Solveeit Logo

Question

Question: Find the value of \( \cos 405 \) . A. \( \sqrt{2} \) B. \( \dfrac{1}{\sqrt{2}} \) C. \( -\sq...

Find the value of cos405\cos 405 .
A. 2\sqrt{2}
B. 12\dfrac{1}{\sqrt{2}}
C. 2-\sqrt{2}
D. 12\dfrac{-1}{\sqrt{2}}

Explanation

Solution

Hint : We explain the process of finding values for associated angles. We find the rotation and the position of the angle for 405{{405}^{\circ }} . We explain the changes that are required for that angle. Depending on those things we find the solution.

Complete step-by-step answer :
We need to find the ratio value for cos405\cos 405 .
For general form of cos(x)\cos \left( x \right) , we need to convert the value of x into the closest multiple of π2\dfrac{\pi }{2} and add or subtract a certain value α\alpha from that multiple of π2\dfrac{\pi }{2} to make it equal to x.
Let’s assume x=k×π2+αx=k\times \dfrac{\pi }{2}+\alpha , kZk\in \mathbb{Z} . Here we took the addition of α\alpha . We also need to remember that απ2\left| \alpha \right|\le \dfrac{\pi }{2} .
Now we take the value of k. If it’s even then keep the ratio as cos and if it’s odd then the ratio changes to sin ratio from cos.
Then we find the position of the given angle as quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angel falls in the first or fourth quadrant then the sign remains positive but if it falls in the second or third quadrant then the sign becomes negative.
The final form becomes cos405=cos(4×π2+45)=cos(45)=12\cos 405=\cos \left( 4\times \dfrac{\pi }{2}+45 \right)=\cos \left( 45 \right)=\dfrac{1}{\sqrt{2}} .
Therefore, the value of cos405\cos 405 is 12\dfrac{1}{\sqrt{2}} . The correct option is B.
So, the correct answer is “Option B”.

Note : We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of π\pi instead of π2\dfrac{\pi }{2} . It makes the multiplied number always even. In that case we don’t have to change the ratio. If x=k×π+α=2k×π2+αx=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha . Value of 2k2k is always even.