Cute Little Trigs

This thread is for sharing all your trigonometry puzzles that are short and sweet..

Ok..here are mine..

1. tan1^o * tan2^o * tan3^o * tan4^o * .....................tan89^o = ???

2. Sin²5^o + Sin²10^o + Sin²15^o + ..... + Sin²85^o + Sin²90^o = ???

You are not supposed to use calculators or any math program and you have to prove the results.

1.
tan(1)=sin(1)/cos(1)=sin(1)/sin(89)
tan(89)=sin(89)/sin(1)
now you have pairs which are each others complements, except tan(45) which is 1. the hole series is 1.
2.brb

Kedaman get a cookie..

2) ??? what's brb ?

2.
sin^2(x)=½(1-cos(2x))
sin^2(5)=½(1-cos(10)) which is compliment to
sin^2(50)=½(1-cos(100))
the whole thing should be 9
brb means be right back, i was getting some breakfast :)

Well..my way of solving 1 is..

tan1^o * tan2^o * tan3^o * tan4^o * .....................tan89^o
can be divided in to three parts..

Part-I
tan1^o * tan2^o *........*tan43^o * tan44^o

Part-II
tan45^o

and

Part-III
tan46^o * tan47^o *.......tan89^o

the third part is..

tan(90^o-44^o) * tan(90^o-43^o) .......tan(90^o-1^o)

which is nothing but
cot44^o * cot43^o * .......cot1^o {by complementary rule}

since tanx * cotx = 1 ...
multiplying part 1 and part 3 results in 1
the second part is tan45 = 1

so the product of all the three parts is 1


Keds, BRB - Be right Back ??

You got 2 wrong...

But 2 is as easy as 1 ..try again :D

yep ;)
and yeah thats what i did but you know it's 4:22 am and i didn't want to got that explicit :)

Go... Brush your teeth...Have a bath ...and have some strong beer ;) and then come back :D

sin^2(x)=½(1-cos(2x))
ok, we divide into four groups
5-40, 45, 50-85,90
drawing the 8 dots 5-40 and 50-90 with the function cos(2x) on a circle, makes compliments of each other (addition compliments=0) 85 and 5, 80 and 10 etc... but (1-a) + (1-(-a) )=2, but ½ of that is 1, so those groups together is 8. ½(1-cos(180)) is 1 and ½(1-cos(90)) is ½ so the whole thing is 9½, correct?

Yup....you are.. :)


It's your turn now..to post some interesting short trigonometry stuff..

My Way solving (2)

Sin²5^o + Sin²10^o + Sin²15^o + ..... + Sin²85^o + Sin²90^o

can be divided in 4 parts like ked..

I

Sin²5^o + Sin²10^o+....Sin²40^o

II

Sin²45^o

III

Sin²50^o + Sin²55^o+....+ Sin²85^o

IV

Sin²90^o


Part 3
can be modified using complementary rule to...

Cos²40^o + Cos²35^o + ......+Cos²5^o

Since Sin²x + Cos²x = 1

From Part I and Part 3
you add up eight 1's = 8

Sin²45^o = 1/2

Sin²90^o = 1

There...you add up 8 + 1 + 1/2 = 9½

Well, i'm no good at coming up with these cute problems :p i guess it's pretty hard to do so anyway, thanks, this was all bracing :)