Question

Homework. Pls halp.

Answer 1

http://imgur.com/6hKV48x

Answer 2

@sauravshakya

Answer 3

\[\large \cos x = \dfrac{2\cos (x-y)\cos(x+y)}{\cos(x-y) + \cos(x-y)}\]

Answer 4

http://imgur.com/QYpTEfb

Answer 5

that looks good ^^
so you're done with #18 ? :)

Answer 6

Wait, am I on the right track?

Answer 7

you're two steps away from final answer... :P

Answer 8

send that cosy to left side,
factor out cos2x

Answer 9

I love you.

Answer 10

something gets cancel out leaving you with the final expression

Answer 11

Wait, hmm, still not done though. I carried 1+cos(2x) to the RHS and took the LCM. Then what?

Answer 12

I cancelled something incorrectly in ur work :o

i was trying like this initialy :

\[\large \cos x = \dfrac{\cos^2x\cos^2y - \sin^2x\sin^2y}{\cos x\cos y}\]

Answer 13

\[\large \cos x = \dfrac{\cos^2x\cos^2y - (1-\cos^2x)(1-\cos^2y)}{\cos x\cos y}\]

Answer 14

\[\large \cos x = \dfrac{-1 +\cos^2x +\cos^2y}{\cos x\cos y}\]

Answer 15

follow ,will check when i come back

Answer 16

let me grab a paper to work the rest

Answer 17

actually we're done :

\[\large \cos^2x \cos y =-1 + \cos^2x + \cos^2y\]

\[\large \cos^2x( \cos y-1) =\cos^2y-1\]

Answer 18

\[\large \cos^2x( \cos y-1) =(\cos y -1)(\cos y+1)\]

Answer 19

\(\cos y \ne 1\) (why ) , so we can cancel it both sides :

\[\large \cos ^2x = \cos y + 1\]

Answer 20

Gotta go, night!

Answer 21

gnite :) for #19 :

\[\large 2\sin \alpha \cos \beta \sin \gamma = \sin \beta \sin(\alpha + \gamma)\]

\[\large \dfrac{2\sin \alpha \sin \gamma }{\sin(\alpha + \gamma)}= \tan \beta \]

dividing numerator and denominator on left side fraction by \(\cos \alpha \cos \gamma \) shows the expression for harmonic mean

Answer 22

nice :)