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Grade: 12th pass

                        

Find the value of 4sin^2 10degree + 4sin^2 50 degree cos 20 degree + cos 80 degree

2 years ago

Answers : (1)

Aarushi Ahlawat
41 Points
							
Transform first terms to cos using 2sin^2theta=cos2theta and then take cos 20 common. Again you can use cos2theta relation and finaly 2cosAcosB relation. In three steps youwill get the solution.
2(2sin^{2}(10) -1 +1)+4sin^{2}(50)cos(20)+cos(80)
=-2cos(20)+2+4sin^{2}(50)cos(20)+cos(80)
=2-2cos(20)\left (1-2sin^{2}(50) \right )+cos(80)
=2-2cos(20)cos(100)+cos(80)
=2-(cos(120)+cos(80))+cos(80)
=2-cos(120)
=2-\left ( -\frac{1}{2} \right )
=\frac{5}{2}
 
 
 
2 years ago
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