Let z be a complex number such that

z=2(cos3∘+icos87∘).

Then z^6 can be expressed as r(sinα∘+icosα∘),

where r is a real number and 0≤α≤90. What is the value of r+α?

i found that r = 64 and α is 18; any one can clarify this ?

You fell into the trap

z^6 = 64 cis18° = 64(cos18° + i sin18°)

and it is thus 64(sin72° + i cos72°)

waaaaaa thx steve

To find the value of z^6, we can use De Moivre's theorem, which states that for any complex number z = r(cosθ + isinθ), the nth power of z can be expressed as z^n = r^n(cos(nθ) + isin(nθ)).

In this case, we have z = 2(cos3∘ + isin87∘). We need to find z^6.

Using De Moivre's theorem, we know that z^6 = 2^6(cos(6 * 3∘) + isin(6 * 3∘)).

Simplifying further, we have z^6 = 64(cos18∘ + isin18∘).

Now, we can express z^6 as r(sinα∘ + icosα∘). Comparing the imaginary parts, we can determine that sinα∘ = cos18∘.

Using the trigonometric identity sin(90 - θ) = cosθ, we can rewrite sinα∘ = sin(90 - 18∘).

Simplifying further, we have sinα = sin72∘.

Since r is a real number and we have cosα∘ = r, we can conclude that cosα∘ = cos72∘.

To find the value of r + α, we need to determine the values of r and α. From the above equations, we have cosα∘ = cos72∘, and sinα∘ = sin72∘.

Using the trigonometric identity sin^2θ + cos^2θ = 1, we can substitute sin^2θ = 1 - cos^2θ.

We have (sin72∘)^2 + (cos72∘)^2 = 1.

Simplifying further, we have 1 - (cos72∘)^2 + (cos72∘)^2 = 1.

Thus, we can conclude that r + α = 1 + 0 = 1.