Question

If sin (A) = 5/13 with A in QII, find sin (A/2).

Answer 1

Firstly you need to find cos(A) also..

Answer 2

\[\cos(A) = \sqrt{1 - \sin^2(A)}\]

Answer 3

Can you find \(\cos(A)\) ??

Answer 4

sorry
im here now

Answer 5

\[\frac{ -\sqrt{26} }{ 26}\]

Answer 6

i think, im stuck here

Answer 7

How you got \(\sqrt{26}\) there in numerator?

Answer 8

multiplied by sqrt? is that wrong?

Answer 9

Let us solve this part only first:
\[1 - \frac{12}{13}\]

Answer 10

Can you tell after solving what will you get?

Answer 11

1/13

Answer 12

Oh, I see now, where are you going.. Your answer is right in my opinion..

Answer 13

but it was wrong

Answer 14

Yeah it is wrong.

Answer 15

your values of cos(A) is wrong I think.

Answer 16

\[\cos(A) = \sqrt{1 - \sin^2(A)} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \pm \frac{12}{13}\]

Answer 17

Up to here, you got?

Answer 18

I want to ask why have you chosen +12/13 for cos(A), As A is in second quadrant, and cos in second quadrant is negative, so don't you think you should use Negative 12/13 there?

Answer 19

mistake

Answer 20

Can you correct it now?

Answer 21

-5\[-5\sqrt{26}/26?\]

Answer 22

still wrong

Answer 23

what do i do?

Answer 24

Choose:
\[\cos(A) = -\frac{12}{13}\]

Answer 25

ok

Answer 26

\[1 + \frac{12}{13} = ??\]

Answer 27

yes i did this already...

Answer 28

\[\sin(\frac{A}{2}) = \sqrt{\frac{25}{26}}\]

Answer 29

Now, tell me will you take positive value or negative value?

Answer 30

+

Answer 31

\[\sin(\frac{A}{2}) = \sqrt{\frac{25}{26}} = \pm \sqrt{\frac{25}{26}}\]

Answer 32

Okay, I agree with you, but why?

Answer 33

sin is y and y is positive in qII