Question

Use the sum or difference identity to search out the precise value of sin 255 degrees.

Answer 1

The sum and difference identities in trigonometry for sine are,
$sin(A + B) = sinAcosB + sinBcosA$
$sin(A - B) = sinAcosB - sinBcosA$
Divide the given angle such that we have A and B as standard angle values (for easier calculations). Put the values of A and B in the appropriate identity to get your final answer.

Complete step by step answer:
The angle in the question can be divided as, $sin255^\circ = sin\left( {75 + 180} \right)^\circ$
Now, we see that $sin\left( {\theta + 2\pi } \right)^\circ = - sin\;\theta ^\circ$
Thus, the question can be rewritten as,
$sin255^\circ = - sin75^\circ$………. (i)
$sin75^\circ$can be further divided into two parts as $sin75^\circ = sin(45 + 30)^\circ$
Using the sum identity, $sin(A + B) = sinAcosB + sinBcosA$.
$sin75^\circ = sin\left( {30 + 45} \right)^\circ = (sin30^\circ \times cos45^\circ ) + (sin45^\circ \times cos30^\circ )$, where $A = 30^\circ ,B = 45^\circ$.
$\Rightarrow sin75^\circ= (\dfrac{1}{2}).(\dfrac{{\sqrt 2 }}{2}) + (\dfrac{{\sqrt 2 }}{2}).(\dfrac{{\sqrt 3 }}{2})$
$\Rightarrow sin75^\circ= (\dfrac{{\sqrt 2 }}{2})(\dfrac{{1 + \sqrt 3 }}{2}) = \dfrac{{\sqrt 2 + \sqrt 6 }}{4}$
Putting the above value in equation (i),
$\therefore - sin75^\circ = - (\dfrac{{\sqrt 2 + \sqrt 6 }}{4})$

Thus, the final answer of $sin255^\circ = - sin75^\circ = - (\dfrac{{\sqrt 2 + \sqrt 6 }}{4})$.

Note: The three main functions in trigonometry are Sine, Cosine and Tangent. Trigonometric identities are equalities that involve trigonometric functions and are true for each value of the occurring variables where each side of the equality is defined. Geometrically, these are identities involving certain functions of 1 or more angles. The Trigonometric Identities are equations that are true for Right Angled Triangles.