Question

How do you find the value of $\sin {435^ \circ }$?

Answer 1

In the given question, we are required to find the value of $\sin {435^ \circ }$. We will use the trigonometric formulae and identities such as $\sin \left( {\theta - {{360}^ \circ }} \right) = \sin \theta$ to find the value of the trigonometric function at a particular angle. We should be clear with the signs of all trigonometric functions in the four quadrants.

Complete step by step solution:
We have to find the value of the trigonometric function $\sin {435^ \circ }$.
Now, we know that the trigonometric function sine is positive in the first and second quadrant. Also, the value of the sine function gets repeated after a regular interval of $2\pi$ radians.
We know that ${435^ \circ } > {360^ \circ }$.
Also, we have, $\sin \left( { {{360}^ \circ }} +\theta \right) = \sin \theta$.
So, we get,
$\sin {435^ \circ } = \sin \left( { {{360}^ \circ }} + {{75}^ \circ } \right)$
Simplifying the expression, we get,
$\Rightarrow \sin {435^ \circ } = \sin {75^ \circ }$
We know that the angle ${75^ \circ }$ lies in the first quadrant. So, the sine of the angle will be a positive value.
Now, we will use the compound angle formula for sine as $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$.
So, we have, $\sin {435^ \circ } = \sin {75^ \circ }$
$\Rightarrow \sin {435^ \circ } \sin {75^ \circ } = \sin \left( {{{45}^ \circ } + {{30}^ \circ }} \right)$
Here, $A = {45^ \circ }$ and $B = {30^ \circ }$.
So, we get,
$\Rightarrow \sin {435^ \circ } = \sin \left( {{{45}^ \circ }} \right)\cos \left( {{{30}^ \circ }} \right) + \sin \left( {{{30}^ \circ }} \right)\cos \left( {{{45}^ \circ }} \right)$
Now, we know the values of trigonometric functions sine and cosine for angles ${45^ \circ }$ and ${30^ \circ }$ as $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, $\sin {30^ \circ } = \dfrac{1}{2}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$.
Substituting these values, we get,
$\Rightarrow \sin {435^ \circ } = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) + \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right)$
Simplifying the expression, we get,
$\Rightarrow \sin {435^ \circ } = \left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)$
Therefore, the value of $\sin {435^ \circ }$ is $\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)$.
This value is approximately equal to $0.966$.

Note:
We can also solve the given problem using the periodicity of the cosine and sine functions. Periodic Function is a function that repeats its value after a certain interval. For a real number $T > 0$, $f\left( {x + T} \right) = f\left( x \right)$ for all x. If T is the smallest positive real number such that $f\left( {x + T} \right) = f\left( x \right)$ for all x, then T is called the fundamental period. The fundamental period of sine is $2\pi$ radians.