Question

How do you evaluate the sine, cosine and tangent of the angle 225 degrees without using a calculator?

Answer 1

In order to solve the question we will require these formulas: $\sin \left( {{180}^{\text{o}}}+\theta \right)=-\sin \left( \theta \right),\cos \left( {{180}^{\text{o}}}+\theta \right)=-\cos \left( \theta \right),\tan \left( {{180}^{\text{o}}}+\theta \right)=\tan \left( \theta \right)$. Here we can clearly notice that the formulas of sine and cosine are negative but for tangent it is positive. This is because of the fact that the given angle ${{225}^{\text{o}}}$ lies in the third quadrant. We will also use $\sin \left( {{45}^{\text{o}}} \right)=\dfrac{1}{\sqrt{2}},\cos \left( {{45}^{\text{o}}} \right)=\dfrac{1}{\sqrt{2}},\tan \left( {{45}^{\text{o}}} \right)=1$.

Complete step by step solution:
First we will discuss the definition of trigonometric functions, which is suitable for all trigonometric angles. One angle in any triangle, taken as a reference for trigonometric function, will be of ${{90}^{\text{o}}}$. The functions with those angles which bond both the sides of a triangle to that angle which is ${{90}^{\text{o}}}$ together are called trigonometric functions. These trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant.
In reference to the given question, it is required to evaluate $\sin \left( {{225}^{\text{o}}} \right),\,\cos \left( {{225}^{\text{o}}} \right),\,\tan \left( {{225}^{\text{o}}} \right)$. By looking at the graph below, we come to know that the angle ${{225}^{\text{o}}}$ lies only in the third quadrant. In this quadrant, we have negative sine and cosine but positive tangent. Moreover, the reference angle can be ${{180}^{\text{o}}}+\theta$. Therefore,

$\begin{aligned} & \sin \left( {{225}^{\text{o}}} \right)=\sin \left( {{180}^{\text{o}}}+{{45}^{\text{o}}} \right) \\\ & \Rightarrow \sin \left( {{225}^{\text{o}}} \right)=-\sin \left( {{45}^{\text{o}}} \right)\,\left( \because \sin \left( {{180}^{\text{o}}}+\theta \right)=-\sin \left( \theta \right) \right) \\\ & \Rightarrow \sin \left( {{225}^{\text{o}}} \right)=-\dfrac{1}{\sqrt{2}}\left( \because \sin \left( {{45}^{\text{o}}} \right)=\dfrac{1}{\sqrt{2}} \right) \\\ \end{aligned}$
$\begin{aligned} & \cos \left( {{225}^{\text{o}}} \right)=\cos \left( {{180}^{\text{o}}}+{{45}^{\text{o}}} \right) \\\ & \Rightarrow \cos \left( {{225}^{\text{o}}} \right)=-\cos \left( {{45}^{\text{o}}} \right)\,\left( \because \cos \left( {{180}^{\text{o}}}+\theta \right)=-\cos \left( \theta \right) \right) \\\ & \Rightarrow \cos \left( {{225}^{\text{o}}} \right)=-\dfrac{1}{\sqrt{2}}\left( \because \cos \left( {{45}^{\text{o}}} \right)=\dfrac{1}{\sqrt{2}} \right) \\\ \end{aligned}$
$\begin{aligned} & \tan \left( {{225}^{\text{o}}} \right)=\tan \left( {{180}^{\text{o}}}+{{45}^{\text{o}}} \right) \\\ & \Rightarrow \tan \left( {{225}^{\text{o}}} \right)=\tan \left( {{45}^{\text{o}}} \right)\,\left( \because \tan \left( {{180}^{\text{o}}}+\theta \right)=\tan \left( \theta \right) \right) \\\ & \Rightarrow \tan \left( {{225}^{\text{o}}} \right)=1\left( \because \tan \left( {{45}^{\text{o}}} \right)=1 \right) \\\ \end{aligned}$
Hence, the values of $\sin \left( {{225}^{\text{o}}} \right)=-\dfrac{1}{\sqrt{2}},\cos \left( {{225}^{\text{o}}} \right)=-\dfrac{1}{\sqrt{2}},\tan \left( {{225}^{\text{o}}} \right)=1$.

Note:
It is important where the given angle is lying. As there are four quadrants, the given angle can lie in any of these quadrants. In this question the angle ${{225}^{\text{o}}}$ lies in the third quadrant. After this we can use the required formulas and complete the answer. We could have also used the reference angle as ${{270}^{\text{o}}}-\theta$ and solved it. For example, $\sin \left( {{225}^{\text{o}}} \right)=\sin \left( {{270}^{\text{o}}}-{{45}^{\text{o}}} \right)$ and solve it further as done above. One should remember all formulas related to trigonometric functions. As here it is required to use $\sin \left( {{45}^{\text{o}}} \right)=\dfrac{1}{\sqrt{2}},\cos \left( {{45}^{\text{o}}} \right)=\dfrac{1}{\sqrt{2}},\tan \left( {{45}^{\text{o}}} \right)=1$.