Question

Given sin A = 3/5 and tan B = -3/4. Also, A and B are in QII. Find the exact values of sin (A+B), cos (A-B), tan (A+B), sin 2A, cos 2B, and tan 2A

Answer 1

please show your solutions

Answer 2

what is Q||

Answer 3

just answer this question if tan B = -3/4 is in quadrant 2 is x=-3 and y=4?

Answer 4

Were you instructed to use trigonometric properties when solving?

Answer 5

Cos A is found by using the Pythagorean Theorem, where x^2 + y^2 = r^2. So \[\sqrt{r^2-y^2} = x\]

\[\rightarrow \sqrt{25-9} = 4\]

Thus, Cos A = -4/5 because cosine is negative in Quadrant II.

Answer 6

Sin B and Cos B are done similarly:

\[\sqrt{x^2 + y^2} = r\]

\[r = \sqrt{(3)^2 + (-4)^2} = \sqrt{25} = 5\]

So, Sin B = 3/5 and Cos B = -4/5

Answer 7

Now that you have all of your trigonometric values for sin, cos and tan of triangles A and B use the trigonometric properties for your equations:

\[\sin (a+b) = \sin a \cos b + \cos a \sin b\]

\[\cos (a - b) = \cos a \cos b + \sin a \sin b\]

\[\tan (a + b) = (\tan a + \tan b)\div(1 - \tan a \tan b)\]