# sinx+cosx=5/4,find sin2x,cos2x,tg2x

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You should raise to square the expression `sin x + cos x = 5/4` such that:

`(sin x + cos x)^2 = (5/4)^2`

`sin^2 x + 2 sin x cos x + cos^2 x = 25/16`

Since `sin^2 x + cos^2 x = 1` yields:

`1 + 2 sin x cos x = 25/16 => 2 sin x cos x = 25/16 - 1`

Using trigonometric identities yields `2 sin x cos x = sin 2x ` such that:

`sin 2x = (25-16)/16 => sin 2x = 9/16`

You may evaluate `cos 2x` using fundamental formula of trigonometry such that:

`sin^2 2x + cos^2 2x = 1 => cos 2x = +-sqrt(1 - sin^2 2x)`

`cos 2x = +-sqrt(1 - 81/256) => cos 2x = +-sqrt175/256 => cos 2x = +-5sqrt7/16`

You may evaluate `tan 2x` using the identity `tan alpha = sin alpha/cos alpha` such that:

`tan 2x = (sin 2x)/(cos 2x) => tan 2x = (9/16)/(+-5sqrt7/16)`

`tan 2x +-9/(5sqrt7) => tan 2x = +-9sqrt7/35`

**Hence, evaluating `sin 2x, cos 2x ` and `tan 2x` , under the given conditions, yields `sin 2x = 9/16 , cos 2x = +-5sqrt7/16, tan 2x = +-9sqrt7/35.` **