Question

Prove that $\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right) = \cos \left( {A - B} \right)$

Answer 1

Here in this question, we have to prove the given trigonometric function by showing the left hand side is equal to the right hand side (i.e., $L.H.S = R.H.S$). To solve this, we have to consider L.H.S separately and simplify by using a formula of compound angle of trigonometric ratios $\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$ to get the required RHS.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the given question:
Prove that
$\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right) = \sin \left( {A - B} \right)$ --------(1)
Consider Left hand side of equation (1) (L.H.S)
$\Rightarrow \,\,\,\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right)$ ----(2)
Let us by the compound angles of trigonometric ratios:
The sum identity of cosine ratio is:
$\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$
On comparing the equation (2) with cosine sum identity
Where, $A = \left( {\dfrac{\pi }{6} - A} \right)$ and $B = \left( {\dfrac{\pi }{3} + B} \right)$, then
By cosine sum identity equation (2) becomes
$= \cos \left( {\left( {\dfrac{\pi }{6} - A} \right) + \left( {\dfrac{\pi }{3} + B} \right)} \right)$
By using a sign conversion
$= \cos \left( {\dfrac{\pi }{6} - A + \dfrac{\pi }{3} + B} \right)$
$=\cos \left( {\dfrac{\pi }{6} + \dfrac{\pi }{3} - \left( {A - B} \right)} \right)$
Take, 6 as LCM between $\dfrac{\pi }{6}$ and $\dfrac{\pi }{3}$, then we have
$=\cos \left( {\dfrac{{\pi + 2\pi }}{6} - \left( {A - B} \right)} \right)$
$= \cos \left( {\dfrac{{3\pi }}{6} - \left( {A - B} \right)} \right)$
On simplification, we get
$= \cos \left( {\dfrac{\pi }{2} - \left( {A - B} \right)} \right)$ ---- (3)
By the ASTC rule $\left( {\dfrac{\pi }{2} - \theta } \right)$ belongs to the first quadrant in that all six ratios are positive and
Let us by the complementary angles of trigonometric ratios:
The angle can be written as
$\sin \left( {90 - \theta } \right) = \cos \theta$
$\cos \left( {90 - \theta } \right) = \sin \theta$
Then equation (3) becomes
$\Rightarrow \,\,\,\sin \left( {A - B} \right)$
$= RHS$
$\therefore \,\,LHS = \,\,RHS$.
$\therefore \,\,\,\,\cos \left( {\dfrac{\pi }{6} - A} \right) \cdot \cos \left( {\dfrac{\pi }{3} + B} \right) - \sin \left( {\dfrac{\pi }{6} - A} \right) \cdot \sin \left( {\dfrac{\pi }{3} + B} \right) = \sin \left( {A - B} \right)$
Hence, proved.

Note:
When solving the trigonometry-based questions, we have to know the definitions of six trigonometric ratios. Remember, the formula of compound angles i.e.,
cosine sum identity: $\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B$ and
cosine difference identity: $\cos \left( {A - B} \right) = \cos A \cdot \cos B + \sin A \cdot \sin B$
Sine sum identity: $\sin \left( {A + B} \right) = \sin A \cdot \cos B + \cos A \cdot \sin B$
Sine difference identity: $\sin \left( {A - B} \right) = \sin A \cdot \cos B - \cos A \cdot \sin B$
Remember, when the sum of two angles is ${90^ \circ }$, then the angles are known as complementary angles at that time the ratios will change like $\sin \leftrightarrow \cos$, $\sec \leftrightarrow cosec$ and $\tan \leftrightarrow \cot$ then should know the some basic formulas of trigonometry like identities, double and half angle formulas, Product to Sum Formulas and Sum to Product Form.