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C3 Trigonometry - Where did I go wrong?

Original post by ps1265A

...

For the tan/sec question,

+\sqrt{3}

in the first line has changed to

-\sqrt{3}

in the second line.

For the other question, your working on the right of your page is fine up until the last line where you wrote

\sin \theta = \frac{5}{3}

instead of the correct version :

\sin \theta = \frac{3}{5}

(edited 9 years ago)

Reply 2

ps1265A

Original post by arkanm

From the first line to the second, you did:

"

\sec^2 \theta -(1+\sqrt{3})\tan \theta +\sqrt{3}=1

\sec^2 \theta-\left [\tan \theta+\sqrt{3}+\sqrt{3}\tan \theta +3\right ] = 1

"

When it should have been:

"

\sec^2 \theta -(1+\sqrt{3})\tan \theta +\sqrt{3}=1

\sec^2\theta - \left[\tan \theta + \sqrt{3} \tan \theta - \sqrt{3}\right] =1

I still don't see how you've formed a "-" square root 3

Reply 3

peterith

How do you guys type sin tan cos theta like that?

Reply 4

wizardmunchahoy

Hi!

Firstly, this should help with regards to typing mathematical notation using LaTeX on TSR. http://www.thestudentroom.co.uk/wiki/LaTex

Re:

5 \cos \theta = 3 \cot \theta

notnek is right,

\sin \theta = \frac{3}{5}

(not 5/3) is the only correction that needs to be made.

Re: other question, don't multiply anything out - just keep it simple!

Hints:
1) Use the formula

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

.
2) Cancel the 1 from both sides.
3) Solve the quadratic equation for

\tan \theta

.
4) Use calculator to arctan the two roots.

Hope this helps. The final answers are nice! :smile:

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C3 Trigonometry - Where did I go wrong?

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