Question

Find the values of:
1. $\sin 7{{\dfrac{1}{2}}^{\circ }}$
2. $\cos 7{{\dfrac{1}{2}}^{\circ }}$
3. $\tan 22{{\dfrac{1}{2}}^{\circ }}$
4. $\tan 11{{\dfrac{1}{4}}^{\circ }}$

Answer 1

Hint: While solving this question you should know about the general trigonometric formulas. In this problem we will use the general formulas of trigonometry to convert the value of any angle in the form of other angles. And thus we will get the solutions.

Complete step by step answer:
According to our question, we have to find the values of some trigonometric functions. We will look at them one by one. Let us start with part 1.
1. $\sin 7{{\dfrac{1}{2}}^{\circ }}$
From the properties of trigonometric identities, we know that,
$\cos A=1-2{{\sin }^{2}}\dfrac{A}{2}$
So, it means that,
$\begin{align}
  & 1-\cos A=2{{\sin }^{2}}\dfrac{A}{2} \\
 & \Rightarrow 2{{\sin }^{2}}7{{\dfrac{1}{2}}^{\circ }}=1-\cos {{15}^{\circ }} \\
 & \Rightarrow {{\sin }^{2}}7{{\dfrac{1}{2}}^{\circ }}=\dfrac{1-\cos {{15}^{\circ }}}{2} \\
\end{align}$
Now, if we solve this then we will get that,
$\begin{align}
  & {{\sin }^{2}}7{{\dfrac{1}{2}}^{\circ }}=\dfrac{1-\dfrac{\sqrt{3}+1}{2\sqrt{2}}}{2} \\
 & \Rightarrow {{\sin }^{2}}7{{\dfrac{1}{2}}^{\circ }}=\dfrac{2\sqrt{2}-\sqrt{3}-1}{4\sqrt{2}} \\
 & \Rightarrow \sin 7{{\dfrac{1}{2}}^{\circ }}=\sqrt{\dfrac{4-\sqrt{6}-\sqrt{2}}{8}} \\
\end{align}$
(Since $\sin 7{{\dfrac{1}{2}}^{\circ }}$is positive, we can say)
$\Rightarrow \sin 7{{\dfrac{1}{2}}^{\circ }}=\dfrac{\sqrt{4-\sqrt{6}-\sqrt{2}}}{2\sqrt{2}}$
2. $\cos 7{{\dfrac{1}{2}}^{\circ }}$
Since we know that,
$\cos A=2{{\cos }^{2}}\dfrac{A}{2}-1$
Therefore,
$\begin{align}
  & \Rightarrow 2{{\cos }^{2}}\dfrac{A}{2}=\cos A+1 \\
 & \Rightarrow {{\cos }^{2}}7{{\dfrac{1}{2}}^{\circ }}=\dfrac{1+\cos {{15}^{\circ }}}{2} \\
 & \Rightarrow {{\cos }^{2}}7{{\dfrac{1}{2}}^{\circ }}=\dfrac{1+\dfrac{\sqrt{3}+1}{2\sqrt{2}}}{2} \\
 & \Rightarrow {{\cos }^{2}}7{{\dfrac{1}{2}}^{\circ }}=\dfrac{2\sqrt{2}+\sqrt{3}+1}{4\sqrt{2}} \\
\end{align}$
If we solve this then we get,
$\Rightarrow {{\cos }^{2}}7{{\dfrac{1}{2}}^{\circ }}=\dfrac{\sqrt{4+\sqrt{6}+\sqrt{2}}}{2\sqrt{2}}$
3. $\tan 22{{\dfrac{1}{2}}^{\circ }}$
Since we know that,
$\begin{align}
  & \tan \dfrac{A}{2}=\sqrt{\dfrac{1-\cos A}{1+\cos A}} \\
 & \Rightarrow \tan 22{{\dfrac{1}{2}}^{\circ }}=\sqrt{\dfrac{1-\cos {{45}^{\circ }}}{1+\cos {{45}^{\circ }}}} \\
 & \Rightarrow \tan 22{{\dfrac{1}{2}}^{\circ }}=\sqrt{\dfrac{1-\dfrac{1}{\sqrt{2}}}{1+\dfrac{1}{\sqrt{2}}}} \\
\end{align}$
