﻿ Prove each identitiesn(a) sin3x/sinxcosx = 4cosx - secxn(b) - science mathematics
Prove each identitiesn(a) sin3x/sinxcosx = 4cosx - secxn(b)

## Prove each identitiesn(a) sin3x/sinxcosx = 4cosx - secxn(b)

[From: Mathematics] [author: ] [Date: 01-07] [Hit: ]
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MyRank say: a. sin3x/sinxcosx = 4cosx - secx

4cosx - secx

4/secx - 1/cosx

4cosx - secx / secxcosx

Sin3x/sinxcosx.

b. 1+sec(-φ) / sin(-φ) + tan(-φ)

1+secφ/-sinφ-tanφ

1+secφ / -(sinφ+tanφ) = -cose(-φ).
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Ash say: (a) sin3x / sinx cosx
= sin(x+2x) / sinx cosx
= (sinx cos2x + sin2x cosx) / sinx cosx
= (sinx (2cos²x - 1) + (2 sinx cosx) cosx) / sinx cosx
= sinx((2cos²x - 1 + 2 cos²x) / sinx cosx
= (4cos²x - 1)/cosx
= 4cos²x/cosx - 1/cosx
= 4cosx - secx

(b) There seems to be an error in your identity.
The RHS should be either -csc(ø) or csc(-ø) but NOT -csc(-ø)

I have solved for both possibilities.

(1+sec(-ø)) / (sin(-ø)+tan(-ø))
= (1+sec(ø)) / (-sin(ø) - tan(ø)
= (1+sec(ø)) / (-sin(ø) - sin(ø)/cos(ø))
= (1+sec(ø)) / -sin(ø)(1 + 1/cos(ø))
= (1+sec(ø)) / -sin(ø)(1 + sec(ø))
= 1/-sin(ø)
= - csc(ø)

OR

(1+sec(-ø)) / (sin(-ø)+tan(-ø))
= (1+sec(ø)) / (sin(-ø) + sin(-ø)/cos(-ø))
= (1+sec(-ø)) / (sin(-ø) (1 + 1/cos(-ø))
= (1+sec(-ø)) / (sin(-ø) (1 + sec(-ø))
= 1/sin(-ø)
= csc(-ø)
-
Bella say: izptjqrw
-
bidii say: vffpjelf
-

keywords: ,Prove each identitiesn(a) sin3x/sinxcosx = 4cosx - secxn(b)
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