If $\sin (A + B) = sinA cosB + cosA sinB$ and $\cos (A - B) = cosA cosB + sinA sinB$ , find the values of $(i) \sin 75 ° and (ii) \cos 15 °$ .

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Solution

$Let A = 45 ° and B = 30 °$

$(i) As, \sin (A + B) = sinA cosB + cosA sinB \Rightarrow \sin (45 ° + 30 °) = \sin 45 ° \cos 30 ° + \cos 45 ° \sin 30 ° \Rightarrow \sin (75 °) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \Rightarrow \sin 75 ° = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} ∴ \sin 75 ° = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

$(ii) As, \cos (A - B) = cosA cosB + sinA sinB \Rightarrow \cos (45 ° - 30 °) = \cos 45 ° \cos 30 ° + \sin 45 ° \sin 30 ° \Rightarrow \cos (15 °) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \Rightarrow \cos 15 ° = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} ∴ \cos 15 ° = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Disclaimer: $\cos 15 °$ can also be calculated by taking $A = 60 ° and B = 45 °$ .

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Similar questions

A = 30^{\circ}

B = 60^{\circ}

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

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

If sin(A - B) = sinA cos B - cosA sinB and cos(A - B)= cos A cos B + sin Asin B then find
sin15 and cos 15 .

A = B = 60^{\circ}

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

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

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

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

A

B

cos 15^{\circ}

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