Question

How do you find the angle of the sum and difference identities to simplify $\sin {15^ \circ }$?

Answer 1

The given question describes the operation of using addition/ subtraction/ multiplication/ division. Also, in this question, we would use a trigonometric formula related to the given question. Also, remind the basic trigonometric table values and related formulae to simplify the given question.

Complete step by step answer:
The given question is shown below,
$\sin {15^ \circ } = ?$
The above equation can also be written as given below,
$\sin {15^ \circ } = \sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = ? \to \left( 1 \right)$
We make this change for easily handling calculations. In the above equation we have $\sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right)$, it is in the form of $\sin (a - b)$.
We know that,
$\sin (a - b) = \sin a\cos b - \cos a\sin b \to (2)$
So, the above equation is compared with the equation $(1)$ we get,
$\sin (a - b) = \sin a\cos b - \cos a\sin b$
$\sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \sin {60^ \circ }\cos {45^ \circ } - \cos {60^ \circ }\sin {45^ \circ } \to (3)$
So, the value of $a$is ${60^ \circ }$ and $b$ is ${45^ \circ }$.
Before we simplify the equation$(3)$, we need to know some basic values $\sin \theta and\cos \theta$. For this, we can use the following table,

| ${0^ \circ }$| ${30^ \circ }$| ${45^ \circ }$| ${60^ \circ }$| ${90^ \circ }$
---|---|---|---|---|---
$\sin \theta$| $0$| $\dfrac{1}{2}$| $\dfrac{1}{{\sqrt 2 }}$| $\dfrac{{\sqrt 3 }}{2}$| $1$
$\cos \theta$| $1$| $\dfrac{{\sqrt 3 }}{2}$| $\dfrac{1}{{\sqrt 2 }}$| $\dfrac{1}{2}$| $0$

By using this table we find the value of $\sin {60^ \circ },\sin {45^ \circ },\cos {60^ \circ },\cos {45^ \circ }$.
Let’s substitute the above-mentioned values in the equation $(3)$ we get,
$\left( 3 \right) \to \sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \sin {60^ \circ }\cos {45^ \circ } - \cos {60^ \circ }\sin {45^ \circ }$
$\sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) - \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right)$
By simplifying the above equation we get,
$\sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \left( {\dfrac{{\sqrt 3 }}{{2\sqrt 2 }}} \right) - \left( {\dfrac{1}{{2\sqrt 2 }}} \right) \to (4)$
We have to eliminate the root term in the denominator. So, let’s multiply the numerator and denominator with $\sqrt 2$, so the equation $\left( 4 \right)$ becomes,
$\left( 4 \right) \to \sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \left( {\dfrac{{\sqrt 3 }}{{2\sqrt 2 }}} \right) - \left( {\dfrac{1}{{2\sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 2 }}{{\sqrt 2 }}} \right)$
So, we get
$\sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \left( {\dfrac{{\sqrt 2 \times \sqrt 3 }}{{2 \times \sqrt 2 \times \sqrt 2 }}} \right) - \left( {\dfrac{{1 \times \sqrt 2 }}{{2 \times \sqrt 2 \times \sqrt 2 }}} \right)$
$\sin ({60^ \circ } - {45^ \circ }) = \left( {\dfrac{{\sqrt {2 \times 3} }}{{2 \times 2}}} \right) - \left( {\dfrac{{\sqrt 2 }}{{2 \times 2}}} \right)$
$\sin ({60^ \circ } - {45^ \circ }) = \dfrac{{\sqrt 6 }}{4} - \dfrac{{\sqrt 2 }}{4}$
When $\sqrt 2$ is multiplied with $\sqrt 2$ the answer will become 2.
In the above equation we can see that the denominator of both terms is the same. So, we can convert the two terms into a single term with small changes. So we get,
$\sin ({60^ \circ } - {45^ \circ }) = \dfrac{{\sqrt 6 - \sqrt 2 }}{4}$

So, the final answer is
$\sin ({15^ \circ }) = \dfrac{{\sqrt 6 - \sqrt 2 }}{4}$

Note: To solve this type of question we would use the operations of addition/ subtraction/ multiplication/ division. Also, note that if the root term is present in the denominator we try to eliminate the root term. When the same root term is multiplied with each other the root will be cancelled.