Question

How do you use the angle difference identity to find the exact value of $\cos 15?$

Answer 1

In the given problem we have to find the value of $\cos 15$ by using the angle of difference identity. There are various problems. Here we have to use the identity of $\cos \left( A-B \right)$ Simplify the answer step wise properly and also you need to know the proper value of sine and cosine of the given angle.

Complete step by step solution: Here we have, $\cos 15{}^\circ$
If we can find the two angles $A$ and $B$ where whose sum and difference is $15{}^\circ$ and also whose sine and cosine we know,
Therefore we have to use difference identity,
$\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B$
We can write $\cos 15{}^\circ$ as,
$\cos 15{}^\circ =\cos \left( 60{}^\circ -45{}^\circ \right)$
Here, $\left( 60-45 \right)=15$ and we know the trigonometric function of $60$ and $45.$
Substitute it in the given identity then the equation becomes,
$\cos 15{}^\circ =\cos 60{}^\circ \cos 45{}^\circ +\sin 60{}^\circ \sin 45{}^\circ$
$=\dfrac{1}{2}\left( \dfrac{1}{\sqrt{2}} \right)+\dfrac{\sqrt{3}}{2}\left( \dfrac{1}{\sqrt{2}} \right)$
$=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}$
$\cos 15{}^\circ =\dfrac{1+\sqrt{3}}{2\sqrt{2}}$
Hence,
The value of $\cos 15{}^\circ =\dfrac{1+\sqrt{3}}{2\sqrt{2}}$

Additional Information:
We have to use the sum and difference identities for sin, cosine and tangent. While solving any trigonometric problem you need to convert all $\sec ,\cos ,\csc$ and to $\sin$ and this helps us to solve it easily. Using identities then check all sums and differences and use identities to remove what you have to remove.
If you have any factored problem so the common factor can be canceled from the numerator and denominator while solving the problem.
You don't always have to use $\sin$ and $\cos$ always but it is easier to compare when both sides are composed of similar functions. If you are practicing let others read through your problem just to check whether they can give a new way. If we stare at an equation for a long time, then we start to miss any ideas that came in our mind.

Note:
Always check properly what is asking in question because if there is $\sin 15$ the students sometimes use $\cos$ identity there. We could use similar property for $\sin 15$ but only we have to change the identity there. We have to use $\sin \left( A-B \right)$ identity. We should use $45-30=15$ instead of $60-45=15$ for $\sin 15.$ Main point is that there is no other way for finding this type of solution.