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Grade: 10 

                        
Find the value of angle A in the following equationcos 2A =cos 60°×cos 30°
3 years ago

Answers : (2)

Ravi malvia
37 Points 
							Cos2A=(√3÷2)×(1÷2)Cos2A= √3÷42CossqureA-1=√3÷42CossqureA=(√3÷4)+1CossqureA=(1÷2)[(√3÷4)+1]CossqureA=(√3÷8)+(1÷2)A=cosinverse[(√3÷8)+(1÷2)]
3 years ago
Arun
24742 Points 
							
cos 2A = cos 60°×cos 30°
           =\frac{1}{2}\times\frac{\sqrt{3}}{2}
          =\frac{1}{2}\times\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}=\frac{1}{2}\times\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \implies 2A=cos^{-1}(\frac{\sqrt{3}}{4})=\frac{1}{2}\times\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \implies 2A=cos^{-1}(\frac{\sqrt{3}}{4}) \implies A=\frac{1}{2} cos^{-1}(\frac{\sqrt{3}}{4})
and A=\frac{1}{2}\times\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \implies 2A=cos^{-1}(\frac{\sqrt{3}}{4}) \implies A=\frac{1}{2} cos^{-1}(\frac{\sqrt{3}}{4}) A= \frac{64.34^{\circ}}{2}=32.17^{\circ}=\frac{1}{2}\times\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \implies 2A=cos^{-1}(\frac{\sqrt{3}}{4}) \implies A=\frac{1}{2} cos^{-1}(\frac{\sqrt{3}}{4}) A= \frac{64.34^{\circ}}{2}=32.17^{\circ}= =\frac{1}{2}\times\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \implies 2A=cos^{-1}(\frac{\sqrt{3}}{4}) \implies A=\frac{1}{2} cos^{-1}(\frac{\sqrt{3}}{4}) cos^{-1}(\frac{\sqrt{3}}{4})= 64.34^{\circ}
So the concept is of inverse trigonometric .It is so simple ,there is one thing in this question which generally not being asked in exam is the values of cos other than 0, 30,45,60,90 
3 years ago
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