Expert: Josh - 7/5/2007

Question
HI! can you help me to verify this trigonometric identity? I've used the 8 basic ones to substitute in many different ways and I can't come up with anything that looks at all equal. The problem is: (1-cos^2Y+sin^2Y)^2+4sin^2Ycos^2Y = 4sin^2Y. I'm so confused, if you can help me then you're my hero. THANX

Answer
Hi Amber,

No need to panic:) The first step I would take is to write the cosine terms as sine.

Since [sin(y)]^2 + [cos(y)]^2 = 1, [cos(y)]^2 = 1-[sin(y)]^2

Immediately, the first term on left hand side (LHS) becomes
([sin(y)]^2+[sin(y)]^2)^2. This simplifies to (2[sin(y)]^2)^2 = 4[sin(y)]^4 ......[#1]
The second term on the LHS becomes
4 [sin(y)]^2 (1-[sin(y)]^2) or 4[sin(y)]^2 - 4[sin(y)]^4 after expansion......[#2]

Adding these two terms on the LHS, we get
4[sin(y)]^4 + 4[sin(y)]^2 - 4[sin(y)]^4
1st and 3rd term cancel, leaving us with 4[sin(y)]^2 which is identical to the RHS.
4 [sin(y)]^2 (1-[cos(y)]^2)