Question

Prove the given trigonometric identity. The identity is:$sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }}\text{ }-\text{ }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ }=\text{ }0$

Answer 1

Hint: Apply the trigonometric identity $\cos \left( s+t \right)=\cos \left( s \right)\cos \left( t \right)-\sin \left( s \right)\sin \left( t \right)$. The sign is reversed and then the given trigonometric expression in the LHS will fit almost in the above expansion form. Then from that we can prove that it is zero. Also, use $cos\text{ 9}{{\text{0}}^{\circ }}=0$ .

Complete step-by-step answer:
It is to prove that $sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }}\text{ - }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ }=\text{ }0$. So here this expression in the left-hand side is similar to the identity $\cos \left( s \right)\cos \left( t \right)-\sin \left( s \right)\sin \left( t \right)$. Also, the expansion formula for the above is $\cos \left( s+t \right)=\cos \left( s \right)\cos \left( t \right)-\sin \left( s \right)\sin \left( t \right)$. Now here the expression in the left-hand side is $sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }}\text{ }-\text{ }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ }$. So, now reversing the sign, we have:
$\text{-( }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ }-sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }})$. Next comparing this expression with $\cos \left( s \right)\cos \left( t \right)-\sin \left( s \right)\sin \left( t \right)$, we have following:
$\Rightarrow -\text{( }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ - }sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }})=-cos({{35}^{\circ }}+{{55}^{\circ }})$
This is because $s={{35}^{\circ }}\,\text{and}\,t={{55}^{\circ }}$ .
So we will simplify the LHS of the given identity $sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }}\text{ }-\text{ }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ }=\text{ }0$, as follows:

& \Rightarrow LHS=-\text{( }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ - }sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }}) \\\ & \Rightarrow LHS=-cos({{35}^{\circ }}+{{55}^{\circ }}) \\\ & \Rightarrow LHS=-cos({{90}^{\circ }}) \\\ & \Rightarrow LHS=0\,\,\,\,\,\,\,\,\,\because cos({{90}^{\circ }})=0 \\\ \end{aligned}$$ And this is same as the right-hand side (RHS) of the identity$$sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }}\text{ }-\text{ }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ }=\text{ }0$$ Hence, we have proven the given identity. Note: The alternate way to prove that $$sin\text{ }{{35}^{\circ }}\text{ }\times sin\text{ }{{55}^{\circ }}\text{ }-\text{ }cos\text{ }{{35}^{\circ }}\text{ }\times \text{ }cos\text{ }{{55}^{\circ }}\text{ }=\text{ }0$$, is as follows: $$\begin{aligned} & \Rightarrow LHS=\sin \left( {{35}^{{}^\circ \;}} \right)\sin \left( {{55}^{{}^\circ \;}} \right)-\cos \left( {{35}^{{}^\circ \;}} \right)\cos \left( {{55}^{{}^\circ \;}} \right) \\\ & \Rightarrow LHS=\sin \left( {{35}^{{}^\circ \;}} \right)\cos \left( {{90}^{{}^\circ \;}}-{{55}^{{}^\circ \;}} \right)-\cos \left( {{35}^{{}^\circ \;}} \right)\cos \left( {{55}^{{}^\circ \;}} \right)\,\,\,\,\,\,\,\,\because \sin \left( {{55}^{{}^\circ \;}} \right)=\cos \left( {{90}^{{}^\circ \;}}-{{55}^{{}^\circ \;}} \right) \\\ & \Rightarrow LHS=\sin \left( {{35}^{{}^\circ \;}} \right)\cos \left( {{35}^{{}^\circ \;}} \right)-\cos \left( {{35}^{{}^\circ \;}} \right)\cos \left( {{55}^{{}^\circ \;}} \right) \\\ & \Rightarrow LHS=\cos \left( {{90}^{{}^\circ \;}}-{{35}^{{}^\circ \;}} \right)\cos \left( {{35}^{{}^\circ \;}} \right)-\cos \left( {{35}^{{}^\circ \;}} \right)\cos \left( {{55}^{{}^\circ \;}} \right)\,\,\,\,\,\,\,\,\,\because \sin \left( {{35}^{{}^\circ \;}} \right)=\cos \left( {{90}^{{}^\circ \;}}-{{35}^{{}^\circ \;}} \right) \\\ & \Rightarrow LHS=\cos \left( {{55}^{{}^\circ \;}} \right)\cos \left( {{35}^{{}^\circ \;}} \right)-\cos \left( {{35}^{{}^\circ \;}} \right)\cos \left( {{55}^{{}^\circ \;}} \right) \\\ & \Rightarrow LHS=0 \\\ & \Rightarrow LHS=RHS \\\ \end{aligned}$$