Question

Trig identity help (again :)

Answer 1

This one's hard!

\[\frac{ 2 }{ \sqrt{3}cosx+sinx } = \sec(\frac{ \pi }{ 6 } - x)\]

Answer 2

Yes, it is a hard one:)

Answer 3

One question before we do this. Are we allowed to perform operations on both sides, or we only get to play with one of the sides?

Answer 4

I think.... both... but I'm not really sure

Answer 5

I have four boxes to fill in and then 4 reasons (like a geometry proof)

Answer 6

I think I can work with both sides

Answer 7

Reason and Statement chart?

Answer 8

I mean the other way around ... Statement and reason, chart ... that?

Answer 9

\(\color{purple}{\displaystyle \frac{2}{\sqrt{3}\cos x+\sin x}=\sec \left(\frac{\pi}{6}-x\right) }\)

\(\color{purple}{\displaystyle 2\cos\left(\frac{\pi}{6}-x\right)=\sqrt{3}\cos x+\sin x }\)

\(\color{purple}{\displaystyle 2\sin\left(\frac{\pi}{6}\right)\sin\left(x\right)+2\cos\left(\frac{\pi}{6}\right)\cos\left(x\right)=\sqrt{3}\cos x+\sin x }\)

Answer 10

zepdrix can you check my stuff real quick?

Answer 11

Wait... how did you go from step 1 to step 2?

Answer 12

this is in essence the proof ... (well, you can find those sine and cosine of π/6 using calculator), but all I used was:

(1) 1/sec(x) = cos(x)
(2) cos(x-y)=cos(x)cos(y)+sin(x)sin(y)

Answer 13

So, from step 1 to step 2,

a. Multiplied both sides by `3√cosx+sinx`

b. Re-wrote `sec(π/6 -x)` as `1/cos(π/6 -x)`,
and multiplied both sides times `cos(π/6 -x)`.

Answer 14

i can write it out if you like ... (let me know, and I will do so)

Answer 15

Wouldn't you get just 2 on the left side if you multiply both sides by sqrt3cosx+sinx ?

Answer 16

\(\color{blue}{\displaystyle \frac{2}{\sqrt{3}\cos x+\sin x}=\sec \left(\frac{\pi}{6}-x\right) }\)

\(\color{blue}{\displaystyle \frac{2}{\sqrt{3}\cos x+\sin x}\color{red}{\times (\sqrt{3}\cos x+\sin x)}=\sec \left(\frac{\pi}{6}-x\right)\color{red}{\times (\sqrt{3}\cos x+\sin x)} }\)

\(\color{blue}{\displaystyle 2=\sec \left(\frac{\pi}{6}-x\right)\cdot (\sqrt{3}\cos x+\sin x) }\)

\(\color{blue}{\displaystyle 2=\color{red}{\frac{1}{\cos \left(\displaystyle\frac{\pi}{6}-x\right)}}\cdot (\sqrt{3}\cos x+\sin x) }\)

\(\color{blue}{\displaystyle \color{red}{\cos \left(\displaystyle\frac{\pi}{6}-x\right)\times}2=\color{red}{ \cos \left(\displaystyle\frac{\pi}{6}-x\right)\times}\frac{1}{\cos \left(\displaystyle\frac{\pi}{6}-x\right)}\cdot (\sqrt{3}\cos x+\sin x) }\)

Answer 17

This is how I get to

\(\color{purple}{\displaystyle 2\cos\left(\frac{\pi}{6}-x\right)=\sqrt{3}\cos x+\sin x }\)

Answer 18

Sorry for skipping that much

Answer 19

That is perfect! Thank you so much. Lots of good stuff for me to chew on... I see what you did.

Answer 20

Nice .... np

Answer 21

That is perfect! Thank you so much. Lots of good stuff for me to chew on... I see what you did.