# Answer to Question #6797 in Trigonometry for David Essien

Question #6797

If cosA 4/5 and cosB 12/13 (A and B are both acute) Find the values of (i) sin (A+B). (ii) cos (A -B).(iii) tan (A+B). Q2. Write an angle in the first quadrant whose tangent is (a)0.8816 (b)1.9496 (c)2.0265.

Expert's answer

Here we need

sinA = sqrt(1-cos²A) = sqrt(1-(4/5)²) = 3/5

tanA = sinA/cosA = (3/5)/(4/5) = 3/4

sinB = sqrt(1-sin²B) = sqrt(1-(12/13)²) = 5/13

tanB = sinB/cosB = (5/13)/(12/13) = 5/12

So,

(i) sin (A+B) = sinA*cosB + cosA*sinB = 3/4*12/13 + 4/5*5/13 = 1;

(ii) cos (A-B) = cosA*cosB + sinA*sinB = 4/5*12/13 + 3/5*5/13 = 63/65;

(iii) tan (A+B) = (tanA+tanB)/(1-tanA*tanB) = (3/4+5/12)/(1-3/4*5/12) = 56/33.

Q2. Write an angle in the first quadrant whose tangent is

(a)0.8816

α = atan(0.8816) = 41.39940060601370911032083150208

(b)1.9496

α = atan(1.9496) = 62.845545364237896213240308213824

(c)2.0265

α = atan(2.0265) = 63.735428584902806386195965140058

sinA = sqrt(1-cos²A) = sqrt(1-(4/5)²) = 3/5

tanA = sinA/cosA = (3/5)/(4/5) = 3/4

sinB = sqrt(1-sin²B) = sqrt(1-(12/13)²) = 5/13

tanB = sinB/cosB = (5/13)/(12/13) = 5/12

So,

(i) sin (A+B) = sinA*cosB + cosA*sinB = 3/4*12/13 + 4/5*5/13 = 1;

(ii) cos (A-B) = cosA*cosB + sinA*sinB = 4/5*12/13 + 3/5*5/13 = 63/65;

(iii) tan (A+B) = (tanA+tanB)/(1-tanA*tanB) = (3/4+5/12)/(1-3/4*5/12) = 56/33.

Q2. Write an angle in the first quadrant whose tangent is

(a)0.8816

α = atan(0.8816) = 41.39940060601370911032083150208

(b)1.9496

α = atan(1.9496) = 62.845545364237896213240308213824

(c)2.0265

α = atan(2.0265) = 63.735428584902806386195965140058

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