Question

Find the value of $\sin {{75}^{\circ }}$.\[\]

Answer 1

Hint: Find out the values of $\sin {{45}^{\circ }},\sin {{30}^{\circ }},\cos {{45}^{\circ }},\cos {{30}^{\circ }}$ and then express ${{75}^{\circ }}$ as the sum of ${{45}^{\circ }}$ and ${{30}^{\circ }}$ . Use the sine sum of two angles formula to calculate the required value.\[\]

Complete step by step answer:

We know from the property of right angled triangles that in a right angled triangle there is an angle otherwise known as an angle with measurement${{90}^{\circ }}$. The side opposite to the right- angle is called hypotenuse and denoted as $h$. The side which is vertical to the ground is called perpendicular and denoted as $p$ .The side which is horizontal or parallel to the ground is called base and denoted as $b$.\[\]
We know from the trigonometric ratios in a right angled triangle the sine of any angle is given by the ratio of side opposite to the angle to the hypotenuse. In the figure the sine of the angle $\theta $ is given by
\[\sin \theta =\dfrac{p}{h}...(1)\]
Similarly the cosine of an angle is the ratio of side adjacent to the angle(excluding hypotenuse) to the hypotenuse
\[\cos \theta =\dfrac{b}{h}...(2)\]
The Pythagoras theorem states that “ The square of hypotenuse is sum of square of base and square of perpendicular. In symbols,
\[\begin{align}
  & {{h}^{2}}={{p}^{2}}+{{b}^{2}} \\
 & \Rightarrow h=\sqrt{{{p}^{2}}+{{b}^{2}}}..(3) \\
\end{align}\]
We assume the right angled triangle to be isosceles with sides $p=b$. Then the angles opposite to them will be equal and sum of all the angles is ${{180}^{\circ }}$.So
\[\begin{align}
  & {{90}^{\circ }}+\theta +\theta ={{180}^{\circ }} \\
 & \Rightarrow {{90}^{\circ }}+2\theta ={{180}^{\circ }} \\
 & \Rightarrow \theta =\dfrac{{{90}^{\circ }}}{2}={{45}^{\circ }} \\
\end{align}\]
Also from (2) we get ,
\[\begin{align}
  & h=\sqrt{{{p}^{2}}+{{p}^{2}}} \\
 & \Rightarrow h=\sqrt{2}p=\sqrt{2}b\left( \because p=b \right) \\
\end{align}\]
We put the values in equation (1) and (2) to get
\[\begin{align}
  & \sin \left( {{45}^{\circ }} \right)=\dfrac{p}{\sqrt{2}p}=\dfrac{1}{\sqrt{2}} \\
 & \cos \left( {{45}^{\circ }} \right)=\dfrac{b}{\sqrt{2}b}=\dfrac{1}{\sqrt{2}} \\
\end{align}\]
Let us now take an equilateral triangle with each side of length $a$ and let anyone angle be bisected. We know that the angle bisector divides the angle into two equal angles of measurement $\dfrac{{{60}^{\circ }}}{2}={{30}^{\circ }}=\theta \left( \text{say} \right)$ and divides the opposite side into two equal segments of length $\dfrac{a}{2}$.

So from equation (1) and (2) we get,
\[\begin{align}
  & \sin \left( {{30}^{\circ }} \right)=\dfrac{\dfrac{a}{2}}{a}=\dfrac{1}{2} \\
 & \cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{{{a}^{2}}-{{\left( \dfrac{a}{2} \right)}^{2}}}}{a}=\dfrac{\sqrt{\dfrac{3}{4}{{a}^{2}}}}{a}=\dfrac{\sqrt{3}}{2} \\
\end{align}\]
We also know from the sine sum of angle formula that if $\theta $ is the sum of angles with measurement say $\alpha $and $\beta $then
$\sin \theta =\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $
We are asked in the question to find the value of $\sin {{75}^{\circ }}$. Let’s take $\theta ={{75}^{\circ }}$, $\alpha ={{45}^{\circ }}$,$\beta ={{30}^{\circ }}$. We can see that $\alpha +\beta ={{30}^{\circ }}+{{45}^{\circ }}={{75}^{\circ }}=\theta $. Now we can use the above formula and put previously obtained values wherever necessary
\[\begin{align}
  & \sin {{75}^{\circ }}=\sin \left( 30+45 \right) \\
 & =\sin 30\cdot \cos 45+\cos 30\cdot \sin 45 \\
 & =\dfrac{1}{2}\cdot \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{2}} \\
 & =\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\
\end{align} \]

So the value of $\sin {{75}^{\circ }}$ is $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$ and rounded up to to 3 decimals is 0.966. \[\]

Note: The question tests your knowledge of trigonometric identities of sum and multiple of angles. The cosine sum of two angles formula is given by $\cos \left( \alpha +\beta \right)=\sin \alpha \sin \beta +\sin \alpha \sin \beta $. The question can also be framed to find the values of $\sin {{15}^{\circ }},\cos {{15}^{\circ }},\cos {{75}^{\circ }}$etc.