How do you solve this trig identity problem without factoring?
I just used @ to represent theta:
sin^4@ + 2sin^2@cos^2@ + cos^4@ = 1
sin^4 + 2sin^2 cos^2 + cos^4 = (sin^2 + cos^2)^2
1^2 = 1
Didn't you just factor that? (perfect square trinomial)
I'm looking for a way to solve it without factoring..
To solve this trig identity problem without factoring, we can use the Pythagorean identity: sin^2@ + cos^2@ = 1. Let's break down the steps:
1. Start with the given trigonometric expression:
sin^4@ + 2sin^2@cos^2@ + cos^4@ = 1
2. Use the Pythagorean identity to replace sin^2@ and cos^2@:
(sin^2@ + cos^2@)^2 - 2sin^2@cos^2@ + cos^4@ = 1
3. Simplify the expression:
1^2 - 2sin^2@cos^2@ + cos^4@ = 1
4. Combine like terms on the left side:
1 - 2sin^2@cos^2@ + cos^4@ = 1
5. Subtract 1 from both sides to isolate the trinomial expression:
- 2sin^2@cos^2@ + cos^4@ = 0
6. Factor out common terms from the trinomial expression:
cos^2@(-2sin^2@ + cos^2@) = 0
7. Apply the zero product property:
cos^2@ = 0 or -2sin^2@ + cos^2@ = 0
8. Solve for cos^2@ = 0:
cos^2@ = 0 implies that @ = 90 degrees or @ = 270 degrees (plus/minus multiples of 180 degrees)
9. Solve for -2sin^2@ + cos^2@ = 0:
-2sin^2@ + cos^2@ = 0
Rearrange the terms:
cos^2@ = 2sin^2@
Divide by cos^2@ (since cos^2@ cannot be zero):
1 = 2tan^2@
10. Solve for tan^2@:
Divide by 2:
1/2 = tan^2@
11. Take the square root of both sides to solve for tan@:
√(1/2) = tan@
Therefore, the solution for this trig identity problem without factoring is @ = 45 degrees or @ = 225 degrees (plus/minus multiples of 180 degrees).