Question

\(\large\tt \begin{align} \color{black}{\normalsize \text{calculate }\hspace{.33em}\\~\\
\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ} \hspace{.33em}\\~\\

}\end{align}\)

Answer 1

\[\cos(90-\theta) = \sin(\theta)\]

Answer 2

oh that was a big hint :)

Answer 3

may i expand on it ?

Answer 4

First regroup the terms like so

cos^(2) + cos(88)^2 + cos(4)^2 + cos(86)^2 + ...

Answer 5

\(\large\tt \begin{align} \color{black}{\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ} \hspace{.33em}\\~\\
=\sin^2 88^{\circ}+\sin^2 86^{\circ}+\sin^2 84^{\circ}+\cdot \cdot \cdot \cdot \cdot+\sin^2 2^{\circ}+\sin^2 0^{\circ}-------\color{red}{(1)} \hspace{.33em}\\~\\
\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ} \hspace{.33em}\\~\\
=45-\sin^2 2^{\circ}+\sin^2 4^{\circ}+\sin^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\sin^2 90^{\circ}-------\color{red}{(2)} \hspace{.33em}\\~\\
}\end{align}\)

Answer 6

First regroup the terms like so
cos^(2) + cos(88)^2 + cos(4)^2 + cos(86)^2 + ...
= cos(2)^2 + sin(2)^2 + cos(4)^2 + sin(4)^2 + ...
= 1 + 1 + ...

Answer 7

cos^2 90 doesnt match with anything, but cos(90) is zero

Answer 8

in the method i did , what is wong

Answer 9

*wrong

Answer 10

how did you get 45

Answer 11

it looks like you went back and forth , changing all cosines to all sines, then back to all cosines

Answer 12

i applied the rule

\(\large\tt \begin{align} \color{black}{1-\sin^2 \theta =\cos^2 \theta \hspace{.33em}\\~\\
}\end{align}\)
and there are \(45\) terms overall

Answer 13

this it the corrected one
\(\Huge \downarrow \)

\(\large\tt \begin{align} \color{black}{\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ} \hspace{.33em}\\~\\
=\sin^2 88^{\circ}+\sin^2 86^{\circ}+\sin^2 84^{\circ}+\cdot \cdot \cdot \cdot \cdot+\sin^2 2^{\circ}+\sin^2 0^{\circ}-------\color{red}{(1)} \hspace{.33em}\\~\\
\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ} \hspace{.33em}\\~\\
=45-(\sin^2 2^{\circ}+\sin^2 4^{\circ}+\sin^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\sin^2 90^{\circ})-------\color{red}{(2)} \hspace{.33em}\\~\\
}\end{align}\)

Answer 14

\(\large\tt \begin{align} \color{black}{2(\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ}) \hspace{.33em}\\~\\
=\sin^2 88^{\circ}+\sin^2 86^{\circ}+\sin^2 84^{\circ}+\cdot \cdot \cdot \cdot +\sin^2 2^{\circ}+\sin^2 0^{\circ}+\hspace{.33em}\\~\\
45-(\sin^2 2^{\circ}+\sin^2 4^{\circ}+\sin^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\sin^2 90^{\circ} \hspace{.33em}\\~\\
=45-\sin^2 90^{\circ} \hspace{.33em}\\~\\
2(\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ})=44 \hspace{.33em}\\~\\
so~~\hspace{.33em}\\~\\
(\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ})=22 \hspace{.33em}\\~\\
}\end{align}\)

is that right ?

Answer 15

let me compare to my solution, one sec

Answer 16

cos^2(2) + cos^2(4) + ... +cos^2 (88) + cos^2(90)
=cos^2(2) + cos^2(4) + ... +cos^2 (88) + 0
=cos^2(2) + cos^2(4) + ... +cos^2 (88)
=(cos^(2) + cos(88)^2) +(cos(4)^2 + cos(86)^2) + ... (cos^2(44)^2 + cos^2(46))
= (cos(2)^2 + sin(2)^2 ) + (cos(4)^2 + sin(4)^2) + ... (cos^2(44)^2 + sin^2(44))
= 1 + 1 + ... 1
= 22*1
= 22

Answer 17

yes we got the same :)

Answer 18

but why is wolfram giving
\(\large 28\)

http://www.wolframalpha.com/input/?i=%5Ccos+2+degree%2B%5Ccos+4+degree%2B%5Ccos+6+degree%2B.......%2B%5Ccos+90+degree

Answer 19

im a little confused about your lines , when you have ____ _ _ _ _ _ _

Answer 20

you didnt square them

Answer 21

wait i forgot squaring lol

Answer 22

you can put latex right into wolfram? thats cool

Answer 23

http://www.wolframalpha.com/input/?i=%5Ccos%5E2+2+degree%2B%5Ccos%5E2+4+degree%2B%5Ccos%5E2+6+degree%2B.......%2B%5Ccos%5E2+90+degree

Answer 24

yes i got the same , so you can use sigma notation

Answer 25

its not accepting \sigma , i have to hack it by some means

Answer 26

http://www.wolframalpha.com/input/?i=Sum+cosd^2%282k%29+%2C+k%3D1..45

Answer 27

oh thats cool :D

Answer 28

your solution is pretty cool, you used the identity sin^2 + cos^2 = 1 , but indirectly

Answer 29

cos^2 = 1 - sin^2

Answer 30

so either way you will end up at the same spot

Answer 31

yes but at first i thought it was wrong

Answer 32

wolfram is a very forgiving math machine , it accepts a lot of forms

Answer 33

i mean , it accepts a lot of different types of syntax

Answer 34

yes

http://www.wolframalpha.com/input/?i=perl

Answer 35

i see now, in your first post you subtracted the two equations , thats what those lines meant _ _ _ _ _ _

Answer 36

:)

Answer 37

wolfram is an encyclopedia?

