Trigonometry Watch

Harry-Pikesley
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#1
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#1
If tan(x) = t, express cos^2(x) in terms of t

My thinking:
tan(x) = sin(x)/cos(x), so sin(x)/cos(x) = t
sin^2(x)/cos^2(x) = t^2
Rearrange for cos^2(x)

But I was wrong... Help?
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Muttley79
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#2
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#2
(Original post by Harry-Pikesley)
If tan(x) = t, express cos^2(x) in terms of t

My thinking:
tan(x) = sin(x)/cos(x), so sin(x)/cos(x) = t
sin^2(x)/cos^2(x) = t^2
Rearrange for cos^2(x)

But I was wrong... Help?
Post your working ...
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Janej77
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#3
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#3
(Original post by Harry-Pikesley)
If tan(x) = t, express cos^2(x) in terms of t

My thinking:
tan(x) = sin(x)/cos(x), so sin(x)/cos(x) = t
sin^2(x)/cos^2(x) = t^2
Rearrange for cos^2(x)

But I was wrong... Help?
Maybe try tan^2(x) +1 = sec^2(x)?
Or replace sin^2(x) with 1-cos^2(x)?
Last edited by Janej77; 4 weeks ago
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Harry-Pikesley
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#4
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#4
(Original post by Muttley79)
Post your working ...
Same as what I initially wrote, but replace sin^2(x) by 1 - cos^2(x)
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Muttley79
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#5
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#5
(Original post by Harry-Pikesley)
Same as what I initially wrote, but replace sin^2(x) by 1 - cos^2(x)
Yes and then you rearranged to get?
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Harry-Pikesley
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#6
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#6
(Original post by Muttley79)
Yes and then you rearranged to get?
The closest I got to was cos^2(x) = (1 - cos^2(x))/t^2
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Muttley79
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#7
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(Original post by Harry-Pikesley)
The closest I got to was cos^2(x) = (1 - cos^2(x))/t^2
OK go back a step and multiply through by cos^2 (x) then expand and rearrange for cos^2 (x).
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Harry-Pikesley
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(Original post by Muttley79)
OK go back a step and multiply through by cos^2 (x) then expand and rearrange for cos^2 (x).
You've lost me here...
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Muttley79
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#9
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#9
(Original post by Harry-Pikesley)
You've lost me here...
I thought you said you'd replaced sin^(x) by 1 - cos^2 (x)?

So you have [1 - cos^2 (x)]/ cos^2 (x) on the left and t^2 on the right?
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jireland
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#10
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#10
You can do this by writing tanx = sinx / cosx. Then writing sinx = sqrt(1-cos^2(x)), and rearranging to find t.
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Harry-Pikesley
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#11
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#11
(Original post by Muttley79)
I thought you said you'd replaced sin^(x) by 1 - cos^2 (x)?

So you have [1 - cos^2 (x)]/ cos^2 (x) on the left and t^2 on the right?
Yes.
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Muttley79
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#12
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#12
(Original post by jireland)
You can do this by writing tanx = sinx / cosx. Then writing sinx = sqrt(1-cos^2(x)), and rearranging to find t.
same method
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Muttley79
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#13
(Original post by Harry-Pikesley)
Yes.
So now read my hint from my previous post
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Harry-Pikesley
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#14
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#14
(Original post by Muttley79)
So now read my hint from my previous post
Don't get how that helps me to express cos^2(x) in terms of t
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Janej77
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#15
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#15
(Original post by Harry-Pikesley)
Don't get how that helps me to express cos^2(x) in terms of t
You factorise so you get cos^2x on its own
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Muttley79
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#16
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#16
(Original post by Harry-Pikesley)
Don't get how that helps me to express cos^2(x) in terms of t
Expand the right hand side and then put the cos^2 (x) terms on the right and take this out as a common factor.
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RDKGames
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#17
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#17
(Original post by Harry-Pikesley)
If tan(x) = t, express cos^2(x) in terms of t

My thinking:
tan(x) = sin(x)/cos(x), so sin(x)/cos(x) = t
sin^2(x)/cos^2(x) = t^2
Rearrange for cos^2(x)

But I was wrong... Help?
You know... why not just draw a right-angled triangle with an angle x, adjacent side as 1 and opposite side as t ... hence use some simple trig to express \cos x in terms of t, then square both sides. This can dodge your issue with identities altogether, but since that's what you're struggling with, I suppose you should try and overcome it.

Alternatively, Janej77's approach of using 1+\tan^2 x \equiv \sec^2 x cleans this up nicely.
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Muttley79
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#18
(Original post by RDKGames)
You know... why not just draw a right-angled triangle with an angle x, adjacent side as 1 and opposite side as t ... hence use some simple trig to express \cos x in terms of t, then square both sides. This can dodge your issue with identities altogether, but since that's what you're struggling with, I suppose you should try and overcome it.

Alternatively, Janej77's approach of using 1+\tan^2 x \equiv \sec^2 x cleans this up nicely.
Yes but why not let the poster carry on with their method first? It's not wrong ...
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RDKGames
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#19
(Original post by Muttley79)
Yes but why not let the poster carry on with their method first? It's not wrong ...
I'm not saying they should stop doing their method... I'm just saying there are better alternatives, and since you're already helping them with their approach I had no comment on it...
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Muttley79
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(Original post by RDKGames)
I'm not saying they should stop doing their method... I'm just saying there are better alternatives, and since you're already helping them with their approach I had no comment on it...
I know there are other ways but I think it's better to let the OP complete their way and THEN advise them.
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