
What is the answer to $\cos \left( \dfrac{3\pi }{8} \right)$?
Answer
513.9k+ views
Hint: To obtain the answer of the given trigonometric function we will use the half-angle formula of cosine.
Firstly we will change the value inside the bracket by taking 2 as the denominator as we have to make it half-angle. Then we will use the cosine half-angle formula in it and simplify it to get the desired answer.
Complete step by step solution:
The trigonometric function is given as:
$\cos \left( \dfrac{3\pi }{8} \right)$…..$\left( 1 \right)$
The cosine half-angle formula is given as below:
$\cos \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1+\cos x}{2}}$…..$\left( 2 \right)$
We know that the value is positive if $\left( \dfrac{x}{2} \right)$ is in first and fourth quadrant and negative if $\left( \dfrac{x}{2} \right)$ is in second and third quadrant.
As we can see $\left( \dfrac{3\pi }{8} \right)$ is in the first quadrant therefore we will use positive signs.
So firstly we will simplify equation (1) as follows:
$\cos \left( \dfrac{3\pi }{8} \right)=\cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)$
Now use formula (2) in above equation and simplify it as below:
$\begin{align}
& \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\sqrt{\dfrac{1+\cos \left( \dfrac{3\pi }{4} \right)}{2}} \\
& \Rightarrow \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\sqrt{\dfrac{1-\dfrac{\sqrt{2}}{2}}{2}} \\
& \Rightarrow \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\sqrt{\dfrac{2-\sqrt{2}}{4}} \\
& \therefore \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\dfrac{\sqrt{2-\sqrt{2}}}{2} \\
\end{align}$
So we got the answer as $\dfrac{\sqrt{2-\sqrt{2}}}{2}$
Hence answer to $\cos \left( \dfrac{3\pi }{8} \right)$ is $\dfrac{\sqrt{2-\sqrt{2}}}{2}$.
Note: Trigonometric functions are those which are obtained by right-angled triangle sides. Six basic trigonometric functions are sine, cosine, tangent, cosecant, secant and cotangent. There are many identities using these functions and they all are related to each other in some or other way. Half angle formula is also one of the identities it is formed by double angle formula and is used in integration also to simplify the term inside the integration sign so that it becomes easy to solve it. Due to periodicity of these functions they are positive in some quadrants and negative in some. Sine, cosine, secant and cosecant have period $2\pi $ while tangent and cotangent have period $\pi $.
Firstly we will change the value inside the bracket by taking 2 as the denominator as we have to make it half-angle. Then we will use the cosine half-angle formula in it and simplify it to get the desired answer.
Complete step by step solution:
The trigonometric function is given as:
$\cos \left( \dfrac{3\pi }{8} \right)$…..$\left( 1 \right)$
The cosine half-angle formula is given as below:
$\cos \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1+\cos x}{2}}$…..$\left( 2 \right)$
We know that the value is positive if $\left( \dfrac{x}{2} \right)$ is in first and fourth quadrant and negative if $\left( \dfrac{x}{2} \right)$ is in second and third quadrant.
As we can see $\left( \dfrac{3\pi }{8} \right)$ is in the first quadrant therefore we will use positive signs.
So firstly we will simplify equation (1) as follows:
$\cos \left( \dfrac{3\pi }{8} \right)=\cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)$
Now use formula (2) in above equation and simplify it as below:
$\begin{align}
& \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\sqrt{\dfrac{1+\cos \left( \dfrac{3\pi }{4} \right)}{2}} \\
& \Rightarrow \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\sqrt{\dfrac{1-\dfrac{\sqrt{2}}{2}}{2}} \\
& \Rightarrow \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\sqrt{\dfrac{2-\sqrt{2}}{4}} \\
& \therefore \cos \left( \dfrac{\dfrac{3\pi }{4}}{2} \right)=\dfrac{\sqrt{2-\sqrt{2}}}{2} \\
\end{align}$
So we got the answer as $\dfrac{\sqrt{2-\sqrt{2}}}{2}$
Hence answer to $\cos \left( \dfrac{3\pi }{8} \right)$ is $\dfrac{\sqrt{2-\sqrt{2}}}{2}$.
Note: Trigonometric functions are those which are obtained by right-angled triangle sides. Six basic trigonometric functions are sine, cosine, tangent, cosecant, secant and cotangent. There are many identities using these functions and they all are related to each other in some or other way. Half angle formula is also one of the identities it is formed by double angle formula and is used in integration also to simplify the term inside the integration sign so that it becomes easy to solve it. Due to periodicity of these functions they are positive in some quadrants and negative in some. Sine, cosine, secant and cosecant have period $2\pi $ while tangent and cotangent have period $\pi $.
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