Question

Convert the following products into the sum or difference. If angles are given in degree, evaluate from tables.
i.$2\sin {{48}^{\circ }}\cos {{12}^{\circ }}$
ii.$2\sin {{54}^{\circ }}\sin {{66}^{\circ }}$
iii.$2\cos 5\theta \cos 3\theta $
iv.$2\cos {{72}^{\circ }}\sin {{56}^{\circ }}$
v.$\cos (\alpha +\beta )\cos (\alpha -\beta )$
vi.$\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$

Answer 1

Hint: We know that this question is purely a trigonometry question. In order to solve this question, all we need to know is the basics of trigonometry, some of its identities (half angle and the double angle formulae), and we will be able to solve it.

Complete step-by-step answer:
Trigonometry is basically the branch of mathematics which, in general, deals with the relations between the length of the sides of the triangle and the angle between them. For example, if we know the length of two of the sides of a triangle, we can practically find out the length of the third side of the triangle and the angle between all of them can also be found with the trigonometry.
Some of the identities that are going to useful for this question are as follows:
$2\sin A\cdot \cos B=\sin (A+B)+\sin (A-B)$
$2\sin A\cdot \sin B=\cos (A-B)-\cos (A+B)$
$2\cos A\cdot \cos B=\cos (A+B)+\cos (A-B)$
$2\cos A\cdot \sin B=\sin (A+B)-\sin (A-B)$

i.$2\sin {{48}^{\circ }}\cos {{12}^{\circ }}$
Looking at this expression, we can see that it is similar to the first identity. Therefore, we can directly apply the first identity to solve this equation:
Applying first identity on this expression, we get:
$2\sin {{48}^{\circ }}\cos {{12}^{\circ }}=\sin ({{48}^{\circ }}+{{12}^{\circ }})+\sin ({{48}^{\circ }}-{{12}^{\circ }})$
$\Rightarrow 2\sin {{48}^{\circ }}\cos {{12}^{\circ }}=\sin {{60}^{\circ }}+\sin {{36}^{\circ }}$
Upon looking at the value table, we get:
$\sin {{60}^{\circ }}+\sin {{36}^{\circ }}=\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{10-2\sqrt{5}}}{4}$
This is the required answer.

ii.$2\sin {{54}^{\circ }}\sin {{66}^{\circ }}$
Looking at this expression, we can see that it is similar to the second identity. Therefore, we can directly apply the second identity to solve this equation:
Applying second identity on this expression, we get:
$2\sin {{54}^{\circ }}\sin {{66}^{\circ }}=\cos ({{54}^{\circ }}-{{66}^{\circ }})-\cos ({{54}^{\circ }}+{{66}^{\circ }})$
$\Rightarrow 2\sin {{54}^{\circ }}\sin {{66}^{\circ }}=\cos {{12}^{\circ }}-\cos {{120}^{\circ }}$
Upon looking at the value table, we get:
$\cos {{12}^{\circ }}-\cos {{120}^{\circ }}=\dfrac{\sqrt{3}}{2}+\sqrt{3}\dfrac{\sqrt{10+2\sqrt{5}}}{8}-\dfrac{\sqrt{5}-{1}}{8}$
This is the required answer.

iii.$2\cos 5\theta \cos 3\theta $
Looking at this expression, we can see that it is similar to the third identity. Therefore, we can directly apply the third identity to solve this equation:
Applying third identity on this expression, we get:
$2\cos 5\theta \cos 3\theta =\cos \left( 5\theta +3\theta \right)+\cos \left( 5\theta -3\theta \right)$
$\Rightarrow 2\cos 5\theta \cos 3\theta =\cos 8\theta +\cos 2\theta $
This is the required answer.

iv.$2\cos {{72}^{\circ }}\sin {{56}^{\circ }}$
Looking at this expression, we can see that it is similar to the fourth identity. Therefore, we can directly apply the fourth identity to solve this equation:
Applying fourth identity on this expression, we get:
$2\cos {{72}^{\circ }}\sin {{56}^{\circ }}=\sin \left( {{72}^{\circ }}+{{56}^{\circ }} \right)-\sin \left( {{72}^{\circ }}-{{56}^{\circ }} \right)$
$\Rightarrow 2\cos {{72}^{\circ }}\sin {{56}^{\circ }}=\sin {{128}^{\circ }}-\sin {{16}^{\circ }}$
This is the required answer.

v.$\cos (\alpha +\beta )\cos (\alpha -\beta )$
Looking at this expression, we can see that it is similar to the third identity. Therefore, we can directly apply the third identity to solve this equation:
Applying third identity on this expression, we get:
$\cos (\alpha +\beta )\cos (\alpha -\beta )=\dfrac{1}{2}\left[ \cos 2\alpha +\cos 2\beta \right]$
This is the required answer.

vi.$\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Looking at this expression, we can see that it is similar to the first identity. Therefore, we can directly apply the first identity to solve this equation:
Applying first identity on this expression, we get:
$\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)=\dfrac{1}{2}\left[ \sin A+\sin B \right]$
This is the required answer.

Note: The trigonometric identities are easy to remember. Even if we can’t remember these identities, we can always derive these simple equations by doing simple manipulation. Nowadays, trigonometry is not just limited to triangles, it is very widely used in real life.