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avatar+201 

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

 Mar 29, 2020

Best Answer 

 #4
avatar+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
avatar+23562 
0

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

 Mar 29, 2020
 #2
avatar+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
avatar
0

Mine gives 45.5 !

Guest Mar 29, 2020
 #4
avatar+483 
+3
Best Answer

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
avatar+201 
0

Thankyou the answer was 45.5 and your explanation was great

FlyEaglesFly  Mar 30, 2020

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