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avatar+140 

Compute \(\sum_{x=0^\circ}^{90} \cos^2 x\)

 Aug 13, 2019
 #1
avatar
+1

~ 45.5     is what I find  

 Aug 14, 2019
 #2
avatar+20216 
0

....I used an online summation calculator to find that....

ElectricPavlov  Aug 14, 2019
 #3
avatar+1694 
0

was that you? 

The guest?

tommarvoloriddle  Aug 14, 2019
 #5
avatar+20216 
0

Yeppers.....I often forget to Log In.....

ElectricPavlov  Aug 14, 2019
 #4
avatar+24043 
+2

Compute \(\sum \limits_{x=0^\circ}^{90^\circ} \cos^2(x)\)

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ \sum \limits_{x=0^\circ}^{90^\circ} \left(\cos^2(x) \right) } \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\cos^2(45^\circ)+\sum \limits_{x=46^\circ}^{90^\circ} \left(\cos^2(x)\right) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=46^\circ}^{90^\circ} \left(\cos^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(90^\circ-x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=0^\circ}^{44^\circ} \left(\sin^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)+\sin^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(1\right)+\cos^2(45^\circ) \\ &=& 45+\cos^2(45^\circ) \quad | \quad \cos(45^\circ) = \dfrac{\sqrt{2}}{2} \\ &=& 45+ \left(\dfrac{\sqrt{2}}{2}\right)^2 \\ &=& 45+ \dfrac{2}{4} \\ &=& 45+ \dfrac{1}{2} \\ &=& \mathbf{45.5} \\ \hline \end{array}\)

 

laugh

 Aug 14, 2019
 #6
avatar+7763 
0

\(\phantom{=} \displaystyle\sum_{x = 0^{\circ}}^{90^{\circ}} \cos^2 x\\ = \cos^2 0^{\circ} + \cos^2 1^{\circ} + \cdots + \cos^2 90^{\circ}\\ = 1 + (\cos^2 1^{\circ} + \cos^2 2^{\circ} + \cos^2 3^{\circ} + \cdots + \cos^2 89^{\circ}) + 0\\ = 1 + (\cos^2 1^{\circ} + \cos^2 89^{\circ}) + (\cos^2 2^{\circ} + \cos^2 88^{\circ}) + \cdots + (\cos^2 44^{\circ} + \cos^2 46^{\circ}) + \cos^2 45^{\circ}\\ \boxed{\text{Note that }\cos^2 n^{\circ} + \cos^2 (90 - n)^{\circ} = 1}\\ = 1 + 1 + 1 + 1 + \cdots + 1 + \dfrac{1}{2}\\ \boxed{\text{There are }\color{red}44\color{black}\text{ 1's}}\\ = 44 \dfrac{1}{2}\\ = \boxed{\dfrac{89}{2}}\)

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 Aug 14, 2019
 #7
avatar+140 
+2

Sorry that's wrong. It should like this: cos^2 0+cos^2 1+cos^2 2 ... cos^2 90 (in degrees). heureka was correct smiley

 Aug 14, 2019
edited by LeoIsTheBest  Aug 14, 2019

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