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Question: Given that \(\sin (A + B) = \sin A\cos B + \cos A\sin B\), find the value of \(\sin {75^o}\)....

Given that sin(A+B)=sinAcosB+cosAsinB\sin (A + B) = \sin A\cos B + \cos A\sin B, find the value of sin75o\sin {75^o}.

Explanation

Solution

Hint: Here, we will split 75o{75^o} into 45o{45^o} and 30o{30^o} then substitute in given equation i.e.., sin(A+B)=sinAcosB+cosAsinB\sin (A + B) = \sin A\cos B + \cos A\sin B to find the value of sin75o\sin {75^o}.

Complete step-by-step answer:
We had been given that sin(A+B)=sinAcosB+cosAsinB(1)\sin (A + B) = \sin A\cos B + \cos A\sin B \to (1)
And we need to find the value of sin75o\sin {75^o}.
Now sin75o\sin {75^o} can be written as sin(45+30)\sin (45 + 30) so using equation 1
sin(45+30)=sin450.cos300+cos450.sin300\sin (45 + 30) = \sin {45^0}.\cos {30^0} + \cos {45^0}.\sin {30^0}
Now, sin450=12,cos450=12,sin300=12,cos300=32\sin {45^0} = \dfrac{1}{{\sqrt 2 }},\cos {45^0} = \dfrac{1}{{\sqrt 2 }},\sin {30^0} = \dfrac{1}{2},\cos {30^0} = \dfrac{{\sqrt 3 }}{2} so putting values
We have
sin(45+30)=12×32+12×12 sin(45+30)=322+122 sin(75)=3+122  \sin (45 + 30) = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} \\\ \sin (45 + 30) = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }} \\\ \sin (75) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \\\
Hence, the value of sin75o=3+122\sin {75^o} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}.

Note: Whenever we come across such questions simply try to change the required angle in the terms of the formula given, then simple substitution and simplification will give you the answer.