Question

Show that: \[\cos \left( {{{35}^ \circ } + A} \right) \cdot \cos \left( {{{35}^ \circ } - B} \right) + \sin \left( {{{35}^ \circ } + A} \right) \cdot \sin \left( {{{35}^ \circ } - B} \right) = \cos \left( {A + B} \right)\].

Answer 1

Hint: 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 and simplify by using a formula of cosine and sum identity and by further simplification we get the required solution.

Complete step by step solution:
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:
show that
\[ \Rightarrow \,\,\cos \left( {{{35}^ \circ } + A} \right) \cdot \cos \left( {{{35}^ \circ } - B} \right) + \sin \left( {{{35}^ \circ } + A} \right) \cdot \sin \left( {{{35}^ \circ } - B} \right) = \cos \left( {A + B} \right)\]--------(1)
Consider Left hand side of equation (1)
\[ \Rightarrow \,\,L.H.S\]
\[ \Rightarrow \,\,\cos \left( {{{35}^ \circ } + A} \right) \cdot \cos \left( {{{35}^ \circ } - B} \right) + \sin \left( {{{35}^ \circ } + A} \right) \cdot \sin \left( {{{35}^ \circ } - B} \right)\]--------(3)
Now, expand each term using a trigonometric formula of sum and difference identity i.e.,
Sine sum identity: \[\sin \left( {A + B} \right) = \sin A \cdot \cos B + \cos A \cdot \sin B\]
Sine difference identity: \[\sin \left( {A - B} \right) = \sin A \cdot \cos B - \cos A \cdot \sin B\]
Cosine sum identity: \[\cos \left( {A + B} \right) = \cos A \cdot \cos B - \sin A \cdot \sin B\]
Cosine difference identity: \[\cos \left( {A - B} \right) = \cos A \cdot \cos B + \sin A \cdot \sin B\]
On substituting the formulas the equation (2) becomes
\[ \Rightarrow \,\,\left( {\cos \left( {{{35}^ \circ }} \right) \cdot \cos \left( A \right) - \sin \left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)} \right) \cdot \left( {\cos \left( {{{35}^ \circ }} \right) \cdot \cos \left( B \right) + \sin \left( {{{35}^ \circ }} \right) \cdot \sin \left( B \right)} \right)\] \[ + \left( {\sin \left( {{{35}^ \circ }} \right) \cdot \cos \left( A \right) + \cos \left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)} \right) \cdot \left( {\sin \left( {{{35}^ \circ }} \right) \cdot \cos \left( B \right) - \cos \left( {{{35}^ \circ }} \right) \cdot \sin \left( B \right)} \right)\]
On multiplication, we have
\[ \Rightarrow \,\,{\cos ^2}\left( {{{35}^ \circ }} \right) \cdot \cos \left( A \right)\cos \left( B \right) + \cos \left( {{{35}^ \circ }} \right)\sin \left( {{{35}^ \circ }} \right) \cdot \cos \left( A \right)\sin \left( B \right)\] \[ - \cos \left( {{{35}^ \circ }} \right)\sin \left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)\cos \left( B \right) - {\sin ^2}\left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)\sin \left( B \right)\] \[ + \sin {\left( {{{35}^ \circ }} \right)^2} \cdot \cos \left( A \right)\cos \left( B \right) - \sin \left( {{{35}^ \circ }} \right)\cos \left( {{{35}^ \circ }} \right) \cdot \cos \left( A \right)\sin \left( B \right)\] \[ + \cos \left( {{{35}^ \circ }} \right)\sin \left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)\cos \left( B \right) - {\cos ^2}\left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)\sin \left( B \right)\]
on simplification and rearranging, we have
\[ \Rightarrow \,\,{\cos ^2}\left( {{{35}^ \circ }} \right) \cdot \cos \left( A \right)\cos \left( B \right) - {\cos ^2}\left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)\sin \left( B \right)\] \[ + \sin {\left( {{{35}^ \circ }} \right)^2} \cdot \cos \left( A \right)\cos \left( B \right) - {\sin ^2}\left( {{{35}^ \circ }} \right) \cdot \sin \left( A \right)\sin \left( B \right)\]
Take out common terms, then
\[ \Rightarrow \,\,{\cos ^2}\left( {{{35}^ \circ }} \right) \cdot \left( {\cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)} \right) + \sin {\left( {{{35}^ \circ }} \right)^2}\left( {\cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)} \right)\]
\[ \Rightarrow \,\,\left( {\cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)} \right)\left( {{{\cos }^2}\left( {{{35}^ \circ }} \right) + \sin {{\left( {{{35}^ \circ }} \right)}^2}} \right)\]
As we now the trigonometric identity i.e., \[{\cos ^2}\theta + {\sin ^2}\theta = 1\], then
\[ \Rightarrow \,\,\left( {\cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)} \right)\left( 1 \right)\]
by using cosine sum identity, the above equation becomes
\[ \Rightarrow \,\,\cos \left( {A + B} \right)\]
\[ \Rightarrow \,\,R.H.S\]
Therefore, \[L.H.S = R.H.S\]
\[\cos \left( {{{35}^ \circ } + A} \right) \cdot \cos \left( {{{35}^ \circ } - B} \right) + \sin \left( {{{35}^ \circ } + A} \right) \cdot \sin \left( {{{35}^ \circ } - B} \right) = \cos \left( {A + B} \right)\]
Hence proved.

Note:
When solving trigonometry based questions, we have to know the definitions of ratios and always remember the standard angles and formulas are useful for solving certain integration problems where a cosine and sum identity may make things much simpler to solve. Thus, in math as well as in physics, these formulae are useful to derive many important identities.