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Question

Prove the following questions please and please discuss it so I can understand it Thanks you.

cos(x+y)secy =sinx (cot x-tan y)


Cos (x+y)+cos(x-y)=2 cos x cos y

Answer 1

1. First recall that:

cos (A + B) = cos(A) cos(B) - sin(A) sin(B)

also, sec(A) = 1/(cos(A))

So, using these properties, we see that the first expression goes to:

[cos(x)cos(y)-sin(x)sin(y)]/cos(y)

which simplifies to

cos(x)-sin(x)tan(y)

However, if we factor out a sin(x), we get

sin(x)[cos(x)/sin(x) - tan(y)] (the sin(x)'s cancel with the tan)

since cot = cos/sin, we get

sin(x)[cot(x)-tan(y)]

2. In addition to the identity shown at the beginning of problem 1, there is a similar one, which is

cos (A - B) = cos(A) cos(B) + sin(A) sin(B)

So, using these two trigonometric identities, we have now

cos(x)cos(y) - sin(x)sin(y) + cos(x)cos(y) + sin(x)sin(y)

Clearly, the second and fourth term cancel, and then the first and third combine to get you

2cos(x)cos(y)

I hope this helped you out. Enjoy Trigonometry!

Answer 2

I'm not 100% sure how to solve it, but I can lead you to the right direction.

For the top equation, multiply the cos (x+y) to get Cos X + Cos Y (Sec Y) = Sin X (Cot X - Tan Y)

Then, replace Sec Y with 1/cos y. So you should have cos x + cos y (1/cos y) = sin x (cot x - tan y)

Cross out the denominator from (1/cos y) and cos y... so you should get something like this

cos x = sin x (cot x - tan y)

Distribute

cos x = (sin x cot x) - sin x tan y

At this point, look at your ref. table to see the Double Angles/ 1/2 angles to see what you can replace sin x cot x and/or sin x tan y with.

Second equation

Distribute cos (x + y) + cos (x - y)

Cos x + Cos y + cos x - cos y = sin x (cot x - tan y)

Since + cos y and - cos y cancel eachother out and you have 2 cos x's.. you should have... 2 cos x = sin x (cot x - tan y)

Distribute the right side...

2 cos x = (sin x cot x) - sin x - tan y

Again, look back to your ref. tables to see which ones you can replace...

Hope this helped...

Answer 3

see the source below for trig identity substatute and solve

1. [cosx cosy - sinx siny] /cosy
cosx - sinx tany
factor out a sinx
sinx(cosx/sinx -tanx)
sinx(cotx-tany)

2.[cosx cosy - sinx siny] +[cosx cosy + sinx siny]
open paranthesis
to get answer after two middle terms cancel

Answer 4

cos(x+y)secy
= (cosx*cosy-sinx*siny)/cosy
= cosx - sinx*tany
= sinx(cotx - tany)

cos (x+y)+cos(x-y)
= cosx*cosy - sinx*siny + cosx*cosy + sinx*siny
= 2cosx*cosy