Question

How do you find the exact value of the sin, cos and tan of the angle -105 degrees?

Answer 1

In the given question, we have been asked to calculate the exact value of a trigonometric ratio in sine, cosine and tangent. The argument (or here, we can say the angle) of the ratio is also given. We just need to calculate the answer of the ratio. The trick here is the sign of the answer. Here, the angle is the one which does not have value in the standard given table. So, first we evaluate the value of $\sin \left( { - 105} \right)$, then use the relation between sine and cosine to calculate the value of cosine and then use the relation between sine and tangent to calculate the value of tangent.

Complete step by step answer:
We have to calculate the value of $\sin \left( { - 105} \right)$. For calculating that, we ignore the negative sign with the angle, calculate the answer and then just attach the ignored negative sign with the answer.
Now, $\sin \left( {a + b} \right) = \sin \left( a \right)\cos \left( b \right) + \cos \left( a \right)\sin \left( b \right)$
Hence, $\sin \left( {105} \right) = \sin \left( {45 + 60} \right)$
Applying the formula,
$\sin \left( {105} \right) = \sin 45\cos 60 + \cos 45\sin 60$
Putting in the values,
$\sin \left( {105} \right) = \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
Now, rationalizing the denominator,
$\sin \left( {105} \right) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}$
Now, $\sin \left( { - 105} \right) = - \sin \left( {105} \right) = - \dfrac{{\sqrt 6 + \sqrt 2 }}{4}$
Now we calculate the value of $\cos \left( { - 105} \right)$.
We know, $\cos \left( {a + b} \right) = \cos \left( a \right)\cos \left( b \right) - \sin \left( a \right)\sin \left( b \right)$.
Hence, $\cos \left( {105} \right) = \cos 45\cos 60 - \sin 45\sin 60$
Putting in the values,
$\cos \left( {105} \right) = \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }} = \dfrac{{\sqrt 2 - \sqrt 6 }}{4}$
Now, we know $\cos \left( { - \theta } \right) = \cos \left( \theta \right)$
Hence, $\cos \left( { - 105} \right) = \cos \left( {105} \right) = \dfrac{{\sqrt 2 - \sqrt 6 }}{4}$
Now, we calculate the value of $\tan \left( { - 105} \right)$ using the identity
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
So,
$\tan \left( { - 105} \right) = \dfrac{{\sin \left( { - 105} \right)}}{{\cos \left( { - 105} \right)}}$
Putting in the values,
$\tan \left( { - 105} \right) = \dfrac{{ - \dfrac{{\sqrt 6 + \sqrt 2 }}{{{4}}}}}{{\dfrac{{\sqrt 2 - \sqrt 6 }}{{{4}}}}} = \dfrac{{\sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }}$
Rationalizing the denominator,
$\tan \left( { - 105} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }} \times \dfrac{{\sqrt 6 + \sqrt 2 }}{{\sqrt 6 + \sqrt 2 }} = \dfrac{{{{\left( {\sqrt 6 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 6 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}$
Applying and solving,
$\tan \left( { - 105} \right) = \dfrac{{6 + 2 + 2 \times \sqrt {12} }}{{6 - 2}} = \dfrac{{8 + 4\sqrt 3 }}{4} = 2 + \sqrt 3$

Note: In the given question, we had to find the value of three trigonometric functions for an equal angle. We did that by using the appropriate formulae. It is important to remember that after calculating the values, if we are given a negative angle, we need to account that. Also, the point where a lot of students make a mistake is not rationalizing the denominator of the calculated value – if the denominator is irrational, it is to be made rational by rationalizing it.