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#1
Hi,

I'm trying
to solve this trig equation but I'm having a lot of difficulty starting it off.

please can someone help start me off!

Thanks!  Last edited by science_geeks; 2 weeks ago
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#2 0
#3
0
2 weeks ago
#4
cos(a) = cos(b)
means (taking inverse cos)
a = +/-b + 360k
where k is an integer. Which answer do you want? Remember cos is symmetric about 0 degrees.

In your working, you've made a mistake not putting (x-20) in brackets when you take inverse cos. Hence the sign of 20 is wrong. Also, the 90 in the right as you can't take the inverse cos of two terms on the left as well.
Last edited by mqb2766; 2 weeks ago
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#5
(Original post by mqb2766)
cos(a) = cos(b)
means (taking inverse cos)
a = +/-b + 360k
where k is an integer. Which answer do you want? Remember cos is symmetric about 0 degrees.

In your working, you've made a mistake not putting (x-20) in brackets when you take inverse cos. Hence the sign of 20 is wrong.

Sorry - I'm not quite following what you mean by a = +/-b + 360k? I use CAST diagrams to find the other solutions because I don't really understand the symmetrical graph stuff
0
2 weeks ago
#6
(Original post by science_geeks)

Sorry - I'm not quite following what you mean by a = +/-b + 360k? I use CAST diagrams to find the other solutions because I don't really understand the symmetrical graph stuff
cos(20) = cos(-20)
Either from cast or symmetry. When you take the inverse cos, the answer could be positive or negative.
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#7
(Original post by mqb2766)
cos(20) = cos(-20)
Either from cast or symmetry. When you take the inverse cos, the answer could be positive or negative. So like this?
0
2 weeks ago
#8
No. Leave the cos on each side of the equation and take inverse cos. You can't (easily) take it to the other side and take the inverse cos of the sum of two cos. You're overcomplicating it.

Note you can easily check whether the answer is correct by subbing it back into the original equation.
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#9
(Original post by mqb2766)
No. Leave the cos on each side of the equation and take inverse cos. You can't (easily) take it to the other side and take the inverse cos of the sum of two cos. You're overcomplicating it.

Note you can easily check whether the answer is correct by subbing it back into the original equation.
Okay, so leaving the cos on both side wouldn't that leave me with:

Cos-1(2x-40) = Cos-1(x-20)
2x-40 = x-20
x = 20

Then cosx = 20? So sorry still pretty confused!
0
2 weeks ago
#10
(Original post by science_geeks)
Okay, so leaving the cos on both side wouldn't that leave me with:

Cos-1(2x-40) = Cos-1(x-20)
2x-40 = x-20
x = 20

Then cosx = 20? So sorry still pretty confused!
One solution is x=20 degrees and cos(20) = ?. You can check as
2*20-40 = 0
20-20 = 0
so you have cos(0) = 1 on both sides. The equation balances.

There are other solutions though as mentioned above. You've not posted the full question, so not sure how much they want?
0
#11
(Original post by mqb2766)
One solution is x=20 degrees and cos(20) = ?. You can check as
2*20-40 = 0
20-20 = 0
so you have cos(0) = 1 on both sides. The equation balances.

There are other solutions though as mentioned above. You've not posted the full question, so not sure how much they want?
Okay, and this is the full question 0
2 weeks ago
#12
so you need all solutions from 0 to 360 degrees or 0 to 2pi radians. So see #4 for the multiple solutions.

Bit of a rubbish question to mix up radians and degrees.
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#13
(Original post by mqb2766)
so you need all solutions from 0 to 360 degrees or 0 to 2pi radians. So see #4 for the multiple solutions.

Bit of a rubbish question to mix up radians and degrees.
I agree, thank you for your help with this! 1
2 weeks ago
#14
you can call x - 20 a letter let's say M

so

cos( 2M ) = cos (M )

now replace the LHS with a function of cos ( M )
0
2 weeks ago
#15
(Original post by the bear)
you can call x - 20 a letter let's say M

so

cos( 2M ) = cos (M )

now replace the LHS with a function of cos ( M )
Even simpler, should have spotted that.
1
#16
(Original post by the bear)
you can call x - 20 a letter let's say M

so

cos( 2M ) = cos (M )

now replace the LHS with a function of cos ( M )
Ah ofc! Thank you, would I not have to use Double angle formulae for this question?
0
2 weeks ago
#17
(Original post by science_geeks)
Ah ofc! Thank you, would I not have to use Double angle formulae for this question?
you should be able to spot which of the 3 versions of Cos (2M) to use 0
#18
(Original post by the bear)
you should be able to spot which of the 3 versions of Cos (2M) to use Awesome, is there any way to do this without using double angles? 0
2 weeks ago
#19
(Original post by science_geeks)
Awesome, is there any way to do this without using double angles? u no likee dubl angol ? 0
#20
(Original post by the bear)
u no likee dubl angol ? Ah no, I haven't been taught it yet!
0
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