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.