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Question: How do you simplify \[\sin {175^ \circ }\cos {55^ \circ } - \cos {175^ \circ }\sin {55^ \circ }\]?...

How do you simplify sin⁡175∘cos⁡55∘−cos⁡175∘sin⁡55∘\sin {175^ \circ }\cos {55^ \circ } - \cos {175^ \circ }\sin {55^ \circ }?

Explanation

Solution

The expanded form of sin⁡(a−b)\sin \left( {a - b} \right) is sin⁡(a)cos⁡(b)−sin⁡(b)cos⁡(a)\sin \left( a \right)\cos \left( b \right) - \sin \left( b \right)\cos \left( a \right)
This expression resembles the expression given in the question. Our aim is to try to fit the given expression in this standard form.

Complete step-by-step solution:
We know that,
sin⁡(a−b)=sin⁡(a)cos⁡(b)−sin⁡(b)cos⁡(a)\sin \left( {a - b} \right) = \sin \left( a \right)\cos \left( b \right) - \sin \left( b \right)\cos \left( a \right)
The given expression is indeed the expanded form of the above mentioned identity.
Here,a=175∘a = {175^ \circ } and b=55∘b = {55^ \circ }
Therefore, substituting a=175∘a = {175^ \circ }and b=55∘b = {55^ \circ } in the identity sin⁡(a−b)\sin \left( {a - b} \right) we will get
sin⁡175∘cos⁡55∘−cos⁡175∘sin⁡55∘\sin {175^ \circ }\cos {55^ \circ } - \cos {175^ \circ }\sin {55^ \circ }
⇒sin⁡(175∘−55∘)\Rightarrow \sin \left( {{{175}^ \circ } - {{55}^ \circ }} \right)
⇒sin⁡(120∘)\Rightarrow \sin \left( {{{120}^ \circ }} \right)
Now, we have to find the value of sin⁡(120∘)\sin \left( {{{120}^ \circ }} \right)
To find the value of sin⁡(120∘)\sin \left( {{{120}^ \circ }} \right), we will try to make it in terms of some other standard values like 180∘{180^ \circ } ,60∘{60^ \circ }etc.
We know that, 120∘=180∘−60∘120^\circ = 180^\circ - 60^\circ
Therefore, we can write sin⁡(120∘)\sin \left( {{{120}^ \circ }} \right) as sin⁡(180∘−60)\sin \left( {{{180}^ \circ } - 60} \right)
Also,
sin⁡(180∘−x)=sin⁡(x)\sin \left( {{{180}^ \circ } - x} \right) = \sin \left( x \right)
Hence, sin⁡(180∘−60)\sin \left( {{{180}^ \circ } - 60} \right) will become sin⁡(60∘)\sin \left( {{{60}^ \circ }} \right)

Hence,sin⁡175∘cos⁡55∘−cos⁡175∘sin⁡55∘\sin {175^ \circ }\cos {55^ \circ } - \cos {175^ \circ }\sin {55^ \circ } =sin⁡(60∘)\sin \left( {{{60}^ \circ }} \right)

Note: Sine or the sine function is one of the three primary functions in trigonometry, the others being cosine, and tan functions. The sine x or sine theta can be defined as the ratio of the opposite side of a right triangle to its hypotenuse
Cos function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse.
Secant is the ratio between the hypotenuses to the shorter side adjacent to an acute angle in a right triangle.
Sin(A+B)=Sin⁡Acos⁡B+Cos⁡A+Sin⁡B(A + B) = \operatorname{Sin} A\cos B + \operatorname{Cos} A + \operatorname{Sin} B
To find the value ofsin⁡(120∘)\sin \left( {{{120}^ \circ }} \right), we will use the addition formula and values of these angles.
sin⁡(120∘)\sin \left( {{{120}^ \circ }} \right)= sin⁡(90+30)\sin \left( {90 + 30} \right)
Now using the formula,
sin⁡(a+b)=sin⁡(a)cos⁡(b)+sin⁡(b)cos⁡(a)\sin \left( {a + b} \right) = \sin \left( a \right)\cos \left( b \right) + \sin \left( b \right)\cos \left( a \right)
We can write;
sin⁡(120∘)=\sin \left( {{{120}^ \circ }} \right) = sin⁡(a+b)=sin⁡(90)cos⁡(30)+sin⁡(30)cos⁡(90)\sin \left( {a + b} \right) = \sin \left( {90} \right)\cos \left( {30} \right) + \sin \left( {30} \right)\cos \left( {90} \right)
Now putting the values sin⁡(90∘)\sin \left( {{{90}^ \circ }} \right), sin⁡(30∘)\sin \left( {{{30}^ \circ }} \right), cos⁡(90∘)\cos \left( {{{90}^ \circ }} \right)and cos⁡(30∘)\cos \left( {{{30}^ \circ }} \right) from the table above, we get;
sin⁡120=(1)(32)−(0)(12)\sin 120 = \left( 1 \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) - \left( 0 \right)\left( {\dfrac{1}{2}} \right)
sin⁡120=(32)\sin 120 = \left( {\dfrac{{\sqrt 3 }}{2}} \right)
This is the standard way of finding the numerical values of the trigonometric ratios. But it is indeed necessary to know some of the standard values before approaching these.