Can anybody solve this simple Trigonometric equation ?

Question

sin[power2]A . cos[power2]B - cos[power2]A . sin[power2]B
The answer is sin[pow2]A - sin[pow2]B.
Can anybody explain.

Answer 1

Here goes. From here on, ^2 equals [power 2] and * equals . or multiply.


1. It is a given rule that

sin^2 A = 1 - cos^2 A

and

cos^2 A = 1 - sin ^2 A (vice versa). With that in mind, and the sum you were given, convert your given equation into the following:

[sin^2 A * (1 - sin^2 B)] - [(1 - sin^2 A) * sin^2 B]



2. Distribute the individual sin^2 into the parentheses.

[sin^2 A - (sin^2 B * sin^2 A)] - [sin^2 B - (sin^2 A * sin^2 B)]



3. Distribute the minus sign in the middle of the brackets, and then remove the brackets to make things simpler.

sin^2 A - (sin^2 B * sin^2 A) - sin^2 B + (sin^2 A * sin^2 B)



4. From here on, it's just cancellation. The two in the parentheses are exactly the same, and because one is positive and the other is negative, they can be canceled out. That leaves you with

sin^2 A - sin^B

which is the answer you gave.



[/end of lecture] That's all. If there were any parts you didn't get, feel free to give your comments and I'll try to elaborate some more, but I think this should be pretty understandable. ^_^ peace

Answer 2

in general (((cos^2 B=1 - sin^2 B)))

soooooooooo:

sin^2 A . cos^2 B - cos^2 A . sin^2 B

sin^2 A . (1 - sin^2 B) - (1 - sin^2 A) . sin^2 B
sin^2 A - sin ^2 A sin^2 B - sin^2 B + sin^2 A sin ^2 B

sin^2 A - sin^2 B

Answer 3

sin^2(A) cos^2(B) - cos^2(A)sin^2(B) = sin^2(A) - sin^2(B)

This is a trig identity. Choose the more complex side (the left hand side).

LHS = sin^2(A) cos^2(B) - cos^2(A)sin^2(B)

The trick here is to "add zero"; what we're going to do is subtract sin^2(B)cos^2(B) and then add sin^2(B)cos^2(B). Why this is adding zero is because we're adding and then subtracting the same term.

LHS = sin^2(A) cos^2(B) - sin^2(B)cos^2(B) + sin^2(B)cos^2(B) - cos^2(A)sin^2(B)

Factor the first two grouped terms and the last two grouped terms.

LHS = cos^2(B) [ sin^2(A) - sin^2(B)] + sin^2(B) [cos^2(B) - cos^2(A)]

But cos^2(B) = 1 - sin^2(B) and cos^2(A) = 1 - sin^2(A).

LHS = cos^2(B) [ sin^2(A) - sin^2(B)] + sin^2(B) [ (1 - sin^2(B)) - (1 - sin^2(A)) ]

LHS = cos^2(B) [ sin^2(A) - sin^2(B)] + sin^2(B) [ -sin^2(B) + sin^2(A) ]

Rearrange the terms in the last set of brackets.

LHS = cos^2(B) [ sin^2(A) - sin^2(B)] + sin^2(B) [ sin^2(A) - sin^2(B) ]

Look at how these two grouped terms have a common factor. Grouping, we get

LHS = [cos^2(B) + sin^2(B)] [sin^2(A) - sin^2(B)]

But cos^2(B) + sin^2(B) = 1, so

LHS = [ 1 ] [sin^2(A) - sin^2(B)]

LHS = sin^2(A) - sin^2(B) = RHS

Answer 4

You need the trig identity: cos^2 = 1-sin^2. Then its straightforward.

sin^2(A)cos^2(B) - cos^(A)sin^2(B)
=sin^2(A)(1-sin^2(B)) - (1-sin^2(A))sin^2(B)
=sin^2(A) - sin^2(A)sin^2(B) - sin^2(B) + sin^2(A)sin^2(B)
=sin^2(A) - sin^2(B)
Since the "unwanted" terms cancel.

Answer 5

I assume that "." means "*", so

Since cos(^2)X = 1 - sin(^2)X......

sin(^2)A*(1 - sin(^2)B) - (1 - sin(^2)A)*sin(^2)B =

= sin(^2)A - sin(^2)A*sin(^2)B - sin(^2)B + sin(^2)A*sin(^2)B=

= sin(^2)A - sin(^2)B

Answer 6

I will assume you know that
cos[power2]C + sin[power2]C = 1 for any C

Therefore

sin[power2]A . cos[power2]B - cos[power2]A . sin[power2]B
= sin[power2]A . (1-sin[power2]B) - (1-sin[power2]A) . sin[power2]B
= sin[power2]A-sin[power2]A.sin[power2]B - sin[power2]B+sin[power2]A . sin[power2]B
= sin[power2]A-sin[power2]B
Done!

Answer 7

sin[pow2]A . cos[pow2]B - cos[pow2]A . sin[pow2]B
=sin[pow2]A(1-sin[pow2]B) - (1-sin[pow2]A)sin[pow2]B
=sin[pow2]A - sin[pow2]A . sin[pow2]B - sin[pow2]B+
sin[pow2]B . sin[pow2]A
=sin[pow2]A - sin[pow2]B (solved)

cos[pow2]B=1-sin[pow2]B
cos[pow2]A=1-sin[pow2]A
hope you can understand.. ;)