Question

use the given information to find the exact value find sin(A-B) using the sine difference identity cos A= 1/3,0 12 years ago

Answer 1

help me plz

Answer 2

help me plz

Answer 3

Trivial. Just use the sin(A-B) = sinAcosB - sinBCosA formula. You can work out sin(A) from cos(A) (and vice versa) using cos^2(A) + sin^2(A) = 1, which implies sin(A) = +/- sqrt(1-cos^2(A). Use bounds on where A and B are to decide whether each is the positive or negative root.

Answer 4

i really dont get this plz explain me more in a board

Answer 5

No, sorry. I gave you everything you nee.d Now do your own homework. <3

Answer 6

plz help me ranga sir plz

Answer 7

sin(A-B) = sinAcosB - sinBCosA
cosA = 1/3
sinB = -1/2
sin(A-B) = sinAcosB - sinBCosA = sin(A-B) = sinAcosB - (-1/2)(1/3)
= sinAcosB + 1/6



Find sinB and cosA from the diagram and substitute.

Answer 8

how do i substitute sir

Answer 9

First find the length of the missing side in the two triangles marked with a question mark. That is, What is PQ in the first triangle and SB in the second triangle?

Answer 10

i dont no how to do that

Answer 11

You should get the basics first before attempting this problem.
The figure shows right triangles.
AQ^2 + PQ^2 = AP^2
1^2 + PQ^2 = 3^2
Find PQ.
(This is the same as using sin^2A + cos^2A = 1. But I showed you the triangle so you can know what sine and cosine represents in a triangle).

Answer 12

2^3

Answer 13

1^2 + PQ^2 = 3^2
1 + PQ^2 = 9
PQ^2 = 9 - 1 = 8
PQ = sqrt(8) = sqrt(4*2) = 2sqrt(2)
sin(A) = opposite / hypotenuse = PQ / AP = 2sqrt(2) / 3.
This is what you will substitute for sin(A).

Use the second triangle to find cos(B) in a similar way.

Answer 14

hmm i got for the pother one sqyrt (2)/2

Answer 15

is it right

Answer 16

hello sir

Answer 17

sqrt(3)/2

Answer 18

yes. And since B is in the fourth quadrant cos(B) is positive.

sin(A-B) = sinAcosB + 1/6
sin(A) = 2sqrt(2) / 3
cos(B) = sqrt(3)/2
Substitute into sin(A-B) = sinAcosB + 1/6

Answer 19

so my answer is 2sqrt(2)/3 and sqrt(3)/2 right sir