Question

Find the value of $\cos {75^\circ }$.

Answer 1

In this problem we are asked to find the cosine of some degree. To proceed this problem mainly we have to write the given degree into some form of addition of two values and the students asked to solve the problem by using some trigonometric identities and trigonometric table values. By following these steps we can get the correct value of the given problem.

Formula used: Compound - angle formulae, $\cos \left( {a + b} \right) = \left( {\cos a} \right)\left( {\cos b} \right) - \left( {\sin a} \right)\left( {\sin b} \right)$

Complete step-by-step solution:
We need to find the value of $\cos 75$.
We can write $\cos 75$ as $\cos \left( {30 + 45} \right)$. Since $75 = 30 + 45$
The values of $\sin 45,\sin 30,\cos 45$ and$\cos 30$ are commonly known.
Now we use the formula which we mentioned above for $\cos \left( {30 + 45} \right)$,
$\cos \left( {45 + 30} \right) = \left( {\cos 30} \right)\left( {\cos 45} \right) - \left( {\sin 30} \right)\left( {\sin 45} \right) - - - - - \left( 1 \right)$
The values of $\cos 30$ and $\sin 30$ also $\cos 45$ and $\sin 45$ are listed below
$\cos 30 = \dfrac{{\sqrt 3 }}{2}$
$\sin 30 = \dfrac{1}{2}$
$\sin 45 = \cos 45 = \dfrac{{\sqrt 2 }}{2}$
We can write $\dfrac{{\sqrt 2 }}{2} = \dfrac{1}{{\sqrt 2 }}$
So, $\sin 45 = \cos 45 = \dfrac{1}{{\sqrt 2 }}$
Substitute these values in the equation $\left( 1 \right)$ we get,
$\cos \left( {30 + 45} \right) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}$
$\cos \left( {30 + 45} \right) = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }}$
Taking LCM we get,
$\cos \left( {30 + 45} \right) = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$

Hence $\cos \left( {75} \right) = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$

Additional Information: The cosine of an angle is defined as the sine of the complementary angle. The complementary angle equals the given angle subtracted from a right angle, ${90^\circ }$. For instance, if the angle is ${30^\circ }$, then its complement is ${60^\circ }$. In trigonometry, the law of cosines related the lengths of the sides of a triangle to the cosine of one of its angles.

Note: In this problem we can also find the answer as decimal value. By substituting the cosine and sine degree values as decimal values we get the answers as decimal value. Also there is an alternate solution for this problem.
$\cos \left( {30 + 45} \right) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
$\Rightarrow \dfrac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{4}$
By substituting the values of $\sqrt 2$ and $\sqrt 3$ we get,
\Rightarrow \dfrac{{\left\\{ {\left( {1.414} \right)\left( {1.732 - 1} \right)} \right\\}}}{4}
Simplifying we get,
$\Rightarrow \dfrac{{\left( {0.707} \right)\left( {0.732} \right)}}{2}$
Divided the term,
$\Rightarrow \left( {0.707} \right)\left( {0.366} \right)$
Multiplying the terms we get,
$\Rightarrow 0.258762$
$\cos \left( {30 + 45} \right) = 0.258762$
So if we substitute the decimal values we get the answer in the form of decimal.