# Maths Trig help please Watch

Announcements

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!

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

0

reply

Report

#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.

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

0

reply

(Original post by

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.

**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

reply

Report

#6

(Original post by

Thanks so much for your reply!

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

**science_geeks**)Thanks so much for your reply!

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

Either from cast or symmetry. When you take the inverse cos, the answer could be positive or negative.

0

reply

(Original post by

cos(20) = cos(-20)

Either from cast or symmetry. When you take the inverse cos, the answer could be positive or negative.

**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

reply

Report

#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.

Note you can easily check whether the answer is correct by subbing it back into the original equation.

0

reply

(Original post by

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.

**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.

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

2x-40 = x-20

x = 20

Then cosx = 20? So sorry still pretty confused!

0

reply

Report

#10

(Original post by

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!

**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!

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

reply

(Original post by

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?

**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?

0

reply

Report

#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.

Bit of a rubbish question to mix up radians and degrees.

0

reply

(Original post by

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.

**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.

1

reply

Report

#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 )

so

cos( 2M ) = cos (M )

now replace the LHS with a function of cos ( M )

0

reply

Report

#15

(Original post by

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 )

**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 )

1

reply

**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 )

0

reply

Report

#17

(Original post by

Ah ofc! Thank you, would I not have to use Double angle formulae for this question?

**science_geeks**)Ah ofc! Thank you, would I not have to use Double angle formulae for this question?

0

reply

(Original post by

you should be able to spot which of the 3 versions of Cos (2M) to use

**the bear**)you should be able to spot which of the 3 versions of Cos (2M) to use

0

reply

Report

#19

(Original post by

Awesome, is there any way to do this without using double angles?

**science_geeks**)Awesome, is there any way to do this without using double angles?

0

reply

0

reply

X

### Quick Reply

Back

to top

to top