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Question

Question: Find the value of \(\sin {75^0}\)...

Find the value of sin750\sin {75^0}

Explanation

Solution

Hint – Since 75 degrees is not a standard angle we can write 75 as summation of two known angles and proceed.

Given equation is
sin750\sin {75^0}
Now, break the angle of sine into two sum of two standard angle
sin750=sin(450+300)\Rightarrow \sin {75^0} = \sin \left( {{{45}^0} + {{30}^0}} \right)
Now, as we know sin(A+B)=sinAcosB+cosAsinB\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B, so use this property we have,
sin750=sin(450+300)=sin450cos300+cos450sin300\sin {75^0} = \sin \left( {{{45}^0} + {{30}^0}} \right) = \sin {45^0}\cos {30^0} + \cos {45^0}\sin {30^0}
Now we know that sin450=cos450=12, sin300=12, cos300=32\sin {45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }},{\text{ sin3}}{{\text{0}}^0} = \dfrac{1}{2},{\text{ }}\cos {30^0} = \dfrac{{\sqrt 3 }}{2}
So substitute these values in the above equation we have,
sin750=sin(450+300)=sin450cos300+cos450sin300 sin750=12×32+12×12=3+122 \begin{gathered} \sin {75^0} = \sin \left( {{{45}^0} + {{30}^0}} \right) = \sin {45^0}\cos {30^0} + \cos {45^0}\sin {30^0} \\\ \Rightarrow \sin {75^0} = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \\\ \end{gathered}
So this is the required value of sin750\sin {75^0}.

Note – whenever we face such types of questions first break the given angle into the sum of two standard angles, then apply the basic trigonometric property which is stated above and also remember all the standard angle values which is written above then apply these values in the given equation and simplify, we will get the required answer.