Ok so I have tan=o/a and cos=a/h with a triangle drawn, how do i go about incorporating the sides of a triangle into my answer though? :/

Don't think Im following what you're saying

Reply 5

SherlockHolmes

Original post by Davelittle

Ok so I have tan=o/a and cos=a/h with a triangle drawn, how do i go about incorporating the sides of a triangle into my answer though? :/

Don't think Im following what you're saying

\displaystyle \tan \theta = \frac{2t}{1-t^2} = \frac{opp}{adj}

Reply 6

Davelittle

Original post by SherlockHolmes

\displaystyle \tan \theta = \frac{2t}{1-t^2} = \frac{opp}{adj}

Ok but how do I then incorporate the hypotenuse? Am I missing something really obvious here? Do I have to use pythagoras?

Reply 7

SherlockHolmes

Original post by Davelittle

Do I have to use pythagoras?

Yes.

Reply 8

Notnek

Original post by Davelittle

Ok so I have tan=o/a and cos=a/h with a triangle drawn, how do i go about incorporating the sides of a triangle into my answer though? :/

Don't think Im following what you're saying

E.g.

\tan \theta = \frac{3}{4}

, you could draw a right-angled triangle with opposite length 3 and adjacent length 4.

Since the hypotenuse is 5,

\cos \theta = \frac{4}{5}

.

Can you see how to use this method for your question?

Alternatively, use

\sec^2 \theta = 1+\tan^2 \theta

to express cos in terms of tan.

Reply 9

Davelittle

Original post by SherlockHolmes

Yes.

\cos \theta = \frac{1-t^2}{1+t^2}

Reply 10

SherlockHolmes

Original post by Davelittle

\cos \theta = \frac{1-t^2}{1+t^2}

Yes that is correct. You can find an expression for sin now too.

Reply 11

Davelittle

Original post by SherlockHolmes

Yes that is correct. You can find an expression for sin now too.

\sin \theta= \frac{2t}{1+t^2}

Reply 12

Notnek

Original post by Davelittle

\cos \theta = \frac{1-t^2}{1+t^2}

That's right.

These t formulae are very useful for finding integrals since making a substitution t=

\tan \frac{x}{2}

gives

\frac{dt}{dx} = \sec^2 \frac{x}{2} = 1+\tan^2\frac{x}{2} = 1+t^2

so you can express sums of sin cos and tan as rational functions of t which can be easily integrated.

This may be explained later in your textbook.

Reply 13

Davelittle

Original post by notnek

That's right.

These t formulae are very useful for finding integrals since making a substitution t=

\tan \frac{x}{2}

gives

\frac{dt}{dx} = \sec^2 \frac{x}{2} = 1+\tan^2\frac{x}{2} = 1+t^2

so you can express sums of sin cos and tan as rational functions of t which can be easily integrated.

This may be explained later in your textbook.

I just checked, it isn't even in the text book but its given on the formula sheet :/

How did they expect us to find cos in part 1 before we found tan (part 3)?

Reply 14

Notnek

Original post by Davelittle

I just checked, it isn't even in the text book but its given on the formula sheet :/

How did they expect us to find cos in part 1 before we found tan (part 3)?

You could start with the cos double angle formula:

\displaystyle \cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \cos^2 \frac{x}{2}\left(1-tan^2 \frac{x}{2}\right)=...

If it's not in your textbook, would you like an example question to see how these t-formulae can be used in integration, something that could be useful in exams?

Reply 15

SherlockHolmes

Original post by Davelittle

\sin \theta= \frac{2t}{1+t^2}

Yes.

Original post by Davelittle

I just checked, it isn't even in the text book but its given on the formula sheet :/

How did they expect us to find cos in part 1 before we found tan (part 3)?

There are other methods of finding cos in terms of t. See below:

Spoiler

(edited 9 years ago)

Reply 16

Davelittle

Original post by notnek

You could start with the cos double angle formula:

\displaystyle \cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \cos^2 \frac{x}{2}\left(1-tan^2 \frac{x}{2}\right)=...

If it's not in your textbook, would you like an example question to see how these t-formulae can be used in integration, something that could be useful in exams?