Question

Cos(L+B)+sin l sinB =cosl

Answer 1

CosL*cosB-sinL*sinB+sinl*sinB=cosl
Cosl*cosB=cosl

Answer 2

Cos(l+b)+sinl sin b=cos l
cos l * cos b-sin l * sin b+sin l - sin b=cos l
cos l * cos b=cos l
cos l(cos b - 1)=0
cos l =0                         cos b = 1
l=π/2+πn, n∈Z               b=2πn, n∈Z

Answer 3

The equation you provided is:

cos(L+B) + sin(L)sin(B) = cos(L)

To analyze this equation, let's break it down and simplify it step by step.

Using the trigonometric identity:

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

We can rewrite the equation as:

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

The sin(L)sin(B) terms cancel out:

cos(L)cos(B) + 0 = cos(L)

Simplifying further, we have:

cos(L)cos(B) = cos(L)

Now, there are two possible scenarios:

1. If cos(L) ≠ 0:

We can divide both sides of the equation by cos(L):

cos(B) = 1

This means that B must be 0 or an even multiple of 2π (i.e., B = 2nπ, where n is an integer). In this case, the original equation holds true for any value of L.

2. If cos(L) = 0:

In this case, the equation becomes:

0 = 0

This means that the equation is satisfied for any values of L and B.

Therefore, the original equation is true for all values of L and B.