## anonymous one year ago Given sec(theta) = -4/3 and 90 degrees < (theta) <180 degrees ; find sin2(theta)

1. UsukiDoll

90 degrees < (theta) <180 degrees tells us that we are restricted in the 2nd quadrant... so all sine values are going to be positive...

2. anonymous

If sec(theta) = -4/3 cos(theta) = -3/4 Use sin^2(x)+cos^2(x)=1 to solve for sin^2(x). Plug in -3/4 for cos(x), square it, and subtract the result from 1.

3. UsukiDoll

$\sin^2x$ or $\sin2x$ for the last part of the question. They have different meanings.

4. UsukiDoll

$\sin^2\theta............ \sin2\theta$

5. anonymous

its Sin2theta

6. UsukiDoll

alright... sin2theta has an identity $2\sin\theta \cos\theta$

7. UsukiDoll

given that sec(theta) = -4/3 sec(theta) =$\frac{1}{-\frac{4}{3}} \rightarrow \frac{-3}{4}$

8. UsukiDoll

it's negative due to the fact that we have the restriction of 90 degrees <theta<180 degrees which means 2nd quadrant and only sine values are positive now let's ignore the - for a bit and find out what our sine really is so cosine is adjacent/hypotenuse so that means cosine 3/4

9. UsukiDoll

|dw:1434340094426:dw| so we need to use the pythagorean theorem a^2+b^2=c^2

10. UsukiDoll

in this case we solve for a. a^2+3^2=4^2 a^2+9=16 a^2 = 16-9 a^2 = 7 a = $\sqrt{7}$ ugh one problem... I know we can't pick the negative result for a because negative sides are nonexistent and doesn't make sense... I don't think decimals can be used either... not sure.

11. UsukiDoll

sec(theta) = -4/3 cos(theta) = -3/4 in 2nd quadrant hmmm... cos(theta) 3/4 = adjacent/hypotenuse and then solve for a... that should be ...why the heck do I have square root of 7 ? sine theta is $\frac{\sqrt{7}}{4}$ ... I thought all sides are supposed to have whole numbers

12. UsukiDoll

@jim_thompson5910

13. jim_thompson5910

It's perfectly possible to have a side length that isn't a whole number example: |dw:1434340852199:dw|

14. jim_thompson5910

and I think you're thinking of this triangle |dw:1434340917667:dw|

15. UsukiDoll

O_O! really? so that sine is $\frac{\sqrt{7}}{4}$ ??

16. jim_thompson5910

yeah sin(theta) = sqrt(7)/4

17. jim_thompson5910

use that to find sin(2*theta)

18. UsukiDoll

ok so all we need to do is solve $\sin(2 \theta) = 2 \sin\theta \cos\theta$

19. UsukiDoll

OP said sin 2 theta! No exponents.

20. UsukiDoll

sine is positive due to that 2nd quadrant restriction so $2\times \frac{\sqrt{7}}{4} \times \frac{-3}{4}$

21. UsukiDoll

$2 \times \frac{-3\sqrt{7}}{16}$

22. UsukiDoll

$\frac{-6\sqrt{7}}{16}$

23. UsukiDoll

$\frac{-3\sqrt{7}}{8}$

24. UsukiDoll

that's it.

25. anonymous

Thank you. I was confused a bit but now I see where I messed up.

26. UsukiDoll

can you give me medal ^_^

27. anonymous

sure :)

28. UsukiDoll

yay