Proving Trigonometry Question

Question

How do I solve: sin(x+y) cos(x-y) = sin x/sec x + cos y/ csc y

And also how do I solve: (csc x - cot x)² = 1 - cos x/1 + cos x

Answer 1

1st problem

sin(x+y)cos(x-y) = sinx/secx + cosy/cscy

for the left side:

use the sum and difference formulas for sines and cosines

sin(x+y)cos(x-y) = (sinxcosy+cosxsiny)(cosxcosy - sinxsiny)

Use FOIL

= cos^2(y)sinxcosx -sin^2(x)sinycosy + cos^2(x)sinycosy + sin^2(y)sinxcosx

group the 1st and last, and the 2nd and 3rd terms

= (sin^2(y) + cos^2(y))(sinxcosx) + (sin^2(x) + cos^2(x))(sinycosy)

replace the sines squared = cosines squared with 1

= (1)sinxcosx + (1)sinycosy

= sinxcosx + sinycosy

replace cosx with 1/secx and replace siny with 1/cscy

=sinx/secx + cosy/cscy which is the right side of the original equation

for the 2nd problem, general approach, if you don't see anything else that looks useful, convert all trig functions to sines and cosines

on the left side is:

(csc x - cot x)^2

= (1/sinx - cosx/sinx)^2

= ((1-cosx)/sinx)^2

= (1/sin^2(x))(1 -cosx)(1-cosx)

the right side is 1- cosx/(1+cosx)

= (1+cosx -cosx)/(1+cosx)

= 1/(1+cosx)

That gives

(1-cosx)(1-cosx)/sin^2(x) = 1/(1+cosx)

cross multiply to get

(1-cosx)(1-cosx)(1+cosx) = sin^2(x)

(1-cosx)(1- cos^2(x)) = sin^2(x)

(1-cosx)(sin^2(x)) = sin^2(x)

cancel the sin^2(x) terms on both sides, leaving

1-cosx = 1

cosx =0 and sin^2(x) = 0, sinx = 0

x = cos^-1(x) and sin^-1(x)

x = 90 +180n where n = any integer

and 180n where n= any integer

x = 0, plus or minus 90, 180, 270, 360, ...

Unless you really meant the right side to be 1-cox/1 + cosx = 1 -cosx +cosx = 1

although that seems unlikely, but if so,

then (cscx - cot)^2 = 1 -

and cscx - cotx = + or - 1

1/sinx - cosx/sinx = 1

multiply by sinx

1 - cosx = sinx

sinx + cosx = 1

sinx + sqr(1-sin^2(x) = 1

sqr(1-sin^2x) = 1-sinx

square both sides

1-sin^2(x) = 1 - 2sinx + sin^2(x)

2sin^2(x) -2sinx = 0

divide by 2, then factor

sin^2x - sinx = 0

sinx(sinx-1) = 0

set each factor = 0

sinx = 0, and sinx = 1

x = 360n where n = any integer

and

x = sin^-1(1) = 90 + 360n where n = any integer

x = 0, plus or minus 90, 180, 270, 360, ....

which is the same solution as the 1st interpretation of the problem

x = pi/2 + 2npi and x= 2npi where n = any integer

or

x = (n/2)pi where n= any integer

or

x = 90n degrees where n= any integer

check the answers

plug in 0 into the original problem

(cscx - cotx)^2 =1 - cosx/1 + cosx

(csc0 - cot0)^2 = 1 -cos0/1 + cos0

(1 /0 - 1/0)^2 = 1 - 1/1 + 1 = 1

left side is undefined

throw out 0 as a solution

when you multiply by a term which includes the variable, you may introduce an extraneous solution that is not in the original equation

try x= 90

(csc90 - cot90)^2 = 1 -cos90/1 + cos90

(1 - 0)^2 = 1 - 0 +0

90 degrees is a valid solution

try x = 180

(csc180 - cot180)^2 = 1 - cos180/1 + cos180

(1/0 - - 1/0)^2 = 1

the left side is undefined.

throw out 180 as a solution

try x = 270

(csc270 -cot270)^2 = 1 -cos270/1 + cos270 = 1-0+0 =1

(-1 - 0)^2 =1

270 is a valid solution

that narrows the solutions down to

x = 90 + 180n degrees where n = any integer

just to make sure this wasn't somehow a trigonmetric identity plug in any other angle such as 45 degrees

(csc45 -cot45)^2 = 1 - cos45/1 + cos45 = 1

(sqr2 -1)^2 = 2-2sqr2 +1 = 3-2sqr2 = about 3-2.828 which does not equal the 1 on the right side

it's not an identity and the only solution is x= 90+180n or pi/2 +npi where n= any integer