If A = 60° and B = 30° , verify that: (i) sin (A + B) = sin A cos

1.	If A = 60° and B = 30° , verify that: (i) sin (A + B) = sin A cos B + cos A sin B (ii) cos (A + B) = cos A cos B - sin A sin B
Answer» A = 60° and B = 30° Now, A + B = 60° + 30° = 90° Also, A – B = 60° – 30° = 30° (i) sin (A + B) = sin 90° = 1 sin A cos B + cos A sin B = sin 60° cos 30° + cos 60° sin 30° = \((\frac{\sqrt{3}}2\times\frac{\sqrt{3}}2+\frac{1}2\times\frac{1}2)\) = \((\frac{3}4+\frac{1}4)\) = 1 ∴ sin (A + B) = sin A cos B + cos A sin B (ii) cos (A + B) = cos 90° = 0 cos A cos B - sin A sin B = cos 60° cos 30° - sin 60° sin 30° = \((\frac{{1}}2\times\frac{\sqrt{3}}2-\frac{{\sqrt3}}2\times\frac{1}2)\) = \((\frac{\sqrt3}4+\frac{\sqrt3}4)\) = 0 ∴ cos (A + B) = cos A cos B - sin A sin B

If A = 60° and B = 30° , verify that: (i) sin (A + B) = sin A cos B + cos A sin B (ii) cos (A + B) = cos A cos B - sin A sin B

Answer»

A = 60° and B = 30°

Now, A + B = 60° + 30° = 90°

Also, A – B = 60° – 30° = 30°

(i) sin (A + B) = sin 90° = 1

sin A cos B + cos A sin B = sin 60° cos 30° + cos 60° sin 30°

= \((\frac{\sqrt{3}}2\times\frac{\sqrt{3}}2+\frac{1}2\times\frac{1}2)\) = \((\frac{3}4+\frac{1}4)\) = 1

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90° = 0

cos A cos B - sin A sin B = cos 60° cos 30° - sin 60° sin 30°

= \((\frac{{1}}2\times\frac{\sqrt{3}}2-\frac{{\sqrt3}}2\times\frac{1}2)\) = \((\frac{\sqrt3}4+\frac{\sqrt3}4)\) = 0

∴ cos (A + B) = cos A cos B - sin A sin B