Question

How do you find the exact value of $\sin 67.5$ degrees?

Answer 1

Hint: We will find the acute angle equivalent to the angle ${{67.5}^{\circ }}={{\left( \dfrac{135}{2} \right)}^{\circ }}.$ Then we will use the sine half-angle identity. We will also use another identity $\cos {{\left( 180-x \right)}^{\circ }}=-\cos {{x}^{\circ }}$ for the cosine function is negative in the second quadrant. We are thorough with the value of cosine of ${{45}^{\circ }}.$ That is $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}.$

Complete step-by-step solution:
We are given with $\sin {{67.5}^{\circ }}.$
We know that $67.5=\dfrac{135}{2}.$
So, we can write $\sin {{67.5}^{\circ }}=\sin {{\left( \dfrac{135}{2} \right)}^{\circ }}.$
We will use the sine half-angle identity.
So, we will get \[\sin {{67.5}^{\circ }}=\pm \sqrt{\dfrac{1-\cos {{135}^{\circ }}}{2}}.\]
Since the sine function is positive in the first quadrant, this will become $\sin {{67.5}^{\circ }}=\sqrt{\dfrac{1-\cos {{135}^{\circ }}}{2}}.$
We know that $135=180-45.$
Now let us consider the cosine function inside the square root. We are going to find the value of cosine of angle ${{135}^{\circ }}.$
That is, $\cos {{135}^{\circ }}=\cos {{\left( 180-45 \right)}^{\circ }}$
After substituting for $135$ in the cosine function, we will use the identity $\cos {{\left( 180-x \right)}^{\circ }}=-\cos {{x}^{\circ }},$ [The cosine function is negative in the second quadrant.]
Thus, we will get the cosine function in this problem as $\cos {{135}^{\circ }}=-\cos {{45}^{\circ }}.$
We know that the value of $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},$ for ${{45}^{\circ }}={{\left( \dfrac{\pi }{2} \right)}^{c}}$ and $\cos {{\left( \dfrac{\pi }{2} \right)}^{c}}=\dfrac{1}{\sqrt{2}}.$
From this we will get $\cos {{135}^{\circ }}=-\cos {{45}^{\circ }}=-\dfrac{1}{\sqrt{2}}.$
In the next step we are going to substitute the value of the cosine function inside the square root.
So, the value of the given sine function will become $\sin {{67.5}^{\circ }}=\sqrt{\dfrac{1-\cos {{135}^{\circ }}}{2}}=\sqrt{\dfrac{1-\left( -\dfrac{1}{\sqrt{2}} \right)}{2}}.$
This will lead us to the step where we get $\sin {{67.5}^{\circ }}=\sqrt{\dfrac{1+\dfrac{1}{\sqrt{2}}}{2}}.$
Now consider the terms in the numerator inside the square root on the right-hand side of the above written equation. That is $1+\dfrac{1}{\sqrt{2}}.$ We want to make the denominators of both the summands the same. For that we will take LCM as $\dfrac{\sqrt{2}+1}{\sqrt{2}}.$ Multiply both the numerator and the denominator with $\sqrt{2}$ to get $\dfrac{2+\sqrt{2}}{2}.$
Now we will get $\sin {{67.5}^{\circ }}=\sqrt{\dfrac{\dfrac{2+\sqrt{2}}{2}}{2}}=\sqrt{\dfrac{2+\sqrt{2}}{4}}=\dfrac{\sqrt{2+\sqrt{2}}}{\sqrt{4}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}.$
Hence the exact value of $\sin {{67.5}^{\circ }}=\dfrac{\sqrt{2+\sqrt{2}}}{2}=0.92388.$

Note: The Sine Half-Angle identity is obtained as follows:
$\Rightarrow \cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x=1-{{\sin }^{2}}x-{{\sin }^{2}}x$
$\Rightarrow \cos 2x=1-2{{\sin }^{2}}x$
$\Rightarrow 2{{\sin }^{2}}x=1+\cos 2x$
$\Rightarrow {{\sin }^{2}}x=\dfrac{1+\cos 2x}{2}$
$\Rightarrow \sin x=\pm \sqrt{\dfrac{1+\cos 2x}{2}.}$
Here we convert the given angle in the standard angles such as $\left( 180, 30, 45, 60, 90 … \right)$ by using the half angle identity to make the problem easier.