Question

How do you use the angle sum identity to find the exact value of $\cos \left( 285{}^\circ \right)$?

Answer 1

We will use the trigonometric identities to solve the above question. We will use the formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ and $\cos \left( 360{}^\circ -\theta \right)=\cos \theta$. At first, we will write $285{}^\circ$ as
$285{}^\circ =360{}^\circ -75{}^\circ$, and then we will use the formula $\cos \left( 360{}^\circ -\theta \right)=\cos \theta$, then we will get $\cos \left( 360{}^\circ -75{}^\circ \right)=\cos 75{}^\circ$. Now, we will write $75{}^\circ =30{}^\circ +45{}^\circ$and we will use the formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$, and then we will write $\cos \left( 30{}^\circ +45{}^\circ \right)=\cos 30{}^\circ \cos 45{}^\circ -\sin 30{}^\circ \sin 45{}^\circ$and since we know the value of $\cos 30{}^\circ ,\sin 30{}^\circ ,\sin 45{}^\circ ,\cos 45{}^\circ$from the trigonometric table and hence find the value.

Complete step-by-step solution:
We can see that the given question is of trigonometric identities so we will use it to solve the above question. We will first try to convert the angle given into less than $90{}^\circ$ and we know their value from the trigonometric table. Then, we will use the trigonometric sum identity of cosine function i.e. $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$.
Since, we can see from the question that $285{}^\circ$is greater than $90{}^\circ$and we also know that it lies in the fourth quadrant. So, we will first convert angle $285{}^\circ$ into angle less than $90{}^\circ$by using circular trigonometric function.
So, we can write $285{}^\circ$as $285{}^\circ =360{}^\circ -75{}^\circ$.
Now, we will take cosine both sides, then we will get:
$\Rightarrow \cos \left( 360{}^\circ -75{}^\circ \right)=\cos 285{}^\circ$
Since, we know that $\cos \left( 360{}^\circ -\theta \right)=\cos \theta$, so we will use it, and we will get:
$\Rightarrow \cos \left( 75{}^\circ \right)=\cos 285{}^\circ$
$\Rightarrow \cos 285{}^\circ =\cos \left( 75{}^\circ \right)$
Now, we will write $75{}^\circ$ as $30{}^\circ +45{}^\circ$ and use the formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$.
$\Rightarrow \cos \left( 75{}^\circ \right)=\cos \left( 30{}^\circ +45{}^\circ \right)$
$\Rightarrow \cos \left( 30{}^\circ +45{}^\circ \right)=\cos 30{}^\circ \cos 45{}^\circ -\sin 30{}^\circ \sin 45{}^\circ$
Since, from standard trigonometric table we know the value of $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2},\sin 30{}^\circ =\dfrac{1}{2},\sin 45{}^\circ =\dfrac{1}{\sqrt{2}},\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}$, so we will use them.
$\Rightarrow \cos \left( 75{}^\circ \right)=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}$
$\Rightarrow \cos \left( 75{}^\circ \right)=\dfrac{\sqrt{3}}{2\sqrt{2}}-\dfrac{1}{2\sqrt{2}}$
$\Rightarrow \cos \left( 75{}^\circ \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
Since, we have seen above that $\cos \left( 75{}^\circ \right)=\cos 285{}^\circ$
$\Rightarrow \cos \left( 285{}^\circ \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
**Hence, the value of $\cos \left( 285{}^\circ \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$.
This is our required solution. **

Note: Students are required to memorize the trigonometric formula, trigonometric table and identities otherwise they will not be able to solve the above question. We can also solve the above question writing $\cos \left( 285{}^\circ \right)=\cos \left( 180{}^\circ +105{}^\circ \right)$ and since we know that formula $\cos \left( 180{}^\circ +\theta \right)=-\cos \theta$, so we can write $\cos \left( 285{}^\circ \right)=\cos \left( 180{}^\circ +105{}^\circ \right)=-\cos 105{}^\circ$. Now, we can write $\cos \left( 105{}^\circ \right)=\cos \left( 60{}^\circ +45{}^\circ \right)$, and then we will use the formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$, and write the value using trigonometric table and simplify similarly as done above.