Question

Evaluate the following \[\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}\]

Answer 1

Hint: we have to evaluate \[\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}\] By applying the formula sum and difference of sines and cosines and then dividing the equations and applying the formulas we will arrive to the final answer.

Complete Step-by-step answer:
Given \[\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
This is in the form of \[\sin \left( a+b \right)\]
\[\sin \left( a+b \right)\]\[=\]\[\sin a\cos b+\cos a\sin b\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Comparing both (a) and (1) we get that to solve (a) we can use the formula (1).
By following this \[\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}\]
\[\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}=\sin \left( 36+9 \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b)
To represent (b) used the formula of (1)
\[\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}=\sin \left( 45 \right)\]
We know that the value of \[\sin \left( 45 \right)\] as \[\dfrac{1}{\sqrt{2}}\]
\[\sin \left( 45 \right)=\dfrac{1}{\sqrt{2}}\]
Hence evaluated \[\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}\]= \[\sin \left( 45 \right)=\dfrac{1}{\sqrt{2}}\]

Note: To solve such types of problems all the related formulas should be handy. Also be careful about the signs in all the formulas. In the above equation division was made to simplify the terms and lead to the final answer. So smart moves lead to easy completion of answers.