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# C4 AQA Further Trig help watch

1. If sin (A) = 4/5 and cos (B) = 12/13 where A is obtuse and B is acute, find the value of sin(A-B)?

My working

Sin(A-B)= sin(A) x sin(B) - cos(A) x cos(B)

(4/5) x (5/13) - (3/5) x (12/13) (triangles)

20/65 - 36/65 = -16/65

3/5 should be a minus? Not sure if it is or why it is
2. (Original post by H3nners)
If sin (A) = 4/5 and cos (B) = 12/13 where A is obtuse and B is acute, find the value of sin(A-B)?

My working

Sin(A-B)= sin(A) x sin(B) - cos(A) x cos(B)

(4/5) x (5/13) - (3/5) x (12/13) (triangles)

20/65 - 36/65 = -16/65

3/5 should be a minus? Not sure if it is or why it is
It should indeed be negative.

Consider the graphs of (red) and (green) where the angle A is obtuse, meaning it is degrees which is the region I have marked blue on the diagram below:

Now sine of an obtuse angle is clearly positive as that portion of us always above 0, however cosine of an obtuse angle is clearly negative as that portion of is below 0.

So, since A is obtuse, and

P.S. Check your compound angle identity that you've used as well, it's not the right one.
3. (Original post by H3nners)
Sin(A-B)= sin(A) x sin(B) - cos(A) x cos(B)

(4/5) x (5/13) - (3/5) x (12/13) (triangles)

20/65 - 36/65 = -16/65

3/5 should be a minus? Not sure if it is or why it is

it should say

Sin(A-B)= sin(A) x cos(B) - cos(A) x sin(B)
4. Yeah my mistake I typed in the formula wrong sorry about that. Cheers rdk makes a load more sense will draw the graphs from now on. Sick explanation!

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Updated: January 26, 2017
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