Question

How do you find the exact values of \[\cos \left( {67.5} \right)\] using the half angle formula?

Answer 1

Hint: In this question we have to find the particular value using half angle formula.
First, we need to know the half angle formula for cosine function and then find out if we can write the given value of cosine function in the form of \[\dfrac{\theta }{2}\] .After that putting the formula, we can find out the value of the given particular.

Formula used:
Half-Angle identity: \[1 + \cos \theta = 2{\cos ^2}\dfrac{\theta }{2}\]

Complete step by step solution:
We need to find out the exact valuus of \[\cos \left( {67.5} \right)\] using the half angle formula.
Since \[67.5\] is in quadrant I, where \[0 < x < 90^\circ \] and \[\cos x\] for \[0 < x < 90^\circ \] is positive, \[\cos \left( {67.5} \right) > 0\]
Also, \[67.5\] is the half of \[135\] and we can take \[\theta = 2 \times 67.5^\circ = 135^\circ \] .
From the half angle identity for cosine function, we can write,
\[1 + \cos \left( {135^\circ } \right) = 2{\cos ^2}\left( {\dfrac{{135^\circ }}{2}} \right) = 2{\cos ^2}\left( {67.5^\circ } \right)\]
\[\therefore ,2{\cos ^2}\left( {67.5^\circ } \right) = 1 + \cos \left( {180^\circ - 45^\circ } \right)\]
\[ = 1 - \cos 45^\circ \]
[using \[\cos \left( {\pi - \theta } \right) = - \cos \theta \] ]
\[ = 1 - \dfrac{1}{{\sqrt 2 }}\] …. [since the value of \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] ]
Taking LCM,
\[ = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}\]
\[ = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\]
\[ = \dfrac{{2 - \sqrt 2 }}{2}\]
Thus, we get, \[2{\cos ^2}\left( {67.5^\circ } \right) = \dfrac{{2 - \sqrt 2 }}{2}\]
\[{\cos ^2}\left( {67.5^\circ } \right) = \dfrac{{2 - \sqrt 2 }}{4}\]
\[\cos \left( {67.5^\circ } \right) = + \sqrt {\dfrac{{2 - \sqrt 2 }}{4}} \]
Positive sign is taken for the square root since,
\[67.5\] is in quadrant I, where \[0 < x < 90^\circ \] and \[\cos x\] for \[0 < x < 90^\circ \] is positive, \[\cos \left( {67.5} \right) > 0\]

Hence, we get, \[\cos \left( {67.5^\circ } \right) = \dfrac{{\sqrt {2 - \sqrt 2 } }}{2}\].

Note: Sine and Cosine formulas are based on sides of the right-angled triangle. Sine and Cosine are basic trigonometric functions along with tangent function, in trigonometry. Sine of angle is equal to the ratio of opposite side and hypotenuse whereas cosine of an angle is equal to ratio of adjacent side and hypotenuse.
\[\sin \theta = \dfrac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}}\]
\[\cos \theta = \dfrac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}}\]
Let us read about the quadrant system.

In the first quadrant ( \[0{\text{ to }}\dfrac{\pi }{2}\] ), all trigonometric functions are positive.
In the second quadrant ( \[\dfrac{\pi }{2}{\text{ to }}\pi \] ) only sine and cosec functions are positive.
In the third quadrant ( \[\pi {\text{ to }}\dfrac{{3\pi }}{2}\] ) only tangent and cotangent functions are positive.
In the fourth quadrant, ( \[\dfrac{{3\pi }}{2}to2\pi \] ) cosine and sec functions are positive.
Thus if \[0 < \theta < \dfrac{\pi }{2}\] then θ lies in the first quadrant where the value of cosine function is positive.