Compute

\(\cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.\)

FlyEaglesFly Mar 29, 2020

#4**+3 **

Use the fact that:

sin(x) = cos(90-x)

We then can rewrite our equation as:

cos^{2}(0) + cos^{2}(1) + cos^{2}(2)........ cos^{2}(44) + sin^{2}(45) + sin^{2}(44) ........ sin^{2}(0).

Next we make use of a basic trig identity:

\(\cos^2\theta + \sin^2\theta = 1\)

Given that the two angles are the same.

We realize we have 0-44 = 45 total such pairs in our rewritten equation(can you see how?).

This gives us 45 + sin^{2}(45) = \(45+(\sqrt2/2)^2 = 45 + 2/4 = 45 + 1/2 = 45.5\)

jfan17 Mar 29, 2020

#1

#2**0 **

You should look up one that includes steps, so you can explain the process to the person in need.

AnExtremelyLongName
Mar 29, 2020

#4**+3 **

Best Answer

Use the fact that:

sin(x) = cos(90-x)

We then can rewrite our equation as:

cos^{2}(0) + cos^{2}(1) + cos^{2}(2)........ cos^{2}(44) + sin^{2}(45) + sin^{2}(44) ........ sin^{2}(0).

Next we make use of a basic trig identity:

\(\cos^2\theta + \sin^2\theta = 1\)

Given that the two angles are the same.

We realize we have 0-44 = 45 total such pairs in our rewritten equation(can you see how?).

This gives us 45 + sin^{2}(45) = \(45+(\sqrt2/2)^2 = 45 + 2/4 = 45 + 1/2 = 45.5\)

jfan17 Mar 29, 2020