Question

How do you find the exact functional value $\cos \dfrac{{23\pi }}{{12}}$ using the cosine sum or difference identity ?

Answer 1

Here in this question we have to find the exact value of a given trigonometric function by using the cosine sum or difference identity. First rewrite the given angle in the form of addition or difference, then the standard trigonometric formula cosine sum i.e., $cos\,(A + B)$or cosine difference i.e., $cos\,(A - B)$ identity defined as $cos\,A.cos\,B - sin\,A.sin\,B$ and $cos\,A.cos\,B + sin\,A.sin\,B$ using one of these we get required value.

Complete step by step solution:
To evaluate the given question by using a formula of cosine addition defined as the cosine addition formula calculates the cosine of an angle that is either the sum or difference of two other angles. It arises from the law of cosines and the distance formula. By using the cosine addition formula, the cosine of both the sum and difference of two angles can be found with the two angles' sines and cosines. Consider the given function
$\cos \dfrac{{23\pi }}{{12}}$-------(1)
The angle $\dfrac{{23\pi }}{{12}}$ can be written as $- \dfrac{\pi }{{12}} + 2\pi$, then
Equation (1) becomes
$\cos \left( { - \dfrac{\pi }{{12}} + 2\pi } \right)$
$\Rightarrow \,\cos \left( {2\pi - \dfrac{\pi }{{12}}} \right)$------(2)
By using the ASTC rule of trigonometry, the angle $2\pi - \dfrac{\pi }{{12}}$ or angle $360 - \theta$ lies in the third quadrant. cosine functions are positive here, hence the angle must be positive. Here we must keep the function as cosine itself.

Then, equation (2) becomes
$\Rightarrow \,\cos \left( {\dfrac{\pi }{{12}}} \right)$----------(3)
Again, the angle $\dfrac{\pi }{{12}}$ can be written as $\dfrac{\pi }{3} - \dfrac{\pi }{4}$, then equation (3) becomes
$\cos \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$
Apply the trigonometric cosine identity of difference cos\,(A - B) = $$$$cos\,A.cos\,B + sin\,A.sin\,B.
Here $A = \,\dfrac{\pi }{3}$ and $B = \,\dfrac{\pi }{4}$
Substitute A and B in formula then
$\cos \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = cos\,\dfrac{\pi }{3}.cos\,\dfrac{\pi }{4} + sin\,\dfrac{\pi }{3}.sin\,\dfrac{\pi }{4}$
By using specified cosine and sine angle i.e., $cos\,\,\dfrac{\pi }{3} = \dfrac{1}{2}$, $cos\,\,\dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$, $sin\,\dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ and $sin\,\dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \,\cos \left( {\dfrac{{23\pi }}{{12}}} \right) = \cos \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = cos\,\dfrac{\pi }{3}.cos\,\dfrac{\pi }{4} + sin\,\dfrac{\pi }{3}.sin\,\dfrac{\pi }{4}$

Substituting the values of $cos\,\,6{0^o}$, $cos\,\,{45^o}$, $\sin \,\,6{0^o}$and $\sin \,\,{45^o}$
$\cos \left( {\dfrac{{23\pi }}{{12}}} \right) = \cos \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \dfrac{1}{2}.\dfrac{1}{{\sqrt 2 }} + \dfrac{{\sqrt 3 }}{2}.\dfrac{1}{{\sqrt 2 }}$
On simplification we get
$\cos \left( {\dfrac{{23\pi }}{{12}}} \right) = \dfrac{1}{{2\sqrt 2 }} + \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
Take $2\sqrt 2$ as LCM in RHS
$\therefore\,\cos \left( {\dfrac{{23\pi }}{{12}}} \right) = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$

Hence, the exact functional value of $\cos \left( {\dfrac{{23\pi }}{{12}}} \right)$ is $\dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$.

Note: While the function is trigonometry we must know about the ASTC rule. Since they have mentioned how to solve the given function by using the cosine sum or difference identity, for this we have a standard formula. To find the value for the trigonometry function we need the table of trigonometry ratios for standard angles.