Question

wel i hav some doubts on trigonometruy wen i did revision can anyone help

Answer 1

what is the formula for cos 2A *cos 4A *cos 8A ETC

Answer 2

cos2A=1-sinA i think..

Answer 3

no

Answer 4

meant 1-sin2A

Answer 5

oh, no? ok

Answer 6

11+cos2a is 2sin^2A

Answer 7

cos 2A= cos^2 A- sin ^2 A

Answer 8

\[\cos^2A=1-\sin^2A\]
but it's not suqared

Answer 9

.........

Answer 10

no formula for this http://www.wolframalpha.com/input/?i=simplify+cos+2A+*cos+4A+*cos+8A

Answer 11

aravind do want the simplification of cos 2A* cos 4A * cos 8A

Answer 12

*you

Answer 13

no i think there is a formula

Answer 14

for n terms

Answer 15

somewat like sin 2^n alpha/(2^n sin alpha)

Answer 16

ah it doesn't end
idk i haven't learn such formula in trig course

Answer 17

its there tomas

Answer 18

you are from different country...

Answer 19

:P but math is universal :P

Answer 20

but we don't learn same things

Answer 21

so as i said i didn't learn it in my trig course

Answer 22

k

Answer 23

aravind I'm trying to think, you're right that there is a formula, but i'm not able to recall

Answer 24

oh thn next doubt u find it and post here later

Answer 25

u knw solutions of triangles?

Answer 26

not very much but I'll try,

Answer 27

:O

Answer 28

in a triangle (a+b+c)(b+c-a)=kbc if 0<k<4 prove

Answer 29

??

Answer 30

ash

Answer 31

yeah aravind i'm thinking

Answer 32

\[Cos2x=1-2\sin^2x\]

Answer 33

\[Cos2x=\cos^2x-\sin^2x\]

Answer 34

\[Cos2x=2\cos^2x-1\]

Answer 35

yeah, i can help with you second question
equation 1....(a+b+c)(b+c-a)= b^2+c^2-a^2+2bc
now we know cosine formula a^2=b^2+c^2-2bc*cos A
so here b^2+c^2-a^2=2bc cos A, subsitute this in equation 1
so
2bc cos A+ 2bc
2bc( cos A+1)
now cos A 's max value is 1 and minimum 0
so the left side can be
0 or 4bc
so k can vary from 0 to 4
0<k<4

Answer 36

wow perfect

Answer 37

http://www.mediafire.com/?t2tdkmznnzq

Answer 38

handbook of mathematics...

Answer 39

it's not mine habit...