Question

How do you find the exact value of the sine, cosine, and tangent of the angle \[285^\circ \]?

Answer 1

Hint: To find the values for an particular angle in trigonometry we have to break the angle into the smaller angle such as zero, thirty, forty five, sixty and ninety, because we know the direct value for these angles, or otherwise we have plot the graph and then see for the value.

Formulae Used:
\[ \Rightarrow \sin (A + B) = \sin A\cos B + \sin B\cos A\]
\[ \Rightarrow \cos (A + B) = \cos A\cos B - \sin B\sin A\]
\[ \Rightarrow \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]

Complete step by step solution:
The given question need to obtain the value of the trigonometric function for the given angle \[285^\circ \]
The given angle is two hundred eighty five degree and needed to be solved by splitting the angle into two hundred twenty five and sixty, we can write as:
\[ \Rightarrow 285 = (225 + 60)\]
Now first solving for “sin” we get:
\[
   \Rightarrow \sin (225 + 60) = \sin 225\cos 60 + \sin 60\cos 225(u\sin g\,\sin (A + B) = \sin A\cos B + \sin B\cos A) \\
   \Rightarrow \sin 225 = \sin (180 + 45) = \sin 180\cos 45 + \sin 45\cos 180 = 0 \times \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} \times ( - 1) = \dfrac{{ - 1}}{{\sqrt 2 }} \\
   \Rightarrow \cos 225 = \cos (180 + 45) = \cos 180\cos 45 - \sin 180\sin 45 = ( - 1) \times \dfrac{1}{{\sqrt 2 }} + 0 \times \dfrac{1}{{\sqrt 2 }} = \dfrac{{ - 1}}{{\sqrt 2 }} \\
   \Rightarrow \sin (225 + 60) = \dfrac{{ - 1}}{{\sqrt 2 }} \times \dfrac{1}{2} + \dfrac{{ - 1}}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} = \dfrac{{ - 1}}{{2\sqrt 2 }} + \dfrac{{ - \sqrt 3 }}{{2\sqrt 2 }} = \dfrac{{ - 2\sqrt 3 }}{{2\sqrt 2 }} = - \sqrt {\dfrac{3}{2}} \\
 \]
Now solving for “cos” we get:
\[
   \Rightarrow \cos (225 + 60) = \cos 225\cos 60 - \sin 225\sin 60(u\sin g\,\cos (A + B) = \cos A\cos B - \sin B\sin A) \\
   \Rightarrow co\operatorname{s} (225 + 60) = \dfrac{{ - 1}}{{\sqrt 2 }} \times \dfrac{1}{2} - \dfrac{{ - 1}}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} = \dfrac{{ - 1}}{{2\sqrt 2 }} + \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
 \]
Now for “tangent” we have to use the relation which is:
\[ \Rightarrow \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
On solving we get:
\[ \Rightarrow \tan (225 + 60) = \dfrac{{\sin (225 + 60)}}{{\cos (225 + 60)}} = \dfrac{{ - \dfrac{{\sqrt 3 }}{{\sqrt 2 }}}}{{\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}}} = \dfrac{{ - 2\sqrt 3 }}{{\sqrt 3 - 1}}\]


Note: Here after finding the value of “sin” we can also use trigonometric formulae to get the value for “cos”, the result from there would be same what we get here, only the thing is for not making any steps or to not recall any other formulae we go through these steps.

The above question can also be solved by plotting the graph for every identity and marking the value for the given angle, but that process needs exact plotting of the graph which is very difficult on normal paper, so this process that we used here is appropriate to solve this question to get the values for the angles.