If A = 60 and B = 30 , show that : (sin A cos B + cos A sin B)2 + (cos A cos B - sin A sin B)2 = 1

Left Hand Side : (sin A cos B + cos A sin B)2 + (cos A cos B - sin A sin B)2 = (sin 60 cos 30 + cos 60 sin 30 )2 + (cos 60 cos 30 - sin 60 sin 30 )2 = = = 1 + 0 = 1. Right Hand Side = 1 Hence, proved that (sin A cos B + cos A sin B)2 + (cos A cos B - sin A sin B)2 = 1.

Mathematics

If A = 60° and B = 30°, show that :
(sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)² = 1

Trigonometrical Ratios

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Answer

Left Hand Side :

(sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)²

= (sin 60° cos 30° + cos 60° sin 30°)² + (cos 60° cos 30° - sin 60° sin 30°)²

= $\Big(\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\times\dfrac{1}{2}\Big)^2 + \Big(\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}\Big)^2$

= $\Big(\dfrac{3}{4} + \dfrac{1}{4})^2 + \Big(\dfrac{\sqrt{3}}{4}- \dfrac{\sqrt{3}}{4}\Big)^2$

= 1 + 0 = 1.

Right Hand Side = 1

Hence, proved that (sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)² = 1.

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Mathematics

If A = 60° and B = 30°, show that :
(sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)² = 1

Trigonometrical Ratios

Answer

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If A = 30°, prove that :
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If A = B = 45°, show that :
(i) sin(A - B) = sin A cos B - cos A sin B
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(iv) tan (A - B) = $\dfrac{\tan A - \tan B}{1 + \tan A \tan B}$

Evaluate : $\dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}$ , when A = 15°.

Mathematics

If A = 60° and B = 30°, show that :(sin A cos B + cos A sin B)2 + (cos A cos B - sin A sin B)2 = 1

Trigonometrical Ratios

Answer

Related Questions

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If A = 60° and B = 30°, show that :
(sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)² = 1

If A = 30°, prove that :
(i) sin 2A = $\dfrac{2 \tan A}{1 + \tan^2A}$
(ii) cos 2A = $\Big(\dfrac{1 - \tan^2A}{1 + \tan^2A}\Big)$

If A = B = 45°, show that :
(i) sin(A - B) = sin A cos B - cos A sin B
(ii) cos(A + B) = cosA cosB - sin A sin B

If A = 60° and B = 30°, prove 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
(iii) cos (A - B) = cos A cos B + sin A sin B
(iv) tan (A - B) = $\dfrac{\tan A - \tan B}{1 + \tan A \tan B}$

Evaluate : $\dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}$ , when A = 15°.