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**Hint:**Use the half angle formula and write down an expression of for sin(67.5) in terms of sine of double of this angle, which is 135 degrees. Then find the value of cosine of 135 degrees and substitute in the half angle formula.

**Complete step by step solution:**

Let us first simplify the given trigonometric ratio, i.e. sin(67.5). Here, let us assume that the angle 67.5 is in degrees. In the given question it is asked us to calculate the value of sin(67.5).

The half angle formula says that

${{\sin }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{1-\cos x}{2}$ …. (1),

where $\dfrac{x}{2}$ is the half angle and x is the double of that angle.

In this case, the half angle is equal to 67.5 degrees. This means that $\dfrac{x}{2}={{67.5}^{\circ }}$

Which further means that $x=2\times {{67.5}^{\circ }}={{135}^{\circ }}$

Substitute the values of the angles x and half of x in equation (1).

This gives us that ${{\sin }^{2}}\left( {{67.5}^{\circ }} \right)=\dfrac{1-\cos ({{135}^{\circ }})}{2}$ …. (i)

Now, let us find the value of $\cos ({{135}^{\circ }})$.

We can write 135 as $180-45$.

Therefore, $\cos ({{135}^{\circ }})$ can be written as $\cos ({{135}^{\circ }})=\cos ({{180}^{\circ }}-{{45}^{\circ }})$.

But we know that $\cos ({{180}^{\circ }}-\theta )=-\cos (\theta )$ …. (ii)

In this case, $\theta ={{45}^{\circ }}$.

Therefore, substitute $\theta ={{45}^{\circ }}$ in equation (ii).

Then, we get that $\cos ({{180}^{\circ }}-{{45}^{\circ }})=-\cos ({{45}^{\circ }})$ ….. (iii).

We know that $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$

After substituting this value in equation (iii) we get that $\cos ({{180}^{\circ }}-{{45}^{\circ }})=-\dfrac{1}{\sqrt{2}}$.

This means that $\cos ({{135}^{\circ }})=\cos ({{180}^{\circ }}-{{45}^{\circ }})=-\dfrac{1}{\sqrt{2}}$.

Now, substitute this value in equation (i).

This gives us that ${{\sin }^{2}}\left( {{67.5}^{\circ }} \right)=\dfrac{1-\left( -\dfrac{1}{\sqrt{2}} \right)}{2}=\dfrac{1+\dfrac{1}{\sqrt{2}}}{2}$

$\Rightarrow {{\sin }^{2}}\left( {{67.5}^{\circ }} \right)=\dfrac{1+\sqrt{2}}{2\sqrt{2}}$

Now, take square roots on both the sides.

Then, this means that $\sin \left( {{67.5}^{\circ }} \right)=\pm \sqrt{\dfrac{1+\sqrt{2}}{2\sqrt{2}}}$.

But here 67.5 degrees is an acute angle and we know that sine of an acute angle is a positive number. Therefore, we discard the negative value, i.e. $\sin \left( {{67.5}^{\circ }} \right)=-\sqrt{\dfrac{1+\sqrt{2}}{2\sqrt{2}}}$ and consider $\sin \left( {{67.5}^{\circ }} \right)=\sqrt{\dfrac{1+\sqrt{2}}{2\sqrt{2}}}$.

**Hence, we found with the help of half angle formula that $\sin \left( {{67.5}^{\circ }} \right)=\sqrt{\dfrac{1+\sqrt{2}}{2\sqrt{2}}}$.**

**Note:**Sometimes in some questions, the formulae may help to solve the questions.

$\sin (2\pi +\theta )=\sin (\theta )$

$\Rightarrow\cos (2\pi +\theta )=\cos (\theta )$

$\Rightarrow\sin (\pi +\theta )=-\sin (\theta )$

$\Rightarrow\cos (\pi +\theta )=-\cos (\theta )$

$\Rightarrow\sin (-\theta )=-\sin (\theta )$

$\Rightarrow\cos (-\theta )=\cos (\theta )$

With the help of these formulae you can find the formulae for the other trigonometric ratios as all the other trigonometric ratios depend on sine and cosine.

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