[RESOLVED] Trigonometric equation help

[astrostar]

Hi all,

I need to solve this equation giving my answers for angle from 0 to 360 degrees.

1-2sin(squared)E + cos(squared)E = 0

My answer

E 1-sin(squared) E = cos2E

cos(sqaured)E + cos2E = 0

(cos E + 2) (cosE + 0) = 0

either cosE = -2 or cosE = 0


E =(inv)cos (-2) impossible

E =(inv)cos (0)

t = 90 or 270 degrees.

Is this correct. Any help appreciated

[Logophobic]

Take a different approach, using the fact that sin² x + cos² x = 1

1 - 2 sin² E + cos² E = 0
1 + sin² E + cos² E = 3 sin² E
2 = 3 sin² E

I'm sure you can take it from there.

[astrostar]

Thanks for the reply

I am completely lost, could you explain further?

[jemidiah]

Are you confused by the sin^2 x + cos^2 x = 1 part? Or if you accept that, what step starts to confuse you?

[astrostar]

Hi,

I don't understand any of it! I don't want someone to answer the question for me, just to explain it in simple terms!

Many thanks for any help!

[jemidiah]

I can't really follow your original equations, since I don't know when by cos2E you mean cos²E and cos(2*E), and I think there are a few other typos in there [like the E floating before the first line of "My answer"].

As for explaining things in simple terms, I'll just give some commentary on Logophobic's response and hope that it's enough:

1 - 2 sin² E + cos² E = 0

---This is given in the problem.

1 + sin² E + cos² E = 3 sin² E

---This is obtained by adding 3 sin² E to both sides of the equation. Why would one do this? Because we have a great simplifying identity, sin² E + cos² E = 1 for any value of E, and we want to try to apply it in this case. To apply this identity, we need to be adding both a sine squared and cosine squared term, which could only be done by adding 3 sin² E to both sides.

2 = 3 sin² E

---We get this when we cancel the sin² E + cos² E (which is equal to one) from the left and combine that "1" with the existing 1 to get 1+1 = 2.

As Logophobic said, from here you really should be able to finish the problem off, since you used inverse trig functions in your original answer.


A note on sin² E + cos² E = 1: this is the same as cos²E = 1 - sin²E, which you're most likely familiar with and have used before (you might have even used it in your initial answer). This formula comes about from the Pythagorean Theorem. When you interpret sinE as the ratio of the opposite side of a triangle to the hypotenuse (sinE = O / H) and cosE similarly as the adjacent side over the hypotenuse (cosE = A / H), you get

sin²E + cos²E = O²/H² + A²/H².

By the Pythagorean Theorem, we know the relationship between O, A, and H to be

O² + A² = H². Dividing by H² and distributing gives

O²/H² + A²/H² = H² / H² = 1, which gives and proves the initial identity sin² E + cos² E = 1.


The best way to figure out where your understanding is faltering is to talk with someone knowledgeable in Trig (probably your teacher) face-to-face. It's really difficult to spot and fix gaps in people's thought processes without body language, and forums sadly don't give us that

[astrostar]

Now I understand.

Thanks for your help