VeroZarate Group Title Verify that (sin x)(tan x cos x – cot x cos x) = 1 – 2 cos2x one year ago one year ago

1. binarymimic Group Title

did you write this down correctly

2. VeroZarate Group Title

yes

3. VeroZarate Group Title

its very confusing

4. binarymimic Group Title

are you sure the last one shouldn't be $1 - 2 \cos^{2}(x)$ instead of 1 - 2 cos 2x

5. VeroZarate Group Title

yes that is what i meant sorry it didnt square it

6. binarymimic Group Title

Ok try starting by rewriting tan x as sin x / cos x and cot x as cos x / sin x

7. binarymimic Group Title

$(\sin x)(\frac{ \sin x }{ \cos x } * \cos x - \frac{ \cos x }{ \sin x } * \cos x) = 1 - 2 \cos^{2} x$

8. VeroZarate Group Title

ok so cos^2x/sinx ?

9. VeroZarate Group Title

at the end ?

10. binarymimic Group Title

it would become $(\sin x)(\sin x - \frac{ \cos^{2} x }{ \sin x }) = 1 - 2 \cos^{2} x$ then $\sin^{2} x - \cos^{2} x = 1 - 2 \cos^{2} x$

11. binarymimic Group Title

last step would just be add 2 cos^2(x) to both sides

12. VeroZarate Group Title

i dont understand where the 2cos^2x came from

13. binarymimic Group Title

there is already a -2 cos^2 x in the problem. the last step is to add 2 cos^2 x to both sides this way the right side becomes: 1 and the left side becomes: sin^2 x + cos^2 x

14. binarymimic Group Title

and we have the identity $\sin^{2} x + \cos^{2} x = 1$

15. VeroZarate Group Title

oh ok i understand! thanks! do you think you can help me out with another one?

16. binarymimic Group Title

sure but they recommend you close this question and start another one

17. binarymimic Group Title

brb though

18. VeroZarate Group Title

ok so ill open another