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Question: How do you find the exact functional value of \(\tan \left( {{{105}^0}} \right)\) using the cosine s...

How do you find the exact functional value of tan(1050)\tan \left( {{{105}^0}} \right) using the cosine sum or difference identity?

Explanation

Solution

Convert the tangent function into sine and cosine and then split the argument such that the functional values of the separated arguments are known to you. Finally use the cosine sum or difference identity to find the required value.Trigonometric identity of cosine sum is given as follows, cos(a+b)=cosacosbsinasinb\cos (a + b) = \cos a\cos b - \sin a\sin b.Whereas Trigonometric identity of sine sum is given as follows, sin(a+b)=sinacosb+sinbcosa\sin (a + b) = \sin a\cos b + \sin b\cos a.

Complete step by step answer:
In order to find the exact value of tan(1050)\tan \left( {{{105}^0}} \right) using the cosine sum or difference identity we will first convert tangent into sine and cosine ratio, that is
tan(1050)=sin(1050)cos(1050)\tan \left( {{{105}^0}} \right) = \dfrac{{\sin \left( {{{105}^0}} \right)}}{{\cos \left( {{{105}^0}} \right)}}
Now we will split the argument in such a manner that we are known to the functional values of the separated arguments, so splitting the given argument 1050{105^0} as the sum of 600  and  450{60^0}\;{\text{and}}\;{45^0}, that is 1050=600+450{105^0} = {60^0} + {45^0}.
Putting this above we will get,
sin(1050)cos(1050)=sin(600+450)cos(600+450)\dfrac{{\sin \left( {{{105}^0}} \right)}}{{\cos \left( {{{105}^0}} \right)}} = \dfrac{{\sin \left( {{{60}^0} + {{45}^0}} \right)}}{{\cos \left( {{{60}^0} + {{45}^0}} \right)}}

Now using the sine sum and cosine sum identity in order to evaluate the above trigonometric expression
sin(600+450)cos(600+450)=sin600cos450+sin450cos600cos600cos450sin600sin450\dfrac{{\sin \left( {{{60}^0} + {{45}^0}} \right)}}{{\cos \left( {{{60}^0} + {{45}^0}} \right)}} = \dfrac{{\sin {{60}^0}\cos {{45}^0} + \sin {{45}^0}\cos {{60}^0}}}{{\cos {{60}^0}\cos {{45}^0} - \sin {{60}^0}\sin {{45}^0}}}
Substituting the respective functional values we will get

\dfrac{{\sin {{60}^0}\cos {{45}^0} + \sin {{45}^0}\cos {{60}^0}}}{{\cos {{60}^0}\cos {{45}^0} - \sin {{60}^0}\sin {{45}^0}}} = \dfrac{{\dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}}}{{\dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}}} \\\ \Rightarrow\dfrac{{\sin {{60}^0}\cos {{45}^0} + \sin {{45}^0}\cos {{60}^0}}}{{\cos {{60}^0}\cos {{45}^0} - \sin {{60}^0}\sin {{45}^0}}} =\dfrac{{\dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}}}{{\dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}}} \\\ \Rightarrow\dfrac{{\sin {{60}^0}\cos {{45}^0} + \sin {{45}^0}\cos {{60}^0}}}{{\cos {{60}^0}\cos {{45}^0} - \sin {{60}^0}\sin {{45}^0}}} = \dfrac{{\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}}{{\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}}} \\\ \therefore\dfrac{{\sin {{60}^0}\cos {{45}^0} + \sin {{45}^0}\cos {{60}^0}}}{{\cos {{60}^0}\cos {{45}^0} - \sin {{60}^0}\sin {{45}^0}}} = \dfrac{{1 + \sqrt 3 }}{{1 - \sqrt 3 }} \\\ $$ **So the required functional of $\tan \left( {{{105}^0}} \right)\;{\text{is}}\;\dfrac{{1 + \sqrt 3 }}{{1 - \sqrt 3 }}$.** **Note:** Denominator of the final result is in irrational form so better you convert it into rational form by multiplying and dividing the number with conjugate of the denominator. Conjugate of a number $(1 + x)$ is equals to $(1 - x)$. So either convert it into rational form or put the value of $\sqrt 3 $ and find the final value in decimals which will come to be approximately $ - 3.732..$ you can check this value with a scientific calculator.