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If sin (A+B) = sin A cos B + cos A sin B and cos (A-B) = cos A cos B + sin A sin B
(i) sin (750)
(ii) cos (150)
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Solution
Let A = 450 and B = 300
(i) As, sin(A + B) = sin A cos B + cos A sin B
⇒ sin (450 + 300) = sin 450 cos 300 + cos 450 sin 300
⇒ sin `(75^0) = 1/sqrt(2) xx sqrt(3)/2 + 1/sqrt(2) xx1/2`
⇒ sin `(75^0) = sqrt(3)/(2sqrt(2)) + 1/(2sqrt(2))`
∴ sin `(75^0) = (sqrt(3) +1)/(2 sqrt(2))`
ii) As, cos (A – B) = cos A cos B + sin A sin B
⇒ cos (450 – 300) = cos 450 cos 300 + sin 450 sin 300
⇒ cos `(15^0) = 1/sqrt(2) xx sqrt(3)/2 + 1/sqrt(2) xx1/2`
⇒ cos `(15^0) = sqrt(3)/(2sqrt(2)) + 1/(2sqrt(2))`
∴ cos `(15^0)=(sqrt(3)+1)/(2sqrt(2))`
Disclaimer: cos 150 can also be written by taking A = 600 and B = 450.
Concept: Trigonometric Ratios and Its Reciprocal
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