+0

# Trig

+1
75
5
+201

Compute
$$\cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.$$

Mar 29, 2020

#4
+483
+3

Use the fact that:

sin(x) = cos(90-x)

We then can rewrite our equation as:

cos2(0) + cos2(1) + cos2(2)........ cos2(44) + sin2(45) + sin2(44) ........ sin2(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 + sin2(45) =  $$45+(\sqrt2/2)^2 = 45 + 2/4 = 45 + 1/2 = 45.5$$

.
Mar 29, 2020

#1
+23562
0

Does any one get 44.5 ?    I used an online  summation calculator.....

Mar 29, 2020
#2
+626
0

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

AnExtremelyLongName  Mar 29, 2020
#3
0

Mine gives 45.5 !

Guest Mar 29, 2020
#4
+483
+3

Use the fact that:

sin(x) = cos(90-x)

We then can rewrite our equation as:

cos2(0) + cos2(1) + cos2(2)........ cos2(44) + sin2(45) + sin2(44) ........ sin2(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 + sin2(45) =  $$45+(\sqrt2/2)^2 = 45 + 2/4 = 45 + 1/2 = 45.5$$

jfan17 Mar 29, 2020
#5
+201
0