Trigonometry

ImageUploadedByStudent Room1428853727.292462.jpg

See the top yellow box in the photo. I understand this concept but how do you know whether e.g. the solutions of cosx=0 satisfy an equation or not? I never know whether I can divide by a constant on both sides and when I shouldn't?

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Reply 1

davros

Original post by anoymous1111

You can always divide by a 'constant' (unless that constant is 0, obviously).

You need to treat each equation on a case by case basis and if you are about to divide by something, ask yourself whether that 'something' can ever be 0.

Reply 2

anoymous1111

Original post by davros

but surely cosx could be 0?

Reply 3

Mr M

Study Forum Helper

Original post by anoymous1111

but surely cosx could be 0?

If cos x = 0 then what would x be? Does this satisfy the original equation? If it doesn't you won't lose part of the solution if you divide.

Reply 4

anoymous1111

Original post by Mr M

If cos x = 0 then what would x be? Does this satisfy the original equation? If it doesn't you won't lose part of the solution if you divide.

ImageUploadedByStudent Room1428856691.475677.jpg

Ahhh ok! Would this be the correct way of figuring out whether cosx=0 satisfies the equation?

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Reply 5

anoymous1111

Original post by Mr M

If cos x = 0 then what would x be? Does this satisfy the original equation? If it doesn't you won't lose part of the solution if you divide.

Or is it just: ImageUploadedByStudent Room1428856909.468775.jpg

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Reply 6

Biomadman

In this case you can always divide by cos to simplfy eqauiton.You do not lose or gain solutions as the cos disaapering from one side is accounted for in the other turning one side to tan.You then solve from there

Reply 7

Mr M

Study Forum Helper

Original post by anoymous1111

...

You are making this far too difficult. You shouldn't be finding sin x = 0.

Solve cos x = 0 for x.

Now substitute your answer into the original equation and spot that it doesn't work.

(edited 9 years ago)

Reply 8

anoymous1111

Original post by Mr M

You are making this far too difficult. You shouldn't be finding sin x = 0.

Solve cos x = 0 for x.

Now substitute your answer into the original equation and spot that it doesn't work.

Can you write out the working for me? Sorry I'm just not getting it :frown:

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Reply 9

anoymous1111

Original post by Mr M

You are making this far too difficult. You shouldn't be finding sin x = 0.

Solve cos x = 0 for x.

Now substitute your answer into the original equation and spot that it doesn't work.

Oh I think I got the wrong end of the stick... Is it because cos(0) doesn't = 0 and so you can divide by cosx. Whereas if it did equal 0 then you can't divide the other side of the equation by 0.

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(edited 9 years ago)

Reply 10

davros

Original post by anoymous1111

The point is that there are NO angles which have cos x = 0 and sin x = 0 simultaneously. So whenever you have an equation like sin x = kcos x it is safe to divide by cos x, because any angle that made cos x = 0 would never satisfy the original equation.