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

Ravi malvia
37 Points
5 years ago
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)]
Arun
25757 Points
5 years ago
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