Answer 38

its a brief encyclopedia

Answer 39

\(\large\tt \begin{align} \color{black}{\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+\cdot \cdot \cdot \cdot \cdot+
\cos^2 90^{\circ} \hspace{.33em}\\~\\
=\sin^2 88^{\circ}+\sin^2 86^{\circ}+\sin^2 84^{\circ}
\hspace{.33em}\\~\\ +\cdot \cdot \cdot \cdot \cdot+\sin^2 2^{\circ}+\sin^2 0^{\circ}-------\color{red}{(1)} \hspace{.33em}\\~\\
\cos^2 2^{\circ}+\cos^2 4^{\circ}+\cos^2 6^{\circ}+ \hspace{.33em}\\~\\
\cdot \cdot \cdot \cdot \cdot+\cos^2 90^{\circ} \hspace{.33em}\\~\\
=45-(\sin^2 2^{\circ}+\sin^2 4^{\circ}+\sin^2 6^{\circ}+ \hspace{.33em}\\~\\
\cdot \cdot \cdot \cdot \cdot+\sin^2 90^{\circ})-------\color{red}{(2)} \hspace{.33em}\\~\\
}\end{align}\)

Answer 40

those lines were these as there was not enough space in the page

Answer 41

and then you added equation (1) and (2)

Answer 42

yes hah

Answer 43

yeah thats nifty

Answer 44

thats weird that wolfram was able to infer the pattern for
cos 2 degrees + cos (4 degrees) + ... + cos ( 90 degrees)

but for the square terms it was stumped

Answer 45

want to do a problem that stumped wolfram, which has a simple solution?

Answer 46

a trig equation

Answer 47

solve
sind(2x) = cosd(3x -10)

sind and cosd only accepts arguments with degrees (so we dont have a problem with radians)

Answer 48

yes that happens wolfram is a machine
and man made machine , machine did'nt made man
so man>wolfram

Answer 49

http://www.wolframalpha.com/input/?i=sind+%282x%29%3Dcosd+%283x-10%29%2C0%5Cleq+x%5Cleq+360

Answer 50

oh i guess it didn't stump it then,

Answer 51

once i had a problem where wolfram was wrong , but i don't remember

Answer 52

hmm, i thought it did a week ago

Answer 53

lolol

Answer 54

you can see the general solutions on the bottom
http://www.wolframalpha.com/input/?i=sind+%282x%29%3Dcosd+%283x-10%29

Answer 55

oh i remember what was the problem, i got a 'wild' solution from mathematica 10

Answer 56

i will if i can attach the output

Answer 57

attach file button is dead :/

Answer 58

haha
http://www.wolframalpha.com/input/?i=who+are+you+%3F

Answer 59

lol that didnt work

Answer 60

the file is not opening

Answer 61

thats creepy

Answer 62

the who are you

Answer 63

its probably a joke

Answer 64

http://www.wolframalpha.com/input/?i=who+invented+you+%3F

Answer 65

LOL

Answer 66

even better
http://www.wolframalpha.com/input/?i=How+many+angels+can+dance+on+the+head+of+a+pin%3F

Answer 67

clever

Answer 68

i uploaded the wild output using pdf, see if this work s

Answer 69

i mean i saved as .pdf

Answer 70

what r u trying to do

Answer 71

im trying to show you what mathematica 10 displays when i try to solve that equation above

Answer 72

wolfram (web version) gives a more logical answer

Answer 73

oh k

Answer 74

http://www.wolframalpha.com/input/?i=tell+me+a+joke

Answer 75

every time it has a new joke lol,

Answer 76

here
http://www.filedropper.com/dd

Answer 77

thats my file

Answer 78

that also isn;t working

Answer 79

click on 'download this file'

Answer 80

ok but at the end it should interpret it as

solve ------- for \(x\)


it displaying \(\cos x ~and~\sin x\)

Answer 81

were you able to open it ?

Answer 82

yes lol

Answer 83

oh for x

Answer 84

usually it solves for x by default

Answer 85

if i say solve x^2 +2x + 1 = 0 , it will solve for x

Answer 86

u have to write it as if

u wanr ,for integers , reals or complex

Answer 87

http://www.wolframalpha.com/input/?i=solve+x%5E2%2Bx%2B1%3D0+over+reals+

Answer 88

http://www.wolframalpha.com/input/?i=solve+x%5E2%2Bx%2B1%3D0+over+complex

Answer 89

i tried using 'over reals', didn't help. i dont know, mathematica 10 acts oddly.
maybe i should go back to mathematica 9 , i liked that version

Answer 90

http://www.wolframalpha.com/input/?i=solve+sind+%282x%29%3Dcosd+%283x-10%29%2C0%3Cx%3C360+over+the+reals

Answer 91

i can try integer solutions

Answer 92

ok now it seems to work when i stipulate integer solutions

Answer 93

yes its working for that

http://www.wolframalpha.com/input/?i=solve+sind+%282x%29%3Dcosd+%283x-10%29%2C0%3Cx%3C360+over+the+integers

Answer 94

ok thanks a lot

Answer 95

it looks like mathematica goes straight to the real 'exact' solutions, thats why there was such a long expression output

Answer 96

I would use this \[\Large \cos^2(2x)=\frac{1}{2}+\frac{1}{2}\cos(4x)\] and change the sum from \[\Large \sum_{n=1}^{45}\cos^2(2x)= \sum_{n=1}^{45}\frac{1}{2} + \frac{1}{2}\cos(4x)\] that way the cosine part goes from almost cos(0) to cos(180) so they will all cancel each other out.