Question

Simplify the trigonometric expression: $ \sin 15^\circ \cos 75^\circ + \cos 15^\circ \sin 75^\circ $

Answer 1

Hint: We are given an expression with sine and cosine functions and two different angles in degrees. We have to simply find the value of the trigonometric expression by using the standard identity of $ \sin \left( {A + B} \right) $ to find its value and then evaluate the sine value for the resultant angle.

Complete step-by-step answer:
Firstly we write down the trigonometric expression given in the question i.e.
 $ \sin 15^\circ \cos 75^\circ + \cos 75^\circ \sin 15^\circ {\text{ - - - - - - (1)}} $
Now as we can see that expression (1) has sine and cosine functions with two different angles in degree so we compare this with standard trigonometric identity i.e.
 $ \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B $
Comparing expression (1) with the above formula we have
 $
  A = 15^\circ \\
  B = 75^\circ \;
  $
Now we put the values of (1) in the formula we have
 $ \sin 15^\circ \cos 75^\circ - \cos 15^\circ \sin 75^\circ = \sin (15^\circ + 75^\circ ) = \sin 90^\circ $
So we have found the result after simplifying the expression to be
 $ \sin 90^\circ $
And we know it is a standard trigonometric function whose value is known to us i.e.
 $ \sin 90^\circ = 1 $
So we have obtained the required value of the given trigonometric expression.
So, the correct answer is “1”.

Note: With the help of right trigonometric identity and value of sine function we solve the question and it is always helpful when you memorize these identities alongside with the values of all six functions described in the table.

θfunction	0 ̊	30 ̊	45 ̊	60 ̊	90 ̊
sin	0	$ \dfrac{1}{2} $	\[\dfrac{1}{{\sqrt 2 }}\]	$ \dfrac{{\sqrt 3 }}{2} $	1
cos	1	$ \dfrac{{\sqrt 3 }}{2} $	\[\dfrac{1}{{\sqrt 2 }}\]	$ \dfrac{1}{2} $	0
tan	0	$ \dfrac{1}{{\sqrt 3 }} $	1	$ \sqrt 3 $	$ \infty $
cot	$ \infty $	$ \sqrt 3 $	1	$ \dfrac{1}{{\sqrt 3 }} $	0
sec	1	$ \dfrac{2}{{\sqrt 3 }} $	$ \sqrt 2 $	2	$ \infty $
cosec	$ \infty $	2	$ \sqrt 2 $	$ \dfrac{2}{{\sqrt 3 }} $	1