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

Given equation is

$\sin {75^0}$

Now, break the angle of sine into two sum of two standard angle

$ \Rightarrow \sin {75^0} = \sin \left( {{{45}^0} + {{30}^0}} \right)$

Now, as we know $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, so use this property we have,

$\sin {75^0} = \sin \left( {{{45}^0} + {{30}^0}} \right) = \sin {45^0}\cos {30^0} + \cos {45^0}\sin {30^0}$

Now we know that $\sin {45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }},{\text{ sin3}}{{\text{0}}^0} = \dfrac{1}{2},{\text{ }}\cos {30^0} = \dfrac{{\sqrt 3 }}{2}$

So substitute these values in the above equation we have,

$\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 $\sin {75^0}$.

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.

Given equation is

$\sin {75^0}$

Now, break the angle of sine into two sum of two standard angle

$ \Rightarrow \sin {75^0} = \sin \left( {{{45}^0} + {{30}^0}} \right)$

Now, as we know $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, so use this property we have,

$\sin {75^0} = \sin \left( {{{45}^0} + {{30}^0}} \right) = \sin {45^0}\cos {30^0} + \cos {45^0}\sin {30^0}$

Now we know that $\sin {45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }},{\text{ sin3}}{{\text{0}}^0} = \dfrac{1}{2},{\text{ }}\cos {30^0} = \dfrac{{\sqrt 3 }}{2}$

So substitute these values in the above equation we have,

$\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 $\sin {75^0}$.

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.