Trigonometry> Find the value of angle A in the followin...

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

Rahul , 7 Years ago

Grade 10

2 Answers

Ravi malvia

Last Activity: 7 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

Last Activity: 7 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