Use Identities to Simplify the expression.

[MEBC01]

Use Identities to Simplify the expression.

sin^2(B)-5cos^2(B)/1-6cos^2(B)

I don't even know where to start

 

[topsquark]

To give us a start is this
[math]\dfrac{sin^2(B) - 5 ~ cos^2(B) }{ 1 - 6 ~ cos^2(B) }[/math]
Here's a hint: [math]1 = sin^2(B) + cos^2(B)[/math]. Use this in the denominator.

-Dan

 

[MEBC01]

To give us a start is this
[math]\dfrac{sin^2(B) - 5 ~ cos^2(B) }{ 1 - 6 ~ cos^2(B) }[/math]
Here's a hint: [math]1 = sin^2(B) + cos^2(B)[/math]. Use this in the denominator.

-Dan

Okay, I believe the answer is

sin^2(B)-5cos^2(B)/6sin^2(B)

 

[Harry_the_cat]

\(\displaystyle 1-6cos^2(B) = 1-cos^2(B) - 5cos^2(B) = ...\)

 

[lookagain]

sin^2(B)-5cos^2(B)/1-6cos^2(B)

You must use grouping symbols:

[sin^2(B) - 5cos^2(B)]/[1 - 6cos^2(B)]

 

[Steven G]

The identity is sin²(x) + cos²(x) = 1 NOT 6sin²(x) + 6cos²(x) = 1. How can you multiply the lhs (which 1) by 6 and think that it still equals the rhs which is 1. It does not matter how complicated we write 1, it is still true that 6*1=6

 

[topsquark]

Okay, I believe the answer is

sin^2(B)-5cos^2(B)/6sin^2(B)

Look more carefully:
[math]\dfrac{sin^2(B) - 5 ~ cos^2(B)}{1 - 6 ~ sin^2(B)} = \dfrac{sin^2(B) - 5 ~ cos^2(B)}{(sin^2(B) + cos^2(B) ) - 6 ~ cos^2(B)}[/math]
And please use parenthesis. (sin^2(B)-5cos^2(B)) / (1-6cos^2(B)) and (sin^2(B)-5cos^2(B)) / (6sin^2(B))

-Dan