Question

Help with identities:
1. ( sin2x/sin x)-(cos2x/cos x)=sec
2. (cos x+cos y)^2 +(sin x-sin y)^2 =2+2cos(x+y)

Answer 1

For question 1. it says the steps should be: apply the double-angle formulas for sine and cosine, then simplify the expression. I got this far.
\[(2\sin x \cos x)/\sin x)-(2 \cos^2x-1)/\cos x)\]
\[2 \cos x=(2 \cos^2 x-1)/\cos x\]

Answer 2

i don't think i am reading question 1. properly,

can you type out the equation again (or draw it) please?

Answer 3

\[\frac{ \sin (2x) }{ sinx }-\frac{ \cos(2x) }{ \cos x }=\sec x\]

Answer 4

ah, i can read it properly now, thakyou

Answer 5

your working so far it good

Answer 6

you have
\[LHS = 2 \cos x-\frac{2 \cos^2 x-1}{\cos x}\]

Answer 7

now break up the fraction like this
\[\frac{a+b}{c}=\frac ac+\frac bc\]

Answer 8

\[\frac{ 2\cos^2 }{ \cos }=2 \cos\]
\[\frac{ -1 }{ \cos}=-\sec\]
Is this right?

Answer 9

right!, (becareful with the extra -sign, outside of the big fraction)

Answer 10

so you've got
\[LHS = 2\cos x-\frac{2 \cos^2 x-1}{\cos x}\\
\qquad=2\cos x-\left(\frac{2 \cos^2 x}{\cos x}+\frac{-1}{\cos x}\right)\\
\qquad=2\cos x-\left(2\cos x-\sec x\right)\\\qquad=\]

Answer 11

2 cos x-2cos x=0
-sec x=sec x ?

Answer 12

be CAREFUL with the minus signs.
\[\qquad=2\cos x-\left(2\cos x-\sec x\right)\\\qquad=2\cos x-2\cos x+\sec x\\\qquad=\]

Answer 13

Ahhhh okay, I got it now!