1. ## Can someone check my trig work please?

2. Well, to start

$-\frac{25\pi}{4} = -\frac{24\pi}{4} - \frac{\pi}{4}$

$= -6\pi - \frac{\pi}{4}$.

So which basic angle do you think $-\frac{25\pi}{4}$ is the same as?

3. It is preferred you post one question per thread.

Consider $-2\pi$ gives the same postition as $0$ so

$\frac{-25\pi}{4} =\frac{(-8-8-8-1)\pi}{4}=\frac{-8\pi}{4}+\frac{-8\pi}{4}+\frac{-8\pi}{4}+\frac{-1\pi}{4}$ $=-2\pi-2\pi-2\pi-\frac{\pi}{4} = -\frac{\pi}{4}$

4. Originally Posted by pickslides
It is preferred you post one question per thread.

Consider $-2\pi$ gives the same postition as $0$ so

$\frac{-25\pi}{4} =\frac{(-8-8-8-1)\pi}{4}=\frac{(-8-8-8-1)\pi}{4}=\frac{-8)\pi}{4}+\frac{-8)\pi}{4}+\frac{(-8)\pi}{4}+\frac{-1\pi}{4}$ $-2/pi-2/pi-2/pi-\frac{\pi}{4} = -\frac{-1\pi}{4}$
Be careful with your equal signs...

$-\frac{25\pi}{4} \neq -\frac{\pi}{4}$.

But you can say that the POSITION of the angle $-\frac{25\pi}{4}$ is the same POSITION as $-\frac{\pi}{4}$.

5. Originally Posted by Prove It

But you can say that the POSITION of the angle $-\frac{25\pi}{4}$ is the same POSITION as $-\frac{\pi}{4}$.
But I did mention the position ;0)

6. Originally Posted by pickslides
But I did mention the position ;0)
Yes, but you're still using equals signs where they shouldn't be used...

Anyway, Question 2 is correct.

In Question 3, you have $\sin{\theta}$ correct. For the rest you could have used the identities:

$\sin^2{\theta} + \cos^2{\theta} = 1$

$\frac{\sin{\theta}}{\cos{\theta}} = \tan{\theta}$

$\tan^2{\theta} + 1 = \sec^2{\theta}$

$\cot{\theta} = \frac{1}{\tan{\theta}}$.

If you were going to use the triangle, you have used Pythagoras' Theorem incorrectly.

It should be $A^2 = H^2 - O^2$. You added them.

7. thanks! correcting.

Sorry about more than one Q per post.

8. Originally Posted by Endymion
thanks! correcting.

Sorry about more than one Q per post.
Your Question 4 and 5 are correct.

Q. 6. (a) is correct but I would have done

$\csc{\frac{\pi}{3}} = \frac{1}{\sin{\frac{\pi}{3}}} = \frac{1}{\sin{60^{\circ}}}$.

(b) is correct.

Q. 7. Remember that $\sec^2{\frac{\pi}{6}} = \frac{1}{\cos^2{\frac{\pi}{6}}} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

So $\sec^2{\frac{\pi}{3}} - 4 = \frac{4}{3} - 4 = \frac{4}{3} - \frac{12}{3} = -\frac{8}{3}$.