Question

How do you use the reference angles to find $$\sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }$$?

Answer 1

Hint: To simplify $$\sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }$$ using the reference angle, we will find the value of given angles step by step. Using the concept of reference angle, we will write
$$\sin {210}^{\circ }=\sin ({180}^{\circ }+{30}^{\circ })$$,
$$\cos {330}^{\circ }=\cos \left({360}^{\circ }-{30}^{\circ }\right)$$ and
$$\tan {135}^{\circ }=\tan \left({180}^{\circ }-{45}^{\circ }\right)$$.
Then using the value of standard angles, we will find the value of $$\sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }$$.

Complete step by step answer:
According to the question, using the reference angles we have to find the value of $$\sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }$$. As we know, the reference angle is the acute angle with the x-axis. Thus, one by one we have to find the value of $$\sin {210}^{\circ }$$, $$\cos {330}^{\circ }$$ and $$\tan {135}^{\circ }$$ using the reference angle.
Let us consider the original angle given by $$\theta$$ and the auxiliary value is given by $$\alpha$$.
For the first quadrant, we have $$\theta =\alpha$$.
For the second quadrant, we have $$\theta ={180}^{\circ }-\alpha$$.
For the third quadrant, we have $$\theta ={180}^{\circ }+\alpha$$.
For the fourth quadrant, we have $$\theta ={360}^{\circ }-\alpha$$.

Consider $$\sin {210}^{\circ }$$. $${210}^{\circ }$$ is in the third quadrant.
Therefore, $$\sin {210}^{\circ }=\sin ({180}^{\circ }+\alpha )$$ i.e., $$\sin {210}^{\circ }=\sin ({180}^{\circ }+{30}^{\circ })$$
In the third quadrant, $$\sin$$ is negative.
So,
$$\Rightarrow \sin ({180}^{\circ }+{30}^{\circ })=-\sin {30}^{\circ }$$
$$\therefore \sin ({210}^{\circ })=-{\displaystyle \frac{1}{2}}$$
Now, consider $$\cos {330}^{\circ }$$. $${330}^{\circ }$$ lies in the fourth quadrant.
Therefore, $$\cos {330}^{\circ }=\cos \left({360}^{\circ }-{30}^{\circ }\right)$$.
In the fourth quadrant, $$\cos$$ is positive.
So,
$$\Rightarrow \cos \left({360}^{\circ }-{30}^{\circ }\right)=\cos {30}^{\circ }$$
$$\therefore \cos \left({330}^{\circ }\right)={\displaystyle \frac{\sqrt{3}}{2}}$$
Now, consider $$\tan {135}^{\circ }$$. $${135}^{\circ }$$ lies in the second quadrant.
Therefore, $$\tan {135}^{\circ }=\tan \left({180}^{\circ }-{45}^{\circ }\right)$$.
In the second quadrant, $$\tan$$ is negative.
So,
$$\Rightarrow \tan \left({180}^{\circ }-{45}^{\circ }\right)=-\tan {45}^{\circ }$$
$$\therefore \tan {135}^{\circ }=-1$$
Putting the values in $$\sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }$$, we get
$$\Rightarrow \sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }=\left(-{\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{\sqrt{3}}{2}}\right)-\left(-1\right)$$
On simplifying, we get
$$\Rightarrow \sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }=1-{\displaystyle \frac{\sqrt{3}}{4}}$$
Therefore, the value of $$\sin {210}^{\circ }\cos {330}^{\circ }-\tan {135}^{\circ }$$ is $$\left(1-{\displaystyle \frac{\sqrt{3}}{4}}\right)$$.

Note:
In the first quadrant, all trigonometric functions are positive. In the second quadrant, $$\sin$$ and $$\cos ec$$ are positive. In the third quadrant, $$\tan$$ and $$\cot$$ are positive. In the fourth quadrant, $$\cos$$ and $$\sec$$ are positive. Also, note that here we have used values of some standard angles i.e., $$\sin {30}^{\circ }={\displaystyle \frac{1}{2}}$$, $$\cos {30}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$$ and $$\tan {45}^{\circ }=1$$.