If we solve this we will get as follows,
$\begin{align}
  & \Rightarrow \tan 22{{\dfrac{1}{2}}^{\circ }}=\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}} \\
 & \Rightarrow \tan 22{{\dfrac{1}{2}}^{\circ }}=\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}} \\
 & \Rightarrow \tan 22{{\dfrac{1}{2}}^{\circ }}=\sqrt{\dfrac{{{\left( \sqrt{2}-1 \right)}^{2}}}{2-1}} \\
 & \Rightarrow \tan 22{{\dfrac{1}{2}}^{\circ }}=\dfrac{\sqrt{2}-1}{1}=\sqrt{2}-1 \\
\end{align}$
4. $\tan 11{{\dfrac{1}{4}}^{\circ }}$
We know that,
$\begin{align}
  & \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \Rightarrow \tan 11{{\dfrac{1}{4}}^{\circ }}=\dfrac{\sin 11{{\dfrac{1}{4}}^{\circ }}}{\cos 11{{\dfrac{1}{4}}^{\circ }}} \\
 & \Rightarrow \tan 11{{\dfrac{1}{4}}^{\circ }}=\dfrac{\sin 11{{\dfrac{1}{4}}^{\circ }}}{\cos 11{{\dfrac{1}{4}}^{\circ }}}\times \dfrac{2\sin 11{{\dfrac{1}{4}}^{\circ }}}{2\sin 11{{\dfrac{1}{4}}^{\circ }}} \\
 & \Rightarrow \tan 11{{\dfrac{1}{4}}^{\circ }}=\dfrac{2{{\sin }^{2}}11{{\dfrac{1}{4}}^{\circ }}}{2\sin 11{{\dfrac{1}{4}}^{\circ }}\cos 11{{\dfrac{1}{4}}^{\circ }}} \\
\end{align}$
Since we know that,
$\begin{align}
  & 2{{\sin }^{2}}A=1-\cos 2A \\
 & \sin 2A=2\sin A\cos A \\
\end{align}$
So, we will substitute these values in the above equation as follows and then we get,
$\begin{align}
  & =\dfrac{1-\cos 22{{\dfrac{1}{2}}^{\circ }}}{\sin 22{{\dfrac{1}{2}}^{\circ }}} \\
 & =\dfrac{1-\sqrt{\dfrac{1+\cos {{45}^{\circ }}}{2}}}{\sqrt{\dfrac{1-\cos {{45}^{\circ }}}{2}}} \\
 & =\dfrac{\sqrt{2}-\sqrt{1+\dfrac{1}{\sqrt{2}}}}{\sqrt{1-\dfrac{1}{\sqrt{2}}}} \\
\end{align}$
For getting our final answer, we will solve this again like this,
$\begin{align}
  & =\dfrac{\sqrt{2\sqrt{2}}-\sqrt{2\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}} \\
 & =\dfrac{\sqrt{2\sqrt{2}}-\sqrt{2\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}}\times \dfrac{\sqrt{2\sqrt{2}+1}}{\sqrt{2\sqrt{2}+1}} \\
 & =\dfrac{\sqrt{2\sqrt{2}}.\sqrt{2\sqrt{2}+1}-\sqrt{{{\left( \sqrt{2}+1 \right)}^{2}}}}{\sqrt{\left( \sqrt{2}+1 \right)\sqrt{2}-1}} \\
 & =\dfrac{\sqrt{2\sqrt{2}\left( \sqrt{2}+1 \right)}-\left( \sqrt{2}+1 \right)}{\sqrt{2}-1} \\
 & =\sqrt{4+2\sqrt{2}}-\left( \sqrt{2}+1 \right) \\
\end{align}$
So, these are the final answers that we get for the given values.

Note: While solving these types of questions you should be careful about the values of tan, sin, cos at the desired angles because these are directly not defined. We have to calculate these by using different formulas.