# TrigonometryWatch

Thread starter 4 weeks ago
#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?
0
reply
4 weeks ago
#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 ...
0
reply
4 weeks ago
#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
0
reply
Thread starter 4 weeks ago
#4
(Original post by Muttley79)
Post your working ...
Same as what I initially wrote, but replace sin^2(x) by 1 - cos^2(x)
0
reply
4 weeks ago
#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?
0
reply
Thread starter 4 weeks ago
#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
0
reply
4 weeks ago
#7
(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).
0
reply
Thread starter 4 weeks ago
#8
(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...
0
reply
4 weeks ago
#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?
0
reply
4 weeks ago
#10
You can do this by writing tanx = sinx / cosx. Then writing sinx = sqrt(1-cos^2(x)), and rearranging to find t.
0
reply
Thread starter 4 weeks ago
#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.
0
reply
4 weeks ago
#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
0
reply
4 weeks ago
#13
(Original post by Harry-Pikesley)
Yes.
So now read my hint from my previous post
0
reply
Thread starter 4 weeks ago
#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
0
reply
4 weeks ago
#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
Posted on the TSR App. Download from Apple or Google Play
0
reply
4 weeks ago
#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.
0
reply
4 weeks ago
#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 , adjacent side as and opposite side as ... hence use some simple trig to express in terms of , 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 cleans this up nicely.
0
reply
4 weeks ago
#18
(Original post by RDKGames)
You know... why not just draw a right-angled triangle with an angle , adjacent side as and opposite side as ... hence use some simple trig to express in terms of , 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 cleans this up nicely.
Yes but why not let the poster carry on with their method first? It's not wrong ...
0
reply
4 weeks ago
#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...
0
reply
4 weeks ago
#20
(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.
0
reply
X

Write a reply...
Reply
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Bath
Undergraduate Virtual Open Day Undergraduate
Sat, 23 Feb '19
• Ravensbourne University London
School of Design, School of Media Further education
Sat, 23 Feb '19
• Leeds Trinity University
PGCE Open Day Further education
Sat, 23 Feb '19

### Poll

Join the discussion

#### Do you have a food intolerance or allergy?

Yes - a food intolerance (52)
13%
Yes - a food allergy (43)
10.75%
Yes - an autoimmune disorder (i.e coeliac, colitis) (11)
2.75%
Yes - I have an intolerance and allergy (9)
2.25%
No (285)
71.25%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.