7 * [(e^(i * 255°) + e^(-i * 255°)) / 2 + i * (e^(i * 255°) - e^(-i * 255°)) / (2i)]
Simplifying this expression by multiplying 7/2 to the real part and getting rid of the complex conjugate in the imaginary part, we have:
= 7/2 * (e^(i * 255°) + e^(-i * 255°)) + i * 7/2 * (e^(i * 255°) - e^(-i * 255°))
Using the laws of exponents, we know that e^(-ix) = 1 / e^(ix). So, we have:
= 7/2 * (e^(i * 255°) + 1 / e^(i * 255°)) + i * 7/2 * (e^(i * 255°) - 1 / e^(i * 255°))
Now, let's simplify the expression further:
= 7/2 * (e^(i * 255°) + e^(-i * 255°)) + i * 7/2 * (e^(i * 255°) - e^(-i * 255°))
= 7/2 * (2cos(255°)) + i * 7/2 * (2isin(255°))
= 7 * cos(255°) + i * 7 * sin(255°)
Finally, we arrive at the standard form:
7 * cos(255°) + i * 7 * sin(255°)
So, the standard form of the given complex number is:
7 * cos(255°) + i * 7 * sin(255°)