How do you simplify $$\cos {35^ \circ }\sin {55^ \circ } + \cos {55^ \circ }\sin {35^ \circ }$$?

Question

Vedantu Content Team · Accepted Answer

Hint: We use the trigonometric identity of $$\sin A\cos B + \cos A\sin B$$ to simplify the given trigonometric equation. Compare the given equation with general trigonometric identity and write the value of A and B. Add the value of angles inside the bracket after obtaining the value from identity. * If A and B are two angles then $$\sin A\cos B + \cos A\sin B = \sin (A + B)$$Complete step-by-step answer: We have to simplify the value of $$\cos {35^ \circ }\sin {55^ \circ } + \cos {55^ \circ }\sin {35^ \circ }$$We can shuffle the terms and write $$\sin {55^ \circ }\cos {35^ \circ } + \cos {55^ \circ }\sin {35^ \circ }$$Since we can see that the given equation matches with the trigonometric identity $$\sin A\cos B + \cos A\sin B$$, then on comparing we get the value of $$A = {55^ \circ };B = {35^ \circ }$$Also, we know that $$\sin A\cos B + \cos A\sin B = \sin (A + B)$$Substituting the value of A and B in right hand side of the identity $$ \Rightarrow \sin {55^ \circ }\cos {35^ \circ } + \cos {55^ \circ }\sin {35^ \circ } = \sin ({55^ \circ } + {35^ \circ })$$Add the value of angles inside the bracket in right hand side of the equation$$ \Rightarrow \sin {55^ \circ }\cos {35^ \circ } + \cos {55^ \circ }\sin {35^ \circ } = \sin {90^ \circ }$$Now substitute the value of $$\sin {90^ \circ } = 1$$ in the right hand side of the equation.$$ \Rightarrow \sin {55^ \circ }\cos {35^ \circ } + \cos {55^ \circ }\sin {35^ \circ } = 1$$ $$\therefore $$The value of $$\cos {35^ \circ }\sin {55^ \circ } + \cos {55^ \circ }\sin {35^ \circ }$$on simplification is equal to 1.  Note:  Many students make the mistake of calculating the value of cosine of the angle given and sine of the angle given using a scientific calculator and then substitute in the equation, then they multiply and add the terms. Keep in mind this is not the appropriate method as the values will be in decimal form and multiplying decimal values with decimal values will again give a long solution, students are advised to avoid this long calculation and make use of the trigonometric identity.Also, many students who don’t remember the value of sine of angle obtained at the end can take help of the table that gives values of some common trigonometric functions at a few angles.Angles \n(in degrees)${0^ \circ }$${30^ \circ }$${45^ \circ }$${60^ \circ }$${90^ \circ }$sin0$\dfrac{1}{2}$$\dfrac{1}{{\sqrt 2 }}$$\dfrac{{\sqrt 3 }}{2}$$1$cos1$\dfrac{{\sqrt 3 }}{2}$$\dfrac{1}{{\sqrt 2 }}$$\dfrac{1}{2}$0tan0$\dfrac{1}{{\sqrt 3 }}$1$\sqrt 3 $Not definedcosecNot defined 2$$\sqrt 2 $$$$\dfrac{2}{{\sqrt 3 }}$$1sec1$$\dfrac{2}{{\sqrt 3 }}$$$$\sqrt 2 $$2Not definedcotNot defined$\sqrt 3 $1$$\dfrac{1}{{\sqrt 3 }}$$0

Angles (in degrees)	${0^ \circ }$	${30^ \circ }$	${45^ \circ }$	${60^ \circ }$	${90^ \circ }$
sin	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0
tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3 $	Not defined
cosec	Not defined	₂	\[\sqrt 2 \]	\[\dfrac{2}{{\sqrt 3 }}\]	1
sec	1	\[\dfrac{2}{{\sqrt 3 }}\]	\[\sqrt 2 \]	₂	Not defined
cot	Not defined	$\sqrt 3 $	1	\[\dfrac{1}{{\sqrt 3 }}\]	0

How do you simplify \[\cos {35^ \circ }\sin {55^ \circ } + \cos {55^ \circ }\sin {35^ \circ }\]?