Question

hello...

need help please!

given: sin A is 4/5 with A in Q2, and cos B = -2/sqrt13 with B in Q3.

Find the Following:

Cos(A-B)

sin2A

tan(B/2)

Answer 1

cos A = -3/5; sinB = -3/sqrt(13)

Answer 2

okay, i got that far, but i just cant figure out how to put them in the equations
been at this for 2 hours :/

Answer 3

You have all the clues from the question and my posts above - just plug them into the equation in my first post in this thread.

Answer 4

cos(A-B) = cosAcosB + sinAsinB
sin2A = 2sinAcosA
tan(B) = [2tan(B/2)]/[1-tan^2(B/2)] (use tanB = sinB/cosB)

Answer 5

so.... say i did sin2A = 2SinACosA...

so it would be 2 * 4/5 *-3/sqrt13

then when you muliply it out, you get -24/5/sqrt13, right?

Answer 6

when i do cos(A-B), i get 306/325, i think im doing the math wrong, could you show me?

Answer 7

In response to your previous question, it should read

\[\frac{ -24 }{ 25 }\], not -24/5/sqrt13.

Answer 8

In response to your later question,
cos(A-B) = (-3/5)(-2/sqrt13) + (4/5){-3/[sqrt(13)]}
= 6/(5sqrt13) - 12/(5sqrt13)
= -6/(5sqrt13) (Answer)

Answer 9

could you explain to me how you got -23/25 on the sin2a answer, because i dont how how you got that

Answer 10

You used the value of sinB instead of cosA - that was the reason for the miscalculation.

Answer 11

cosA = -3/5, not -3/sqrt(13) (which is the value of sinB instead) ...

Answer 12

so... for 2sinA its 2 * 4/5 * -3/5? i thought you had to use one of each values since it gave you two functions

Answer 13

You meant sin(2A). For this, we have:
sin(2A) = 2sin(A)cos(A)
= 2 * (4/5) * (-3/5)
= -24/25 (Answer)

Answer 14

Can you try tan(B/2) and show me how you do it? Hint: You'd first need to find tan(B).

Answer 15

is tan B 3/2?

Answer 16

Yes, you're correct. :) Now substitute this value of tanB into the formula in my previous post to find tan(B/2).

Answer 17

did i enter it in right? i put 2(3/2) over 1 - (3/2)^2

Answer 18

help? ;(

Answer 19

can anybody help me? im stuck

Answer 20

@GhostShip99 : No, you have to sub. tan(B)=3/2, not tan(B/2) = 3/2 (as you have done). So the expression becomes:
2tan(B/2)/[1-tan^2(B/2)] = 3/2
Then evaluate this expression to find tan(B/2).