Question

How do you evaluate sine, cosine, tangent of$-{{405}^{\circ }}$without using a calculator?

Answer 1

Hint: In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratio of two side lengths. The above given functions are trigonometric functions. The angles of sine, cosine, and tangent are the primary classification of functions of trigonometry. The trigonometric functions are the periodic functions. The smallest periodic cycle is $2\pi $ but for tangent and the cotangent it is$\pi $.

Complete step by step answer:
Now, we have to calculate the$\sin \left( -405 \right)$ ,$\cos \left( -405 \right)$ and$\tan \left( -405 \right)$ . Since they are periodic functions, it means they can be written as:
$\begin{align}
  & \Rightarrow \sin \left( x+2n\pi \right)=\sin x \\
 & \Rightarrow \cos \left( x+2n\pi \right)=\cos x \\
 & \Rightarrow \tan \left( x+n\pi \right)=\tan x \\
\end{align}$
Where n is an integer.
To find the value of$\sin \left( -405 \right)$, as we know that period of sin is$2\pi $. So we write ${{405}^{\circ }}$ in terms of$\sin \left[ -\left( 360+45 \right) \right]$, we get
$\Rightarrow \sin \left[ -\left( 360+45 \right) \right]=\sin \left( -{{45}^{\circ }} \right)$
As we know very well$\sin \left( -\theta \right)=-\sin \theta $ , so we get
$\Rightarrow \sin \left( -{{45}^{\circ }} \right)=-\sin {{45}^{\circ }}$
And we also the value of $\sin {{45}^{\circ }}$ is$\dfrac{1}{\sqrt{2}}$ , then
\[\Rightarrow \sin \left( -{{405}^{\circ }} \right)=-\sin {{45}^{\circ }}=-\dfrac{1}{\sqrt{2}}\]
Similarly, to find the value of$\cos \left( -405 \right)$, as we know that period of cosine is $2\pi $.so we write${{405}^{\circ }}$in term of $\cos \left[ -\left( 360+45 \right) \right]$ , we get
$\Rightarrow \cos \left[ -\left( 360+45 \right) \right]=\cos \left( -{{45}^{\circ }} \right)$
And we know that $\cos \left( -\theta \right)=\cos \theta $ thus, we get
$\Rightarrow \cos \left( -{{405}^{\circ }} \right)=\cos \left( -{{45}^{\circ }} \right)=\cos {{45}^{\circ }}$
And we know the value of$\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ , we get
$\Rightarrow \cos \left( -{{405}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}$
Again to find the value of$\tan \left( -405 \right)$, since the period of tan is $n\pi $ where n is a constant.
And we also know we can write$\tan \left( -\theta \right)=\dfrac{\sin \left( -\theta \right)}{\cos \left( -\theta \right)}$ and we have been given that $\theta ={{405}^{\circ }}$and we already calculated the value of $\sin \left( -405 \right)$and$\cos \left( -405 \right)$, so now simply putting the value of these, we get
$\begin{align}
  & \Rightarrow \tan \left( -{{405}^{\circ }} \right)=\dfrac{\sin \left( -{{405}^{\circ }} \right)}{\cos \left( -{{405}^{\circ }} \right)} \\
 & \Rightarrow \tan \left( -{{405}^{\circ }} \right)=\dfrac{\left( -\dfrac{1}{\sqrt{2}} \right)}{\dfrac{1}{\sqrt{2}}} \\
 & \Rightarrow \tan \left( -{{405}^{\circ }} \right)=-1 \\
\end{align}$

Note: To solve these types of questions we should always need to remember the trigonometric formulas and properties and the basic values of all the trigonometric functions. We should know the all the values of trigonometric function on some basic which usually use in any trigonometric question are${{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }},{{0}^{\circ }}$ .