Find sin(A−B) given that cosA=13, with A in quadrant I, and sinB=−12, with B in quadrant IV.

Answer 1

Find sin(A−B) given that cosA=13, with A in quadrant I, and sinB=−12, with B in quadrant IV.

acos13=β=3.256.° quad 1 Your label is A

asin(-12)=α=3.1761.° quad 1 " " B

or: -1.5079........° quad 4

sin(3.256-(-1.508))=sin4.746=.0827438°

Answer 2

Oh no, the sine and cosine functions MUST be between -1 and 1,

so the statement of your problem is incorrect...

However, judging by the values given, the Pythagorean triple

of {5,12,13} will work just fine, so you can see how the problem

is solved.

So here is the RE-STATEMENT of the problem:

Given: cos A = 5/13 with A in quadrant 1 ;

sin B = -12/13 with B in quadrant 4;

The goal is to calculate sin(A-B)

The given information is consistent with the rule

of signs for trig functions on the unit circle.

Remember "Add Sugar To Coffee"..

A S T C

So ALL trig functions are positive in quadrant 1;

SINE is positive in quadrant 2;

Tangent is positive in quadrant 3;

Cosine is positive in quadrant 4; <--- given sine is negative, so we're good

Regarding angle A:

cosine is 5/13 --> adjacent = 5 and hypotenuse = 13;

then opposite is 12

So then sin(A) = 12/13, cos(A) = 5/13, and tan(A) = 12/5

Regarding angle B:

sine is -12/13 --> opposite is -12 and hypotenuse =13;

then adjacent is -5

So then sin(B) = -12/13, cos(B) = -5/13, tan(B) = 12/5

By angle subtraction formula, sin(A-B) = sinA cos B + sin B cos A

= (12/13)(-5/13) + (-12/13)(5/13)

= -60/169 + -60/169

= -120/169