How do you find the exact functional value tan 165 using the cosine sum or difference identity?
Answer
626.7k+ views
Hint:Every trigonometric function and formulae are designed on the basis of three primary ratios. Sine, Cosine and tangents are these ratios in trigonometry based on Perpendicular, Hypotenuse and Base of a right triangle . In order to calculate the angles sin , cos and tan functions .
According to the formula of tan = $\dfrac{{\sin \theta }}{{\cos \theta }}$ , we will find the value of cosine and sine by putting the same angle $\theta = {165^{^ \circ }}$. Also by using the cosine sum or difference identity as required by the question.
Complete Step by step solution :
We will first find $\sin ({165^ \circ })$ by two methods. First by splitting the angle into two angles from which the values of the trigonometric functions are already known. We can split 165 into 120 and 45 as the trigonometric function .
Here we are using the second method by applying the formula of \[si{n^2}(x) = \dfrac{{[1 -
cos(2x)]}}{2}\] and \[{\cos ^2}(x) = \dfrac{{[1 + cos(2x)]}}{2}\]
Now, first we are going to calculate the $\sin ({165^ \circ })$, we are going to split the angles in such a way that the value is known to us corresponding to that angle of trigonometric function .
Like here we have split the angle into 180 and 15 and whenever there is 180 degree angle the other angle remains the same of which we are doing difference as,
\[\;{\mathbf{sin}}{\text{ }}\left({{\mathbf{180}}^\circ {\text{ }} - {\text{ }}{\mathbf{\theta }}} \right){\text{ }} = {\text{}}{\mathbf{sin}}{\text{ }}{\mathbf{\theta }}\] \[sin(165) = sin(180 - 15) = sin(15) = \sqrt {\dfrac{{1 - cos2(15)}}{2}} \]
Now we have to calculate for angle 30 for which value is known to us ,
\[\sqrt {\dfrac{{1 - \cos 30}}{2}} = \sqrt {\dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{2}} = \dfrac{{\sqrt {2 -\sqrt 3 } }}{2}\]
Now , we will find the $\cos ({165^ \circ })$by applying the formula \[{\cos ^2}(x) = \dfrac{{[1 + cos(2x)]}}{2}\]and applying the same strategy as we applied for $\sin ({165^ \circ })$ ,
we are going to get= \[cos(165) = cos(180 - 15) = - cos(15) = - \sqrt {\dfrac{{1 + cos2(15)}}{2}} \]
\[ - \sqrt {\dfrac{{1 + \cos 30}}{2}} = - \sqrt {\dfrac{{1 + \dfrac{{\sqrt 3 }}{2}}}{2}} = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}\]
By getting the values of $\sin ({165^ \circ })$ and $\cos ({165^ \circ })$
we are going to fetch the value of tangent by $\dfrac{{\sin \theta }}{{\cos \theta }}$ \[tan(165) = \dfrac{{sin(165)}}{{cos(165)}} = \dfrac{{\dfrac{{\sqrt {2 - \sqrt 3 } }}{2}}}{{ -
\dfrac{{\sqrt {2 + \sqrt 3 } }}{2}}} = \dfrac{{\sqrt {2 - \sqrt 3 } }}{{ - \sqrt {2 + \sqrt 3 } }}\]
This is the final answer.
Note : Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
It should be very clear that \[sin\left( {A + B} \right)\]is not equal to $\sin A + \sin B$.
No sine or cosine can ever be more than 1 as the ratio has the hypotenuse as its denominator.
Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its
domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its
domain.
Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T >
0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the
fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
According to the formula of tan = $\dfrac{{\sin \theta }}{{\cos \theta }}$ , we will find the value of cosine and sine by putting the same angle $\theta = {165^{^ \circ }}$. Also by using the cosine sum or difference identity as required by the question.
Complete Step by step solution :
We will first find $\sin ({165^ \circ })$ by two methods. First by splitting the angle into two angles from which the values of the trigonometric functions are already known. We can split 165 into 120 and 45 as the trigonometric function .
Here we are using the second method by applying the formula of \[si{n^2}(x) = \dfrac{{[1 -
cos(2x)]}}{2}\] and \[{\cos ^2}(x) = \dfrac{{[1 + cos(2x)]}}{2}\]
Now, first we are going to calculate the $\sin ({165^ \circ })$, we are going to split the angles in such a way that the value is known to us corresponding to that angle of trigonometric function .
Like here we have split the angle into 180 and 15 and whenever there is 180 degree angle the other angle remains the same of which we are doing difference as,
\[\;{\mathbf{sin}}{\text{ }}\left({{\mathbf{180}}^\circ {\text{ }} - {\text{ }}{\mathbf{\theta }}} \right){\text{ }} = {\text{}}{\mathbf{sin}}{\text{ }}{\mathbf{\theta }}\] \[sin(165) = sin(180 - 15) = sin(15) = \sqrt {\dfrac{{1 - cos2(15)}}{2}} \]
Now we have to calculate for angle 30 for which value is known to us ,
\[\sqrt {\dfrac{{1 - \cos 30}}{2}} = \sqrt {\dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{2}} = \dfrac{{\sqrt {2 -\sqrt 3 } }}{2}\]
Now , we will find the $\cos ({165^ \circ })$by applying the formula \[{\cos ^2}(x) = \dfrac{{[1 + cos(2x)]}}{2}\]and applying the same strategy as we applied for $\sin ({165^ \circ })$ ,
we are going to get= \[cos(165) = cos(180 - 15) = - cos(15) = - \sqrt {\dfrac{{1 + cos2(15)}}{2}} \]
\[ - \sqrt {\dfrac{{1 + \cos 30}}{2}} = - \sqrt {\dfrac{{1 + \dfrac{{\sqrt 3 }}{2}}}{2}} = - \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}\]
By getting the values of $\sin ({165^ \circ })$ and $\cos ({165^ \circ })$
we are going to fetch the value of tangent by $\dfrac{{\sin \theta }}{{\cos \theta }}$ \[tan(165) = \dfrac{{sin(165)}}{{cos(165)}} = \dfrac{{\dfrac{{\sqrt {2 - \sqrt 3 } }}{2}}}{{ -
\dfrac{{\sqrt {2 + \sqrt 3 } }}{2}}} = \dfrac{{\sqrt {2 - \sqrt 3 } }}{{ - \sqrt {2 + \sqrt 3 } }}\]
This is the final answer.
Note : Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
It should be very clear that \[sin\left( {A + B} \right)\]is not equal to $\sin A + \sin B$.
No sine or cosine can ever be more than 1 as the ratio has the hypotenuse as its denominator.
Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its
domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its
domain.
Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T >
0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the
fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
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