Help with a few trig problems. :)

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layzer

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Joined
Nov 3, 2013
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Hello everyone i seem to be having trouble with the following problems

1. Find secA/2 with CosA=1/2

i use the formula for cosA/2 and just use the inverse.
= - sqrt(1+1/2)/(2)
=-sqrt(3)/(4)
= -sqrt(3)/(2)

correct? and sec would just be 2/sqrt3? but for some reason that is not the right answer.

2. Use the quadratic formula to find all degree solutions and θ if 0° ≤ θ < 360°.
 Use a calculator to approximate all answers to the nearest tenth of a degree. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)2 cos2 θ − 2 cos θ − 1 = 0 I dont even know where to start with this one.

Thanks!
 
Last edited:
layzer said:
Hello everyone i seem to be having trouble with the following problems

1. Find secA/2 with CosA=1/2

i use the formula for cosA/2 and just use the inverse.
= - sqrt(1+1/2)/(2).................... Why do you have negative sign?
=-sqrt(3)/(4)
= -sqrt(3)/(2)

correct? and sec would just be 2/sqrt3? but for some reason that is not the right answer.

2. Use the quadratic formula to find all degree solutions and θ if 0° ≤ θ < 360°.
 Use a calculator to approximate all answers to the nearest tenth of a degree. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)2 cos2 θ − 2 cos θ − 1 = 0

Do you know the quadratic formula?

If not - use Google to find out about it and if still cannot proceed - write back...


I dont even know where to start with this one.

Thanks!
Click to expand...
.
 
layzer said:
because cos is in QII so it would be negative
Click to expand...

In that case cos(A) would have been negative!!

But you posted cos(A) = 1/2..... a positive number (either QI or QIV)
 
For the following, assume that all the given angles are in simplest form, so that if A is in QIV you may assume that 270° < A < 360°.


i suppose it would have helped if i posted this.
 
layzer said:
Hello everyone i seem to be having trouble with the following problems

1. Find secA/2 with CosA=1/2

i use the formula for cosA/2 and just use the inverse.
= - sqrt(1+1/2)/(2)
=-sqrt(3)/(4)
= -sqrt(3)/(2)

correct? and sec would just be 2/sqrt3?

Now where did the negative sign go??

but for some reason that is not the right answer.

2. Use the quadratic formula to find all degree solutions and θ if 0° ≤ θ < 360°.
 Use a calculator to approximate all answers to the nearest tenth of a degree. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)2 cos2 θ − 2 cos θ − 1 = 0 I dont even know where to start with this one.

Thanks!
Click to expand...
.
 
alright so it would be -2/sqrt3?

for the second one. i used the quadratic formula and came up with 2sqrt12/4 which simplified is sqrt12/2 but now i dont know where to go from there.
 
layzer said:
alright so it would be -2/sqrt3?

for the second one. i used the quadratic formula and came up with 2sqrt12/4 which simplified is sqrt12/2 but now i dont know where to go from there.
Click to expand...

That is incorrect.

Please share your work.
 
plug in 2 sqrt(2^2-4(2)(-1))/2(2) = 2sqrt(4-(-8))/4 = 2sqrt(12)/4 = cancel 2 and simply 4 to 2 =sqrt12/2
 
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