Answer to Question #126023 in Trigonometry for Habeeb goat

Question

Suppose sin(A)=3/5 and A is in Q2, and cos(B)=−5/13 and B is in Q3. Find the exact value of cos(A+B).

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Answer 1

Given sin(A)=3/5 and A is in Q2, and cos(B)=−5/13 and B is in Q3.

Now, cos(A) = $\pm \sqrt{1-sin^2(A)} = \pm \sqrt{1-\frac{9}{25}} = \pm \frac{4}{5}$ .Now, since A is in quadrant 2, so cos(A) is negative in second quadrant. Hence, cos(A) = -4/5.

Similarly, sin(B) = $-\sqrt{1-cos^2(B)}$ because B is in Quadrant 3. $\implies sin(B) = - \frac{12}{13}$

cos(A+B) = cos(A) cos(B) - sin(A) sin(B) = $(-\frac{4}{5})( -\frac{5}{13}) - (\frac{3}{5})(-\frac{12}{13}) = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$ .