The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

trig question

solve:
sec(2θ - 15) = cosec 135

idk if im meant to be doing this kinda trig at AS cos this is just extra work i found this online but i'd like to know how to solve it anyway pls
what's the interval you need to solve it in?
this is defo not AS btw bc I'm pretty sure you don't learn these identities until A2
Original post by _princessxox
what's the interval you need to solve it in?
this is defo not AS btw bc I'm pretty sure you don't learn these identities until A2


0<x<360
LOL thought so but show me anyway I self taught trig so I kinda get it just no idea with the bracket
Original post by chanel_666
0<x<360
LOL thought so but show me anyway I self taught trig so I kinda get it just no idea with the bracket


you need to change the bounds first and then solve and do the inverse of the bracket
are the answers: 30, 165, 210 and 345?
Original post by _princessxox
you need to change the bounds first and then solve and do the inverse of the bracket
are the answers: 30, 165, 210 and 345?


yeh those are the answers !!!! I get as far as - cos(2θ - 15) =√2/2
Reply 5
Avatar for simon0
simon0
13
In basic form, you have:

cos⁡(t)=a, \cos(t) = a,

where t is the variable and a is a constant.

What can we do to obtain t?
Original post by simon0
In basic form, you have:

cos⁡(t)=a, \cos(t) = a,

where t is the variable and a is a constant.

What can we do to obtain t?


Cos-1(a) = t
Yeh I get that bit but how do I solve to find just theta
Reply 7
Avatar for simon0
simon0
13
What if I said:

cos⁡(t)=22, \cos(t) = \dfrac{\sqrt{2}}{2}, where t=2θ−15∘. t = 2\theta - 15^{\circ}.

Does this help?
(edited 6 years ago)
Original post by simon0
What if I said:

cos⁡(t)=22, \cos(t) = \dfrac{\sqrt{2}}{2}, where t=2θ−15∘. t = 2\theta - 15^{\circ}.

Does this help?


I don't think u get what I'm trying to say but thanks anyway
Reply 9
Avatar for simon0
simon0
13
Try:

2θ−15∘=arccos⁡(22). 2 \theta - 15^{\circ} = \arccos \Big( \dfrac{ \sqrt{2}}{2} \Big) .
Reply 10
Avatar for davros
davros
16
Original post by chanel_666
I don't think u get what I'm trying to say but thanks anyway


Do you know how sec and cosec relate to cos and sin? You can relate your original equation to an equation involving more basic trig functions.
Original post by simon0
Try:

2θ−15∘=arccos⁡(22). 2 \theta - 15^{\circ} = \arccos \Big( \dfrac{ \sqrt{2}}{2} \Big) .


Original post by chanel_666
I don't think u get what I'm trying to say but thanks anyway



once you get a value for arccos((sqrt2)/2) (plug it into your calculator), find the other values for 2(theta)-15.

once you have found all of those values you can treat each one as a simple algebraic equation where 2(theta)-15 = the first solution, rearrange to give theta, then do the same for the next solution, etc
Original post by chanel_666
I don't think u get what I'm trying to say but thanks anyway


So you've got: cos(2θ - 15) =√2/2

now you need to calculate: cos-1(√2/2)

the answer should be 45. Now find the rest of the solutions using the cos graph (remember the bounds changed) and once you have them set each one equal to 2θ - 15 then solve to find θ
Original post by _princessxox
So you've got: cos(2θ - 15) =√2/2

now you need to calculate: cos-1(√2/2)

the answer should be 45. Now find the rest of the solutions using the cos graph (remember the bounds changed) and once you have them set each one equal to 2θ - 15 then solve to find θ


ohhh i get it now thanks b
Original post by simon0
Try:

2θ−15∘=arccos⁡(22). 2 \theta - 15^{\circ} = \arccos \Big( \dfrac{ \sqrt{2}}{2} \Big) .


Original post by davros
Do you know how sec and cosec relate to cos and sin? You can relate your original equation to an equation involving more basic trig functions.


Original post by Jules kemps
once you get a value for arccos((sqrt2)/2) (plug it into your calculator), find the other values for 2(theta)-15.

once you have found all of those values you can treat each one as a simple algebraic equation where 2(theta)-15 = the first solution, rearrange to give theta, then do the same for the next solution, etc


thank u

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you