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1 vote
If sin A = 1/√10 with A in QII find cos 2A.

User Betelgeuse
by
7.9k points

1 Answer

5 votes

Answer:

cos(2A) = 4/5

Explanation:

Since A is in QII, then
\cos A < 0


\displaystyle \sin A=\frac{\text{Opposite}}{\text{Hypotenuse}}=(1)/(√(10))\\\\\cos A=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{\sqrt{√(10)^2-1^2}}{√(10)}=(√(10-1))/(√(10))=(√(9))/(√(10))=(-3)/(√(10))

Therefore:


\displaystyle \cos 2A=\cos^2A-\sin^2 A=\biggr((-3)/(√(10))\biggr)^2-\biggr((1)/(√(10))\biggr)^2=(9)/(10)-(1)/(10)=(8)/(10)=(4)/(5)

User DerGral
by
8.5k points
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