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Question: How to prove that: \[\cos {80^ \circ } + \cos {40^ \circ } - \cos {20^ \circ } = 0\]?...

How to prove that: cos⁡80∘+cos⁡40∘−cos⁡20∘=0\cos {80^ \circ } + \cos {40^ \circ } - \cos {20^ \circ } = 0?

Explanation

Solution

To solve this question, we will use the formula of cos⁡A+cos⁡B\cos A + \cos B. We pair the first two terms given in the question and then apply the formula. Further we will solve and put the values of known trigonometric angles.
Formula used:
cos⁡A+cos⁡B=2cos⁡(A+B2)cos(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)cos\left( {\dfrac{{A - B}}{2}} \right)
cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}

Complete step by step solution:
LHS=cos⁡80∘+cos⁡40∘−cos⁡20∘ = \cos {80^ \circ } + \cos {40^ \circ } - \cos {20^ \circ }
Now we will group the first two terms. So, we have;
=(cos⁡80∘+cos⁡40∘)−cos⁡20∘= \left( {\cos {{80}^ \circ } + \cos {{40}^ \circ }} \right) - \cos {20^ \circ }
Now we will apply the formula:
cos⁡A+cos⁡B=2cos⁡(A+B2)cos(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)cos\left( {\dfrac{{A - B}}{2}} \right)
So, we get;
=2cos⁡(80∘+40∘2)cos⁡(80∘−40∘2)−cos⁡20∘= 2\cos \left( {\dfrac{{{{80}^ \circ } + {{40}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{80}^ \circ } - {{40}^ \circ }}}{2}} \right) - \cos {20^ \circ }
On simplification we get;
=2cos⁡(120∘2)cos⁡(40∘2)−cos⁡20∘= 2\cos \left( {\dfrac{{{{120}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{40}^ \circ }}}{2}} \right) - \cos {20^ \circ }
=2cos⁡(60∘)cos⁡(20∘)−cos⁡20∘= 2\cos \left( {{{60}^ \circ }} \right)\cos \left( {{{20}^ \circ }} \right) - \cos {20^ \circ }
Now we will put the value of cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}.
=2×12cos⁡(20∘)−cos⁡20∘= 2 \times \dfrac{1}{2}\cos \left( {{{20}^ \circ }} \right) - \cos {20^ \circ }
On simplification we get;
=cos⁡(20∘)−cos⁡20∘= \cos \left( {{{20}^ \circ }} \right) - \cos {20^ \circ }
=0= 0
And that is equal to the RHS.

Additional details:
Trigonometry is a branch of mathematics that deals with the relationship between the side lengths and angles of triangles. There are a total six trigonometric functions. They are: - sine function, cosine function, tangent function, cotangent function, secant function, cosecant function. Out of the six there are three primary functions: - sine, cosine, tangent function. The reciprocal of sine is cosecant. The reciprocal of cosine is a secant function. The reciprocal of tangent is cotangent function. Sine function is defined as the ratio of perpendicular to hypotenuse. Cosine function is defined as the ratio of base to hypotenuse. Tangent function is defined as the ratio of perpendicular and base. With these we can obtain the ratios for the other three trigonometric functions.
Hypotenuse is the side opposite to the 90∘{90^ \circ } angle in a right triangle. Perpendicular is the side opposite to the angle for which we are finding the trigonometric ratios and base is the side adjacent to that angle.

Note: One important formula, in which most of the students commit mistakes is of cos⁡A−cos⁡B\cos A - \cos B, which is given as;
cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
We could also use this formula but here we get the value in sin ratio and the remaining ratio is cos so we have to use the compound angle formula which would be a long approach. Hence there we have to choose the use of formulas wisely.
We can also note some other formula like:
cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)