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Expert: - 12/2/2008


Question
Hey, My name is Ben and my class has a test coming up on trig identities. I wanted to do some extra work in order to prepare for my test but got stumped on these questions. Wondered if you could help me.

1)2sin5xcos4x-sinx = sin9x

2)tan3A-tan2A-tanA = tan3A . tan2A . tanA

3)(cotA-tanA)sinA=2cosA-secA

I need to be able to show on the test how left side equals right side. Any helyoucould give me would be greatly appriciated.


Answer
1) It is known that sin(9x) = sin(5x)cos(4x) + sin(4x)cos(5x).

What we are given is 2sin(5x)cos(4x) - sin(x) = sin(9x).

Setting both of these equal gives

2sin(5x)cos(4x) - sin(x) = sin(5x)cos(4x) + sin(4x)cos(5x).

This reduces to
sin(5x)cos(4x) - sin(x) = sin(4x)cos(5x), which the same as
sin(5x)cos(4x) - sin(x) - sin(4x)cos(5x) = 0.

Note that sin(5x) = sin(4x)cos(x) + sin(x)cos(4x) and that
cos(5x) = cos(4x)cos(x) - sin(4x)sin(x), so now we have

(sin(4x)cos(x) + sin(x)cos(4x))cos(4x) - sin(x) -
sin(4x)(cos(4x)cos(x) - sin(4x)sin(x)) = 0.

Multiplying through gives
sin(4x)cos(x)cos(4x) + sin(x)cosē(4x) - sin(x) -
sin(4x)cos(4x)cos(x) + sinē(4x)sin(x)) = 0.

Note that the first term on each line is the same and the second term on each line is sin(x)(sinē(x)+cosē(x)), which is the same as sin(x) since the other part is 1.  What we are left with is
sin(x) - sin(x) = 0.

They all cancel, so all that need to be done is to turn one side into the other.

2) I'll assume this is
tan(3A) - tan(2A) - tan(A) = tan(3A)tan(2A)tan(A).

It is known that tan(3A) = [tan(2A) + tan(A)]/[1 - tan(2A)tan(A)].
Putting this in gives us

[tan(2A) + tan(A)]/[1 - tan(2A)tan(A)] - tan(2A) - tan(A) =
{[tan(2A) + tan(A)]/[1 - tan(2A)tan(A)]}tan(2A)tan(A).

Multiplying by the denominator gives us

[tan(2A) + tan(A)] - (tan(2A) + tan(A))(1 - tan(2A)tan(A)) =
[tan(2A) + tan(A)]tan(2A)tan(A).

Getting rid of parenthises gives us

tan(2A) + tan(A) - tan(2A) - tan(A) + tanē(2A)tan(A) + tan(2A)tanē(A)
= tanē(2A)tan(A) + tan(2A)tanē(A).

Note that after further inspection, this is always true.  To do this properly, take one side and invert it into the other side.

3) The last problem is (cot(A)-tan(A))sin(A) = 2cos(A) - sec(A).
Noting that cot(A) is cos(A)/sin(A), tan(A) is sin(A)/cos(A), and the sec(A) is 1/cos(A), the problem can be transformed into

(cos(A)/sin(A) - sin(A)/cos(A))sin(A) - 2cos(A) + 1/cos(A) = 0.

This become
cos(A) - sinē(A)/cos(A) - 2cos(A) + 1/cos(A) = 0.

Combining the cos(A) terms and rearranging the rest gives us
1/cos(A) - sinē(A)/cos(A) - cos(A) = 0.

Multiplying the entire equation by cos(A) gives
1 - sinē(A) - cosē(A) = 0.  This too is seen to be an identity.
Again, to do it properly, one side needs to be transformed into the other side.

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