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प्रश्न
If sin A = `3/5`, find the values of cos 3A and tan 3A.
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उत्तर
Given sin A = `3/5`
cos A = `"Adjacent side"/"Hypotenuse" = 4/5`
and tan A = `Opposite side/"Adjacent side" = 3/4`
We know that cos 3A = 4 cos3 A – 3 cos A
`= 4(4/5)^3 - 3(4/5)`
`= (4/5)[4 xx (4/5)^2 - 3] = 4/5[4 xx 16/25 - 3]`
`= 4/5[(64 - 3 xx 25)/(25)]`
`= 4/5((64 - 75)/25)`
`= 4/5 xx (-11)/25 = (-44)/125`
tan 3A = `(3 tan "A" - 4tan^3 "A")/(1 - 3 tan^2"A")`
`= (3(3/4) - (3/4)^3)/(1 - 3(3/4)^2)`
`= (3/4[3 - (3/4)^2])/(1 - 3 xx 9/16)`
`= (3/4 [(48 - 9)/16])/((16 - 27)/16)`
`= 3/4[39/16 xx 16/(-11)]`
`= (- 117)/44`
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