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Question: How do you use the half angle formula to find sin(67.5) ?...

How do you use the half angle formula to find sin(67.5) ?

Explanation

Solution

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(x2)=1−cos⁡x2{{\sin }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{1-\cos x}{2} …. (1),
where x2\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 x2=67.5∘\dfrac{x}{2}={{67.5}^{\circ }}
Which further means that x=2×67.5∘=135∘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(67.5∘)=1−cos⁡(135∘)2{{\sin }^{2}}\left( {{67.5}^{\circ }} \right)=\dfrac{1-\cos ({{135}^{\circ }})}{2} …. (i)
Now, let us find the value of cos⁡(135∘)\cos ({{135}^{\circ }}).
We can write 135 as 180−45180-45.
Therefore, cos⁡(135∘)\cos ({{135}^{\circ }}) can be written as cos⁡(135∘)=cos⁡(180∘−45∘)\cos ({{135}^{\circ }})=\cos ({{180}^{\circ }}-{{45}^{\circ }}).
But we know that cos⁡(180∘−θ)=−cos⁡(θ)\cos ({{180}^{\circ }}-\theta )=-\cos (\theta ) …. (ii)
In this case, θ=45∘\theta ={{45}^{\circ }}.
Therefore, substitute θ=45∘\theta ={{45}^{\circ }} in equation (ii).
Then, we get that cos⁡(180∘−45∘)=−cos⁡(45∘)\cos ({{180}^{\circ }}-{{45}^{\circ }})=-\cos ({{45}^{\circ }}) ….. (iii).
We know that cos⁡45∘=12\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}
After substituting this value in equation (iii) we get that cos⁡(180∘−45∘)=−12\cos ({{180}^{\circ }}-{{45}^{\circ }})=-\dfrac{1}{\sqrt{2}}.
This means that cos⁡(135∘)=cos⁡(180∘−45∘)=−12\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(67.5∘)=1−(−12)2=1+122{{\sin }^{2}}\left( {{67.5}^{\circ }} \right)=\dfrac{1-\left( -\dfrac{1}{\sqrt{2}} \right)}{2}=\dfrac{1+\dfrac{1}{\sqrt{2}}}{2}
⇒sin⁡2(67.5∘)=1+222\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⁡(67.5∘)=±1+222\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⁡(67.5∘)=−1+222\sin \left( {{67.5}^{\circ }} \right)=-\sqrt{\dfrac{1+\sqrt{2}}{2\sqrt{2}}} and consider sin⁡(67.5∘)=1+222\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⁡(67.5∘)=1+222\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⁡(2π+θ)=sin⁡(θ)\sin (2\pi +\theta )=\sin (\theta )
⇒cos⁡(2π+θ)=cos⁡(θ)\Rightarrow\cos (2\pi +\theta )=\cos (\theta )
⇒sin⁡(π+θ)=−sin⁡(θ)\Rightarrow\sin (\pi +\theta )=-\sin (\theta )
⇒cos⁡(π+θ)=−cos⁡(θ)\Rightarrow\cos (\pi +\theta )=-\cos (\theta )
⇒sin⁡(−θ)=−sin⁡(θ)\Rightarrow\sin (-\theta )=-\sin (\theta )
⇒cos⁡(−θ)=cos⁡(θ)\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.