Question

How do you find the exact functional value $\cos \,{105^ \circ }$ using the cosine sum or difference identity?

Answer 1

Hint : To find the exact functional value cos 105o, by Here we use 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 answer :
We solve this by two methods
Method:1
Here in this question, we have to find the exact value of given $\cos \,{105^ \circ }$ by using cosine sum identity
$\cos \,{105^ \circ }$ can be written as $cos\,\left( {60 + 45} \right)$
We know the formula $cos\,(A + B) =$ $cos\,A.cos\,B - sin\,A.sin\,B$
Here $A = \,6{0^o}$ and $B = \,4{5^o}$
Substitute A and B in formula then
$\Rightarrow cos\,\left( {60 + 45} \right) = cos\,6{0^o}.cos\,4{5^o} - sin\,6{0^o}.sin\,4{5^o}$
By using specified cosine and sine angle i.e., $cos\,\,6{0^o} = \dfrac{1}{2}$ , $cos\,\,4{5^o} = \dfrac{1}{{\sqrt 2 }}$ , $sin\,6{0^o} = \dfrac{{\sqrt 3 }}{2}$ and $sin\,4{5^o} = \dfrac{1}{{\sqrt 2 }}$
$\therefore \,\,cos\,\left( {10{5^o}} \right) = cos\,6{0^o}.cos\,4{5^o} - sin\,6{0^o}.sin\,4{5^o}$
Substituting the values of $cos\,\,6{0^o}$ , $cos\,\,{45^o}$ , $\sin \,\,6{0^o}$ and $\sin \,\,{45^o}$
$\Rightarrow cos\,\left( {10{5^o}} \right) = \dfrac{1}{2}.\dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2}.\dfrac{1}{{\sqrt 2 }}$
On simplification we get
$\Rightarrow cos\,\left( {10{5^o}} \right) = \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
Take $2\sqrt 2$ as LCM in RHS
$\therefore cos\,\left( {10{5^o}} \right) = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
Hence, the exact functional value of $\cos \,{105^ \circ }$ is $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
Or
Method:2
Otherwise, we can also find the exact value of given $\cos \,{105^ \circ }$ by using cosine difference identity
$\cos \,{105^ \circ }$ can be written as $cos\,\left( {180 - 75} \right)$
We know the formula $cos\,(A - B) = cos\,A.cos\,B + sin\,A.sin\,B$
Here $A = \,18{0^o}$ and $B = \,7{5^o}$
Substitute A and B in formula then
$\therefore \,\,cos\,\left( {180 - 75} \right) = cos\,18{0^o}.cos\,7{5^o} + sin\,18{0^o}.sin\,7{5^o}$
We know the specified angle $cos 18{0^ \circ } = - 1$ and $\sin \,18{0^ \circ } = 0$
But we don’t know the value of $cos\,{75^o}$ and $sin\,{75^o}$ to find this by using formula of cosine and sine sum identity i.e., $cos\,(A + B) = cos\,A.cos\,B - sin\,A.sin\,B$ and $\sin \,(A + B) = sin\,A.cos\,B + cos\,A.sin\,B$
$\Rightarrow \,cos{75^o} = \cos \left( {45 + 30} \right) = cos\,{45^o}.cos\,{30^o} - sin\,{45^o}.sin\,{30^o}$
$\sin {75^o} = \sin \,(45 + 30) = sin\,{45^o}.cos\,{30^o} + cos\,{45^o}.sin\,{30^o}$
We know the value of $cos\,\,4{5^o} = \dfrac{1}{{\sqrt 2 }}$ , $cos\,\,{30^o} = \dfrac{{\sqrt 3 }}{2}$ , $sin\,4{5^o} = \dfrac{1}{{\sqrt 2 }}$ and $sin\,{30^o} = \dfrac{1}{2}$
$\Rightarrow \,cos{75^o} = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2} = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
$\,\sin {75^o} = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
$\,\therefore \cos {105^o} = \,\,cos\,\left( {180 - 75} \right) = - 1.\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right) + 0.\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)$
$\,\therefore \cos {105^o} = \,\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
Hence, the exact functional value of $\cos \,{105^ \circ }$ is $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
So, the correct answer is “ $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$ ”.

Note : The value of cosine can be determined by using several methods like double angle formula, half angle formula. Here we have found the exact value of $\cos \,{105^ \circ }$ by applying the cosine sum formula and cosine difference formula. It is defined as $cos\,(A + B) = cos\,A.cos\,B - sin\,A.sin\,B$ and $cos\,(A - B) = cos\,A.cos\,B + sin\,A.sin\,B$ . Here we have used the value of trigonometry ratios of standard angles. Hence, we can determine the solution for the question.