Math Destroyer Test 7 Q 2

[wired202808]

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sin a * (csc A - cosA/cotA)

can someone explain why the answer is cos^2A? I would have thought after reduction it would be something like sin^2A?

 

[herkulease]

do you understand how he came to 1-sin^2A?

I'm sorry if that comes off as making sound like you're dumb or something. not my intention. I'm having trouble try to see where could make the mistake and get just sin^2A.

Anyhow all he did was convert the cscA to 1/SinA and Cot A to Cos A/Sin A then simplified.

Sin A * [1/SinA - Cos A/(CosA/SinA) ]

Sin A * [1/SinA - (SinA/1) ] or just Sin A * [1/SinA - SinA]

Now just distribute.

SinA/Sin A - SinA(SinA) = 1 - Sin^2 A

1-Sin^2 A = Cos^2A

Its just one of those trig identities you should know. there's really no trick to it that I know of to solve it if you don't know the identities.

 

[KyoPhan]

You probably made a mistake when you were working on the problem.

Here's a neater workout of the problem.

 

[wired202808]

You probably made a mistake when you were working on the problem.

Here's a neater workout of the problem.

thanks so much for the work up kyo! this helps a lot I see what you did. I simplified incorrectly and then did not apply that formula.