Simplfiying trig expressions=/

[KristenD77]

---

Members don't see this ad.

Hey everyone so I get the basics of simplifying these expressions. I know that I derive eevrything from this formula Sin^(2)X+Cos^(2)X=1. However I came across this one problem and the answer isn't given.

2SinX/ 1/sec * 1/csc
I know that Sec is 1/cos and Csc is 1/sin
Therefore what I did was
2sinX *Cos/1*1/sin= Sin^(2)*Cos^(2)
( I did keep change flip for the division and then canceled stuff out)
Is this correct?

 

[Krentist_72]

I think your mistake is that you converted 1/csc incorrectly. You're right in saying csc = 1/sin. However, the problem was initially 1/csc, and you incorrectly simplified this to 1/sin. It should be 1/(1/sin) which turns out as just sin.

If 1/csc->1/sin, as you did, then csc=sin, which is wrong, although it would make trig a lot easier! Hope this helps!

 

[KristenD77]

I think your mistake is that you converted 1/csc incorrectly. You're right in saying csc = 1/sin. However, the problem was initially 1/csc, and you incorrectly simplified this to 1/sin. It should be 1/(1/sin) which turns out as just sin.

If 1/csc->1/sin, as you did, then csc=sin, which is wrong, although it would make trig a lot easier! Hope this helps!

Thank you so much! I see that now!

 

[KristenD77]

so then the answer should be tan I think

 

[Krentist_72]

Thank you so much! I see that now!

Also now that I'm looking at it again, I think you did the same mistake for 1/sec(x)... If 2sin(x)/(1/sec(x)) and sec(x)=1/cos(x), then 1/(1/cos(x)) is simply cos(x). Plug that back in and it should be 2sin(x)/cos(x)... Not 2sin(x)*cos(x)/1, as you put.

I can't really tell what the correct answer would be based off your typing, because where the functions are placed in relation to the fraction totally changes the answers. Heads up for the future, parentheses are vitally important for typing out math problems.
i.e. 2sin(x)/(1/sec(x)*1/csc(x)) -> 2sin(x)/(cos(x)*sin(x)) -> 2sec(x)
in comparison to
2sin(x)/(1/sec(x)) *1/csc(x) -> 2tan(x) * 1/csc(x) -> 2tan(x)sin(x)

 

[aaa127]

https://www.mathsisfun.com/algebra/trig-magic-hexagon.html

 

[Krentist_72]

The problem can be looked at 2 different ways depending on how the expression is written.
See attached picture.

Feel free to correct me if I'm wrong, but I think that converting 1/sec(x) -> cos(x) leaves it as a denominator under 2sin(x), making it 2sin(x)/cos(x) *sin(x) and the final answer being 2tan(x)sin(x)?

 

[DAT Destroyer]

Feel free to correct me if I'm wrong, but I think that converting 1/sec(x) -> cos(x) leaves it as a denominator under 2sin(x), making it 2sin(x)/cos(x) *sin(x) and the final answer being 2tan(x)sin(x)?

sorry was going to fast will respond again in a bit.

 

[DAT Destroyer]

Hey everyone so I get the basics of simplifying these expressions. I know that I derive eevrything from this formula Sin^(2)X+Cos^(2)X=1. However I came across this one problem and the answer isn't given.

2SinX/ 1/sec * 1/csc
I know that Sec is 1/cos and Csc is 1/sin
Therefore what I did was
2sinX *Cos/1*1/sin= Sin^(2)*Cos^(2)
( I did keep change flip for the division and then canceled stuff out)
Is this correct?

 

[KristenD77]

View attachment 204592

Thank you guys!