## anonymous one year ago 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)

1. anonymous

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. $(2\sin x \cos x)/\sin x)-(2 \cos^2x-1)/\cos x)$ $2 \cos x=(2 \cos^2 x-1)/\cos x$

2. UnkleRhaukus

i don't think i am reading question 1. properly, can you type out the equation again (or draw it) please?

3. anonymous

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

4. UnkleRhaukus

ah, i can read it properly now, thakyou

5. UnkleRhaukus

your working so far it good

6. UnkleRhaukus

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

7. UnkleRhaukus

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

8. anonymous

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

9. UnkleRhaukus

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

10. UnkleRhaukus

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

11. anonymous

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

12. UnkleRhaukus

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

13. anonymous

Ahhhh okay, I got it now!